Difference between revisions of "מלאכת המספר"

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|The author said: When I saw the lengthiness of the discussions of the ancient scholars, the composers of the books of number, and that the necessary methods and teachings in astronomy, geometry, and the ratios of music are explained in those books merely in great difficulty; likewise the seven types of [operations] with integers as well as with fractions, the roots and the proportions of numbers; and since the teaching that is [based] on brief and comprehensive methods is better chosen for the student as well as for the teacher, for every lengthiness of words ''and loquacity are exhaustion of the body'' [Ecclesiastes 12, 12].

|The author said: When I saw the lengthiness of the discussions of the ancient scholars, the composers of the books of number, and that the necessary methods and teachings in astronomy, geometry, and the ratios of music are explained in those books merely in great difficulty; likewise the seven types of [operations] with integers as well as with fractions, the roots and the proportions of numbers; and since the teaching that is [based] on brief and comprehensive methods is better chosen for the student as well as for the teacher, for every lengthiness of words ''and loquacity are exhaustion of the body'' [Ecclesiastes 12, 12].

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|Therefore, I Yiẓḥaq b. R. Moshe ʽEli ha-Sefaradi [= the Spanish] from the city of Oriola of the kingdom of Aragon, at the request of my friends, who studied astronomy and geometry, because they used those methods with a great difficulty and bother, I shook out my lap and wrote this short treatise that encompasses all that is necessary for this science, arithmetic, that is called arishmetika, by the grace of God to me [Psalms 57, 2], and ''according to the good hand of my God upon me'' [Nehemiah 2, 8].

|Therefore, I Yiẓḥaq b. R. Moshe ʽEli ha-Sefaradi [= the Spanish] from the city of Oriola of the kingdom of Aragon, at the request of my friends, who studied astronomy and geometry, because they used those methods with a great difficulty and bother, I shook out my lap and wrote this short treatise that encompasses all that is necessary for this science, arithmetic, that is called arishmetika, by the grace of God to me [Psalms 57, 2], and ''according to the good hand of my God upon me'' [Nehemiah 2, 8].

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|I have written in it comprehensive and short ways in all seven types of [operations] of number, its properties and the existence of the ratios in all that is possible in this science.

|I have written in it comprehensive and short ways in all seven types of [operations] of number, its properties and the existence of the ratios in all that is possible in this science.

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|This is by demonstrative ways that are loved by the intellectuals, in a manner that anyone who endeavors in studying this short book will comprehend all that is necessary for this science, will be saved from the non-useful lengthiness and from the loss of time, and will attain in it whatever his heart desires; since it is for mathematics ''as the light of dawn shines ever brighter until the perfect'' [Proverbs 4, 18].

|This is by demonstrative ways that are loved by the intellectuals, in a manner that anyone who endeavors in studying this short book will comprehend all that is necessary for this science, will be saved from the non-useful lengthiness and from the loss of time, and will attain in it whatever his heart desires; since it is for mathematics ''as the light of dawn shines ever brighter until the perfect'' [Proverbs 4, 18].

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|With the help of God, Blessed is He, I will begin and say:

|With the help of God, Blessed is He, I will begin and say:

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::*Chapter seven: knowing the ratio of the six proportional numbers [= the proportional hexad].

::*Chapter seven: knowing the ratio of the six proportional numbers [= the proportional hexad].

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|style="text-align:right;"|<big>מלאכת</big> המספר היא מלאכה תורה למנות הרבה אחדים והבדליהם וסגולותם ובנקלה יקויימו בזכירה{{#annotend:weza}}

|style="text-align:right;"|<big>מלאכת</big> המספר היא מלאכה תורה למנות הרבה אחדים והבדליהם וסגולותם ובנקלה יקויימו בזכירה{{#annotend:weza}}

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|style="text-align:right;"|<big>אחדות</big> הוא יסוד וחלק ראשון מהמספר וכל מספר יורכב ממנו אבל הוא חוץ לכל מספר{{#annotend:dYUi}}

|style="text-align:right;"|<big>אחדות</big> הוא יסוד וחלק ראשון מהמספר וכל מספר יורכב ממנו אבל הוא חוץ לכל מספר{{#annotend:dYUi}}

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|style="text-align:right;"|ולכן יגדר המספר בשהוא קבוץ אחדים{{#annotend:I7x1}}

|style="text-align:right;"|ולכן יגדר המספר בשהוא קבוץ אחדים{{#annotend:I7x1}}

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!<span style=color:Green>The written ranks [= decimal places] and their writing order</span>

!<span style=color:Green>The written ranks [= decimal places] and their writing order</span>

|style="text-align:right;"|<big>הפרק</big> הראשון במין הראשון מהמספר והוא הקבוץ

|style="text-align:right;"|<big>הפרק</big> הראשון במין הראשון מהמספר והוא הקבוץ

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|style="text-align:right;"|<big>קבוץ</big> הוא חבור שני מספרים או יותר במספר אחד כולל לכולם{{#annotend:qlUB}}

|style="text-align:right;"|<big>קבוץ</big> הוא חבור שני מספרים או יותר במספר אחד כולל לכולם{{#annotend:qlUB}}

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*Then all the numerals are summed as units.

*Then all the numerals are summed as units.

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|This sum necessarily belongs to one of three categories, as you know: either it is units, or tens, or consists of both.

|This sum necessarily belongs to one of three categories, as you know: either it is units, or tens, or consists of both.

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*If it is tens, we write a zero and shift the 10 to the first rank next to it, which is the rank of the tens.

*If it is tens, we write a zero and shift the 10 to the first rank next to it, which is the rank of the tens.

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:<math>\scriptstyle5243+8962</math>

:<math>\scriptstyle5243+8962</math>

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|The proof of this is that we cast out the nines from the sum and keep the remainder. We do the same with the addend. If the remainders from both are equal, then the addition we made is correct, otherwise it is not.

|The proof of this is that we cast out the nines from the sum and keep the remainder. We do the same with the addend. If the remainders from both are equal, then the addition we made is correct, otherwise it is not.

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|This is enough for the first type.

|This is enough for the first type.

|style="width:45%; text-align:right;"|<big>הפרק</big> הב' במין הב' שהוא חסור

|style="width:45%; text-align:right;"|<big>הפרק</big> הב' במין הב' שהוא חסור

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!<span style=color:Green>Description of the procedure:</span>

!<span style=color:Green>Description of the procedure:</span>

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*If they are equal, we write zero beneath them, as a sign that nothing remains.

*If they are equal, we write zero beneath them, as a sign that nothing remains.

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:<math>\scriptstyle4282-2432</math>

:<math>\scriptstyle4282-2432</math>

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:*Next, we add 1 to the 2 that is in the fourth rank; it is 3. We subtract it from the 4, which is the upper digit; 1 remains beneath.

:*Next, we add 1 to the 2 that is in the fourth rank; it is 3. We subtract it from the 4, which is the upper digit; 1 remains beneath.

|style="text-align:right;"|ואחר כך בב' שהיא במדרגה הד' {{#annot:term|178,1206|0QE7}}נוסיף{{#annotend:0QE7}} א' ויהיו ג' ונחסרם מהד' שהיא האות העליון וישאר למטה א&#x202B;'

|style="text-align:right;"|ואחר כך בב' שהיא במדרגה הד' {{#annot:term|178,1206|0QE7}}נוסיף{{#annotend:0QE7}} א' ויהיו ג' ונחסרם מהד' שהיא האות העליון וישאר למטה א&#x202B;'

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|428<span style="color:red>2</span>||rowspan="3"|<math>\scriptstyle\xrightarrow{{\color{red}{2-2}}={\color{blue}{0}}}</math>||42<span style="color:red>8</span>2||rowspan="3"|<math>\scriptstyle\xrightarrow{{\color{red}{8-3}}={\color{blue}{5}}}</math>||4<span style="color:red>2</span>82||rowspan="3"|<math>\scriptstyle\xrightarrow{{\color{red}{2+\left(10-4\right)=2+6}}={\color{blue}{8}}}</math>||<span style="color:red>4</span>282||rowspan="3"|<math>\scriptstyle\xrightarrow{{\color{red}{4-\left(2+1\right)=4-3}}={\color{blue}{1}}}</math>||4282

|428<span style="color:red>2</span>||rowspan="3"|<math>\scriptstyle\xrightarrow{{\color{red}{2-2}}={\color{blue}{0}}}</math>||42<span style="color:red>8</span>2||rowspan="3"|<math>\scriptstyle\xrightarrow{{\color{red}{8-3}}={\color{blue}{5}}}</math>||4<span style="color:red>2</span>82||rowspan="3"|<math>\scriptstyle\xrightarrow{{\color{red}{2+\left(10-4\right)=2+6}}={\color{blue}{8}}}</math>||<span style="color:red>4</span>282||rowspan="3"|<math>\scriptstyle\xrightarrow{{\color{red}{4-\left(2+1\right)=4-3}}={\color{blue}{1}}}</math>||4282

| ||&#8199;&#8199;&#8199;<span style="color:#0000FF>0</span>||&#8199;&#8199;<span style="color:#0000FF>5</span>0||&#8199;<span style="color:#0000FF>8</span>50||<span style="color:#0000FF>1</span>850

| ||&#8199;&#8199;&#8199;<span style="color:#0000FF>0</span>||&#8199;&#8199;<span style="color:#0000FF>5</span>0||&#8199;<span style="color:#0000FF>8</span>50||<span style="color:#0000FF>1</span>850

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|By this order it should be done, even if the ranks are numerous.

|By this order it should be done, even if the ranks are numerous.

|style="width:45%; text-align:right;"|<big>הפרק</big> הג' במין הג' והוא הכפול ודרכים כוללים למצאות המספרים השלמים

|style="width:45%; text-align:right;"|<big>הפרק</big> הג' במין הג' והוא הכפול ודרכים כוללים למצאות המספרים השלמים

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|style="text-align:right;"|<big>כפול</big> הוא קבוץ איזה ב' מספרים שיהיו שוים{{#annotend:jEGw}}

|style="text-align:right;"|<big>כפול</big> הוא קבוץ איזה ב' מספרים שיהיו שוים{{#annotend:jEGw}}

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::<math>\scriptstyle2\times5372</math>

::<math>\scriptstyle2\times5372</math>

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|style="text-align:right;"|ובדרך ה{{#annot:term|159,1348|XTTg}}כפול{{#annotend:XTTg}} הזה ימצאו {{#annot:term|75,1268|cqAo}}המספרים השלמים{{#annotend:cqAo}}

|style="text-align:right;"|ובדרך ה{{#annot:term|159,1348|XTTg}}כפול{{#annotend:XTTg}} הזה ימצאו {{#annot:term|75,1268|cqAo}}המספרים השלמים{{#annotend:cqAo}}

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|style="text-align:right;"|וגדר המספר השלם הוא כל מספר שיבנה מקבוץ כל חלקיו שבשילקח כל אחד מחלקיו ויקובצו יבנו אותו לא פחות ולא יתר{{#annotend:1ldm}}

|style="text-align:right;"|וגדר המספר השלם הוא כל מספר שיבנה מקבוץ כל חלקיו שבשילקח כל אחד מחלקיו ויקובצו יבנו אותו לא פחות ולא יתר{{#annotend:1ldm}}

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|style="text-align:right;"|ואם יהיה מספר ראשון אז {{#annot:term|185,1255|I01K}}נכה{{#annotend:I01K}} אותו הכפל שלקחנו עם כפלו פחות א' והעולה מהכאה זו הוא {{#annot:term|75,1268|K1qA}}מספר שלם{{#annotend:K1qA}}

|style="text-align:right;"|ואם יהיה מספר ראשון אז {{#annot:term|185,1255|I01K}}נכה{{#annotend:I01K}} אותו הכפל שלקחנו עם כפלו פחות א' והעולה מהכאה זו הוא {{#annot:term|75,1268|K1qA}}מספר שלם{{#annotend:K1qA}}

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|style="width:45%; text-align:right;"|<big>הפרק הרביעי</big> במין הרביעי שהוא חלוק באמצע

|style="width:45%; text-align:right;"|<big>הפרק הרביעי</big> במין הרביעי שהוא חלוק באמצע

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|style="text-align:right;"|&#x202B;[חלוק באמצע]&#x202B;<ref>P1095 om.</ref> והוא חלוק איזה מספר שיהיה בשני חלקים שוים{{#annotend:MyS6}}

|style="text-align:right;"|&#x202B;[חלוק באמצע]&#x202B;<ref>P1095 om.</ref> והוא חלוק איזה מספר שיהיה בשני חלקים שוים{{#annotend:MyS6}}

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:<math>\scriptstyle262144\div2</math>

:<math>\scriptstyle262144\div2</math>

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:<math>\scriptstyle1048876\div2</math>

:<math>\scriptstyle1048876\div2</math>

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|The proof of this is doubling: if after it is doubled it becomes equal to the halved, then it is correct, otherwise it is incorrect.

|The proof of this is doubling: if after it is doubled it becomes equal to the halved, then it is correct, otherwise it is incorrect.

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|This is enough for the type of halving.

|This is enough for the type of halving.

|style="width:45%; text-align:right;"|<big>פרק</big> ה' במין הה' והוא הרבוע וקצת מסגולותיו

|style="width:45%; text-align:right;"|<big>פרק</big> ה' במין הה' והוא הרבוע וקצת מסגולותיו

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!<span style=color:Green>Multiplication table</span>

!<span style=color:Green>Multiplication table</span>

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|10||9||8||7||6||5||4||3||2||1

|10||9||8||7||6||5||4||3||2||1

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|י||ט||ח||ז||ו||ה||ד||ג||ב||א

|י||ט||ח||ז||ו||ה||ד||ג||ב||א

|ק||צ||פ||ע||ס||נ||מ||ל||כ||י

|ק||צ||פ||ע||ס||נ||מ||ל||כ||י

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|What is received from one of these two tables, is received from the other, there is no difference between them, except for the extensiveness versus brevity.

|What is received from one of these two tables, is received from the other, there is no difference between them, except for the extensiveness versus brevity.

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!<span style=color:Green>Description of the procedure:</span>

!<span style=color:Green>Description of the procedure:</span>

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::<math>\scriptstyle\sum_{i=1}^{12} i=1+\ldots+12</math>

::<math>\scriptstyle\sum_{i=1}^{12} i=1+\ldots+12</math>

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::We take a half of 12, which 6, multiply it by 13, which is the last number plus 1; it is 78 and this is the sum from 1 to 12.

::We take a half of 12, which 6, multiply it by 13, which is the last number plus 1; it is 78 and this is the sum from 1 to 12.

::<math>\scriptstyle{\color{blue}{\sum_{i=1}^{12} i=1+\ldots+12=\left(\frac{1}{2}\sdot12\right)\sdot\left(12+1\right)=6\sdot13=78}}</math>

::<math>\scriptstyle{\color{blue}{\sum_{i=1}^{12} i=1+\ldots+12=\left(\frac{1}{2}\sdot12\right)\sdot\left(12+1\right)=6\sdot13=78}}</math>

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::<math>\scriptstyle\sum_{i=1}^{13} i=1+\ldots+13</math>

::<math>\scriptstyle\sum_{i=1}^{13} i=1+\ldots+13</math>

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::We take 6 and a half plus one half, which is 7, and multiply it by 13; the result is 78 and this is the sum of all of them.

::We take 6 and a half plus one half, which is 7, and multiply it by 13; the result is 78 and this is the sum of all of them.

|style="text-align:right;"|נקח ו' וחצי וחצי יותר שהם ז' ונכה אותם על י"ג ויעלו צ"א וכך הוא המקובץ מכולם

|style="text-align:right;"|נקח ו' וחצי וחצי יותר שהם ז' ונכה אותם על י"ג ויעלו צ"א וכך הוא המקובץ מכולם

::<math>\scriptstyle{\color{blue}{\sum_{i=1}^{13} i=1+\ldots+13=\left[\left(\frac{1}{2}\sdot13\right)+\frac{1}{2}\right]\sdot13=\left[\left(6+\frac{1}{2}\right)+\frac{1}{2}\right]\sdot13=7\sdot13=91}}</math>

::<math>\scriptstyle{\color{blue}{\sum_{i=1}^{13} i=1+\ldots+13=\left[\left(\frac{1}{2}\sdot13\right)+\frac{1}{2}\right]\sdot13=\left[\left(6+\frac{1}{2}\right)+\frac{1}{2}\right]\sdot13=7\sdot13=91}}</math>

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::<math>\scriptstyle\sum_{i=1}^6 2i=2+\ldots+12</math>

::<math>\scriptstyle\sum_{i=1}^6 2i=2+\ldots+12</math>

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::We take its half, which is 6, and multiply it by 7, which is half the number plus 1; the result is 42 and this is the sum of all of them.

::We take its half, which is 6, and multiply it by 7, which is half the number plus 1; the result is 42 and this is the sum of all of them.

|style="text-align:right;"|נקח חציו שהוא ו' ונכהו על ז' שהוא חצי המספר בתוספת א' ויעלה למ"ב וכך הוא הקבוץ של כולם

|style="text-align:right;"|נקח חציו שהוא ו' ונכהו על ז' שהוא חצי המספר בתוספת א' ויעלה למ"ב וכך הוא הקבוץ של כולם

::<math>\scriptstyle{\color{blue}{\sum_{i=1}^6 2i=2+\ldots+12=\left(\frac{1}{2}\sdot12\right)\sdot\left[\left(\frac{1}{2}\sdot12\right)+1\right]=6\sdot\left(6+1\right)=6\sdot7=42}}</math>

::<math>\scriptstyle{\color{blue}{\sum_{i=1}^6 2i=2+\ldots+12=\left(\frac{1}{2}\sdot12\right)\sdot\left[\left(\frac{1}{2}\sdot12\right)+1\right]=6\sdot\left(6+1\right)=6\sdot7=42}}</math>

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::<math>\scriptstyle\sum_{i=1}^8 \left(2i-1\right)=1+\ldots+15</math>

::<math>\scriptstyle\sum_{i=1}^8 \left(2i-1\right)=1+\ldots+15</math>

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::We take half the last number, which is 7 and a half, plus one half; it is 8. We multiply it by itself; the result is 64 and this is the sum of all of them.

::We take half the last number, which is 7 and a half, plus one half; it is 8. We multiply it by itself; the result is 64 and this is the sum of all of them.

|style="text-align:right;"|נקח חצי האחרון שהוא ז' וחצי וחצי יותר ויהיו ח' ונכהו בעצמו ויעלו ס"ד וכך הוא המקובץ מכולם

|style="text-align:right;"|נקח חצי האחרון שהוא ז' וחצי וחצי יותר ויהיו ח' ונכהו בעצמו ויעלו ס"ד וכך הוא המקובץ מכולם

::<math>\scriptstyle{\color{blue}{\sum_{i=1}^8 \left(2i-1\right)=1+\ldots+15=\left[\left(\frac{1}{2}\sdot15\right)+\frac{1}{2}\right]^2=\left[\left(7+\frac{1}{2}\right)+\frac{1}{2}\right]^2=8^2=64}}</math>

::<math>\scriptstyle{\color{blue}{\sum_{i=1}^8 \left(2i-1\right)=1+\ldots+15=\left[\left(\frac{1}{2}\sdot15\right)+\frac{1}{2}\right]^2=\left[\left(7+\frac{1}{2}\right)+\frac{1}{2}\right]^2=8^2=64}}</math>

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::<math>\scriptstyle8\times9</math>

::<math>\scriptstyle8\times9</math>

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::We find that the greater number, which is 9, is less than 10 by one. So, we subtract 8 once from 80, which is its product by ten; 72 remains and this is the product of 8 by 9.

::We find that the greater number, which is 9, is less than 10 by one. So, we subtract 8 once from 80, which is its product by ten; 72 remains and this is the product of 8 by 9.

|style="text-align:right;"|ומצאנו שהמספר היותר גדול שהוא ט' הוא פחות מי' אחד ולכן נחסר ח' פעם אחת מפ' שהוא עשיריתו וישארו ע"ב וזאת היא ההכאה מח' בט&#x202B;'

|style="text-align:right;"|ומצאנו שהמספר היותר גדול שהוא ט' הוא פחות מי' אחד ולכן נחסר ח' פעם אחת מפ' שהוא עשיריתו וישארו ע"ב וזאת היא ההכאה מח' בט&#x202B;'

::<math>\scriptstyle{\color{blue}{8<9<10\longrightarrow8\times9=\left(10\sdot8\right)-\left[\left(10-9\right)\sdot8\right]=80-\left(1\sdot8\right)=80-8=72}}</math>

::<math>\scriptstyle{\color{blue}{8<9<10\longrightarrow8\times9=\left(10\sdot8\right)-\left[\left(10-9\right)\sdot8\right]=80-\left(1\sdot8\right)=80-8=72}}</math>

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:*Example: as we said regarding the product of 8 by 9: by how much is the 9 less than 10, so we do when we examine by how much is 8 less than 10, which is by 2. So we subtract 2 times 9 from 90; 72 remains.

:*Example: as we said regarding the product of 8 by 9: by how much is the 9 less than 10, so we do when we examine by how much is 8 less than 10, which is by 2. So we subtract 2 times 9 from 90; 72 remains.

::<math>\scriptstyle{\color{blue}{8<9<10\longrightarrow8\times9=\left(10\sdot9\right)-\left[\left(10-8\right)\sdot9\right]=90-\left(2\sdot9\right)=72}}</math>

::<math>\scriptstyle{\color{blue}{8<9<10\longrightarrow8\times9=\left(10\sdot9\right)-\left[\left(10-8\right)\sdot9\right]=90-\left(2\sdot9\right)=72}}</math>

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::<math>\scriptstyle10\times12=120</math>

::<math>\scriptstyle10\times12=120</math>

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::<math>\scriptstyle20\times5=\left(2\sdot5\right)0=100</math>

::<math>\scriptstyle20\times5=\left(2\sdot5\right)0=100</math>

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::<math>\scriptstyle30\times20=\left(3\sdot20\right)0=600</math>

::<math>\scriptstyle30\times20=\left(3\sdot20\right)0=600</math>

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|style="width: 45%; text-align:right;"|<big>פרק שישי</big> במין השישי שהוא ה{{#annot:term|157,1223|W8wP}}חלוק{{#annotend:W8wP}}

|style="width: 45%; text-align:right;"|<big>פרק שישי</big> במין השישי שהוא ה{{#annot:term|157,1223|W8wP}}חלוק{{#annotend:W8wP}}

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|In this type [of operation] one should start from the highest rank.

|In this type [of operation] one should start from the highest rank.

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:<math>\scriptstyle144\div8</math>

:<math>\scriptstyle144\div8</math>

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:Hence, we write 0 above the 6 and 0 above the 4.

:Hence, we write 0 above the 6 and 0 above the 4.

|style="text-align:right;"|ולכן כתבנו על ו' ספרא ו0' על הד&#x202B;'

|style="text-align:right;"|ולכן כתבנו על ו' ספרא ו0' על הד&#x202B;'

::<span style=color:Green>[Illustration of the procedure:]</span>

::<span style=color:Green>[Illustration of the procedure:]</span>

::{|

::{|

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::<math>\scriptstyle\longrightarrow{\color{Violet}{18\ the\ result}}</math>

::<math>\scriptstyle\longrightarrow{\color{Violet}{18\ the\ result}}</math>

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|This way for all the ranks, even if they are very many.

|This way for all the ranks, even if they are very many.

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::We multiply the 2, which is the quotient by the 2, which is the right [digit] of the divisor; the product is 4. We subtract it from the 7; 3 remain above the 7.

::We multiply the 2, which is the quotient by the 2, which is the right [digit] of the divisor; the product is 4. We subtract it from the 7; 3 remain above the 7.

|style="text-align:right;"|עוד נכה הב' שהוא החלק על הב' שהוא ימנית המחולק ויעלו ד' ונחסרם מהז' וישאר ג' על הז&#x202B;'

|style="text-align:right;"|עוד נכה הב' שהוא החלק על הב' שהוא ימנית המחולק ויעלו ד' ונחסרם מהז' וישאר ג' על הז&#x202B;'

::<span style=color:Green>[Illustration of the procedure:]</span>

::<span style=color:Green>[Illustration of the procedure:]</span>

::{|

::{|

|<span style="color:red>12</span>&#8199;&#8199;||12&#8199;&#8199;||&#8199;<span style="color:red>12</span>&#8199;||&#8199;12&#8199;

|<span style="color:red>12</span>&#8199;&#8199;||12&#8199;&#8199;||&#8199;<span style="color:red>12</span>&#8199;||&#8199;12&#8199;

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|style="text-align:right;"|<big>וקודם</big> שנדבר [מהם]&#x202B;<ref>P1095 om.</ref> צריך שנאמר מהו שבר ואיך יסודר בכתיבתו ובאיזה חשבון ישבר

|style="text-align:right;"|<big>וקודם</big> שנדבר [מהם]&#x202B;<ref>P1095 om.</ref> צריך שנאמר מהו שבר ואיך יסודר בכתיבתו ובאיזה חשבון ישבר

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:<math>\scriptstyle\frac{3}{4}-\frac{2}{3}</math>

:<math>\scriptstyle\frac{3}{4}-\frac{2}{3}</math>

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:<math>\scriptstyle2\times\frac{3}{4}</math>

:<math>\scriptstyle2\times\frac{3}{4}</math>

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|In this operation you only need to double the digit that is beneath the line and the fraction is the halved.

|In this operation you only need to double the digit that is beneath the line and the fraction is the halved.

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:<math>\scriptstyle\frac{1}{4}\div2</math>

:<math>\scriptstyle\frac{1}{4}\div2</math>

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:<math>\scriptstyle4\times\frac{2}{3}</math>

:<math>\scriptstyle4\times\frac{2}{3}</math>

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:<math>\scriptstyle\left(4+\frac{1}{2}\right)\times\frac{2}{3}</math>

:<math>\scriptstyle\left(4+\frac{1}{2}\right)\times\frac{2}{3}</math>

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:<math>\scriptstyle\left(4+\frac{1}{3}\right)\times6</math>

:<math>\scriptstyle\left(4+\frac{1}{3}\right)\times6</math>

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:<math>\scriptstyle\left(2+\frac{1}{2}\right)\times\left(4+\frac{3}{4}\right)</math>

:<math>\scriptstyle\left(2+\frac{1}{2}\right)\times\left(4+\frac{3}{4}\right)</math>

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|<span style=color:Green>denominator:</span> First of all, we extract the denominator, which is the product of the digits that are beneath the fraction lines as follows:

|<span style=color:Green>denominator:</span> First of all, we extract the denominator, which is the product of the digits that are beneath the fraction lines as follows:

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|If we divide the seconds by 60, they become minutes.

|If we divide the seconds by 60, they become minutes.

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|The same for the others, when we divide them by 60 they are converted to the preceding rank.

|The same for the others, when we divide them by 60 they are converted to the preceding rank.

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|style="text-align:right;"|כנראה בצורה זו

|style="text-align:right;"|כנראה בצורה זו

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!44||26||42||21||7

!44||26||42||21||7

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!רביעיים||שלישיים||שניים||ראשונים||מעלות

!רביעיים||שלישיים||שניים||ראשונים||מעלות

|In order to add a further explanation, I write this second diagram which involves a longer procedure and study.  

|In order to add a further explanation, I write this second diagram which involves a longer procedure and study.  

|style="text-align:right;"|&#x202B;[וכדי להוסיף לך ביאור כתבתי זאת הצורה השנית שיש בה יותר מלאכה ועיון&#x202B;]&#x202B;<ref>P1095 om.:</ref>

|style="text-align:right;"|&#x202B;[וכדי להוסיף לך ביאור כתבתי זאת הצורה השנית שיש בה יותר מלאכה ועיון&#x202B;]&#x202B;<ref>P1095 om.:</ref>

{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none; color: blue; text-align:center;"

{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none; color: blue; text-align:center;"

!36||32||32||49||57||35||1

!36||32||32||49||57||35||1

|}

|}

{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none; text-align:center;"

{|class="wikitable" style="margin-left: auto; margin-right: auto; border: none; text-align:center;"

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:<math>\scriptstyle\frac{3}{4}\div\frac{2}{3}</math>

:<math>\scriptstyle\frac{3}{4}\div\frac{2}{3}</math>

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:<math>\scriptstyle\frac{3}{8}\div2</math>

:<math>\scriptstyle\frac{3}{8}\div2</math>

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:<math>\scriptstyle\frac{2}{3}\div\left(4+\frac{1}{2}\right)</math>

:<math>\scriptstyle\frac{2}{3}\div\left(4+\frac{1}{2}\right)</math>

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:<math>\scriptstyle\left(9+\frac{1}{2}\right)\div\left(3+\frac{4}{5}\right)</math>

:<math>\scriptstyle\left(9+\frac{1}{2}\right)\div\left(3+\frac{4}{5}\right)</math>

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:<math>\scriptstyle12\div\frac{4}{9}</math>

:<math>\scriptstyle12\div\frac{4}{9}</math>

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:<math>\scriptstyle\left(4+\frac{1}{3}\right)\div\frac{5}{6}</math>

:<math>\scriptstyle\left(4+\frac{1}{3}\right)\div\frac{5}{6}</math>

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:<math>\scriptstyle\left(3+\frac{1}{3}\right)\div4</math>

:<math>\scriptstyle\left(3+\frac{1}{3}\right)\div4</math>

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:<math>\scriptstyle50\div\left(2+\frac{1}{2}\right)</math>

:<math>\scriptstyle50\div\left(2+\frac{1}{2}\right)</math>

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|style="text-align:right;"|וראשונה נדבר מהשרש המרובע

|style="text-align:right;"|וראשונה נדבר מהשרש המרובע

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|style="text-align:right;"|ולכן ב' הם שרש ד' בעבור כשיוכה בעצמו יוליד ארבעה

|style="text-align:right;"|ולכן ב' הם שרש ד' בעבור כשיוכה בעצמו יוליד ארבעה

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|Hence, every number that is multiplied by itself generates a number that has a root and this number is called a square.

|Hence, every number that is multiplied by itself generates a number that has a root and this number is called a square.

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|style="text-align:right;"|ובעבור שלא נשאר כלום נכתוב 0'0' על ב' ועל ה' וזה מה שרצינו

|style="text-align:right;"|ובעבור שלא נשאר כלום נכתוב 0'0' על ב' ועל ה' וזה מה שרצינו

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::<span style=color:Green>[Illustration of the procedure:]</span>

::<span style=color:Green>[Illustration of the procedure:]</span>

::{|

::{|

| ||&#8199;<span style="color:#0000FF>1</span>&#8199;&#8199;||&#8199;1&#8199;&#8199;

| ||&#8199;<span style="color:#0000FF>1</span>&#8199;&#8199;||&#8199;1&#8199;&#8199;

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*For, the first numerical rank, which is the rank of units, has a cube root. Because, the one that is cubically multiplied generates one, which is a cube number.

*For, the first numerical rank, which is the rank of units, has a cube root. Because, the one that is cubically multiplied generates one, which is a cube number.

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:<math>\scriptstyle\sqrt[3]{12167}</math>

:<math>\scriptstyle\sqrt[3]{12167}</math>

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:Therefore, we know that this number is a [perfect] cube and its root is 23.

:Therefore, we know that this number is a [perfect] cube and its root is 23.

|style="text-align:right;"|וא"כ ידענו שהמספר הזה הוא מעוקב ושורשו כ"ג

|style="text-align:right;"|וא"כ ידענו שהמספר הזה הוא מעוקב ושורשו כ"ג

::<span style=color:Green>[Illustration of the procedure:]</span>

::<span style=color:Green>[Illustration of the procedure:]</span>

::{|

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::<math>\scriptstyle\longrightarrow{\color{Violet}{the\ result\ \sqrt[3]{12167}=23}}</math>

::<math>\scriptstyle\longrightarrow{\color{Violet}{the\ result\ \sqrt[3]{12167}=23}}</math>

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:<math>\scriptstyle\sqrt[3]{571787}</math>

:<math>\scriptstyle\sqrt[3]{571787}</math>

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:Therefore, nothing remains after the extraction of this root, so it is deduced that this number is a perfect cube and its root is 83.

:Therefore, nothing remains after the extraction of this root, so it is deduced that this number is a perfect cube and its root is 83.

|style="text-align:right;"|וא"כ לא נשאר דבר אחר הוצאת השרש הזה יראה המספר הזה הוא מעוקב שלם ושורשו הוא פ"ג

|style="text-align:right;"|וא"כ לא נשאר דבר אחר הוצאת השרש הזה יראה המספר הזה הוא מעוקב שלם ושורשו הוא פ"ג

::<span style=color:Green>[Illustration of the procedure:]</span>

::<span style=color:Green>[Illustration of the procedure:]</span>

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| ||rowspan="6"|<math>\scriptstyle\xrightarrow{\begin{align}&\scriptstyle{\color{red}{571-{\color{blue}{8}}^3=571-512=}}{\color{green}{59}}\\&\scriptstyle{\color{red}{3\times8=}}{\color{blue}{24}}\\\end{align}}</math>||&#x200B;||rowspan="6"|<math>\scriptstyle\xrightarrow{\begin{align}&\scriptstyle{\color{red}{59-\left(8\times2\times{\color{blue}{3}}\right)=59-\left(8\sdot6\right)=59-48=}}{\color{green}{11}}\\&\scriptstyle{\color{red}{117-\left[\left(8\times4\times{\color{blue}{3}}\right)+\left(3\times2\times{\color{blue}{3}}\right)\right]=}}\\&\scriptstyle{\color{red}{117-\left[\left(8\sdot12\right)+\left(3\sdot6\right)\right]=}}\\&\scriptstyle{\color{red}{117-\left(96+18\right)=117-114=}}{\color{green}{3}}\\&\scriptstyle{\color{red}{38-\left(3\times4\times{\color{blue}{3}}\right)=38-\left(3\sdot12\right)=38-36=}}{\color{green}{2}}\\&\scriptstyle{\color{red}{27-{\color{blue}{3}}^3=27-27=}}{\color{green}{0}}\\\end{align}}</math>||&#8199;<span style="color:LimeGreen>00</span>&#8199;&#8199;&#8199;

| ||rowspan="6"|<math>\scriptstyle\xrightarrow{\begin{align}&\scriptstyle{\color{red}{571-{\color{blue}{8}}^3=571-512=}}{\color{green}{59}}\\&\scriptstyle{\color{red}{3\times8=}}{\color{blue}{24}}\\\end{align}}</math>||&#x200B;||rowspan="6"|<math>\scriptstyle\xrightarrow{\begin{align}&\scriptstyle{\color{red}{59-\left(8\times2\times{\color{blue}{3}}\right)=59-\left(8\sdot6\right)=59-48=}}{\color{green}{11}}\\&\scriptstyle{\color{red}{117-\left[\left(8\times4\times{\color{blue}{3}}\right)+\left(3\times2\times{\color{blue}{3}}\right)\right]=}}\\&\scriptstyle{\color{red}{117-\left[\left(8\sdot12\right)+\left(3\sdot6\right)\right]=}}\\&\scriptstyle{\color{red}{117-\left(96+18\right)=117-114=}}{\color{green}{3}}\\&\scriptstyle{\color{red}{38-\left(3\times4\times{\color{blue}{3}}\right)=38-\left(3\sdot12\right)=38-36=}}{\color{green}{2}}\\&\scriptstyle{\color{red}{27-{\color{blue}{3}}^3=27-27=}}{\color{green}{0}}\\\end{align}}</math>||&#8199;<span style="color:LimeGreen>00</span>&#8199;&#8199;&#8199;

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::<math>\scriptstyle\longrightarrow{\color{Violet}{the\ result\ \sqrt[3]{571787}=83}}</math>

::<math>\scriptstyle\longrightarrow{\color{Violet}{the\ result\ \sqrt[3]{571787}=83}}</math>

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:<math>\scriptstyle\sqrt[3]{12812904}</math>

:<math>\scriptstyle\sqrt[3]{12812904}</math>

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:*So, we multiply the 2 cubically; it yields 8.

:*So, we multiply the 2 cubically; it yields 8.

::<math>\scriptstyle{\color{blue}{2^3=8}}</math>

::<math>\scriptstyle{\color{blue}{2^3=8}}</math>

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::We shift the 2 also, and write it in the next rank, so it is placed beneath the upper 4.

::We shift the 2 also, and write it in the next rank, so it is placed beneath the upper 4.

|style="text-align:right;"|והב' ג"כ נסיעה למקום הסמוך אליה ויעמוד תחת הד' העליונה

|style="text-align:right;"|והב' ג"כ נסיעה למקום הסמוך אליה ויעמוד תחת הד' העליונה

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| ||rowspan="6"|<math>\scriptstyle\xrightarrow{\begin{align}&\scriptstyle{\color{red}{12-{\color{blue}{2}}^3=12-8=}}{\color{green}{4}}\\&\scriptstyle{\color{red}{3\times2=}}{\color{blue}{6}}\\\end{align}}</math>||&#x200B;||rowspan="6"|<math>\scriptstyle\xrightarrow{\begin{align}&\scriptstyle{\color{red}{48-\left(2\times6\times{\color{blue}{3}}\right)=48-\left(2\sdot18\right)=48-36=}}{\color{green}{12}}\\&\scriptstyle{\color{red}{121-\left(3\times6\times{\color{blue}{3}}\right)=121-\left(3\sdot18\right)=121-54=}}{\color{green}{67}}\\&\scriptstyle{\color{red}{72-{\color{blue}{3}}^3=72-27=}}{\color{green}{45}}\\&\scriptstyle{\color{red}{3\times3=}}{\color{blue}{9}}\\\end{align}}</math>||&#8199;<span style="color:LimeGreen>0</span>&#8199;&#8199;&#8199;&#8199;&#8199;&#8199;

| ||rowspan="6"|<math>\scriptstyle\xrightarrow{\begin{align}&\scriptstyle{\color{red}{12-{\color{blue}{2}}^3=12-8=}}{\color{green}{4}}\\&\scriptstyle{\color{red}{3\times2=}}{\color{blue}{6}}\\\end{align}}</math>||&#x200B;||rowspan="6"|<math>\scriptstyle\xrightarrow{\begin{align}&\scriptstyle{\color{red}{48-\left(2\times6\times{\color{blue}{3}}\right)=48-\left(2\sdot18\right)=48-36=}}{\color{green}{12}}\\&\scriptstyle{\color{red}{121-\left(3\times6\times{\color{blue}{3}}\right)=121-\left(3\sdot18\right)=121-54=}}{\color{green}{67}}\\&\scriptstyle{\color{red}{72-{\color{blue}{3}}^3=72-27=}}{\color{green}{45}}\\&\scriptstyle{\color{red}{3\times3=}}{\color{blue}{9}}\\\end{align}}</math>||&#8199;<span style="color:LimeGreen>0</span>&#8199;&#8199;&#8199;&#8199;&#8199;&#8199;

|style="text-align:right;"|ולכן נכתוב ספרא על הד' העליונה ועל הו' העליונה

|style="text-align:right;"|ולכן נכתוב ספרא על הד' העליונה ועל הו' העליונה

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::<span style=color:Green>[Illustration of the procedure:]</span>

::<span style=color:Green>[Illustration of the procedure:]</span>

::{|

::{|

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::<math>\scriptstyle\longrightarrow{\color{Violet}{the\ result\ \sqrt[3]{12812904}=234}}</math>

::<math>\scriptstyle\longrightarrow{\color{Violet}{the\ result\ \sqrt[3]{12812904}=234}}</math>

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:<math>\scriptstyle2:3=5:X</math>

:<math>\scriptstyle2:3=5:X</math>

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:<math>\scriptstyle5:\left(7+\frac{1}{2}\right)=2:X</math>

:<math>\scriptstyle5:\left(7+\frac{1}{2}\right)=2:X</math>

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:<math>\scriptstyle\frac{2}{3}:\left(7+\frac{4}{9}\right)=\frac{4}{13}:X</math>

:<math>\scriptstyle\frac{2}{3}:\left(7+\frac{4}{9}\right)=\frac{4}{13}:X</math>

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:<math>\scriptstyle\frac{2}{3}:\left(4+\frac{1}{2}\right)=6:X</math>

:<math>\scriptstyle\frac{2}{3}:\left(4+\frac{1}{2}\right)=6:X</math>

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:<math>\scriptstyle\frac{2}{3}:8=9:X</math>

:<math>\scriptstyle\frac{2}{3}:8=9:X</math>

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:<math>\scriptstyle\left(5+\frac{2}{3}\right):\left(6+\frac{3}{4}\right)=\left(8+\frac{5}{12}\right):X</math>

:<math>\scriptstyle\left(5+\frac{2}{3}\right):\left(6+\frac{3}{4}\right)=\left(8+\frac{5}{12}\right):X</math>

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:<span style=color:Green>denominator</span>: To find the denominator, we multiply 5 by 3 that is beneath the 2 next to it ; the result is 15. We add the 2 that is above the 3; the result is 17. We multiply it by 4 that is beneath the 3 of the second number; the result is 68. We multiply it by 12 that is beneath the 5 of the third number; the result is 816 and this is the denominator.

:<span style=color:Green>denominator</span>: To find the denominator, we multiply 5 by 3 that is beneath the 2 next to it ; the result is 15. We add the 2 that is above the 3; the result is 17. We multiply it by 4 that is beneath the 3 of the second number; the result is 68. We multiply it by 12 that is beneath the 5 of the third number; the result is 816 and this is the denominator.

:<math>\scriptstyle{\color{blue}{\left[\left[\left(5\sdot3\right)+2\right]\sdot4\right]\sdot12=\left[\left(15+2\right)\sdot4\right]\sdot12=\left(17\sdot4\right)\sdot12=68\sdot12=816}}</math>

:<math>\scriptstyle{\color{blue}{\left[\left[\left(5\sdot3\right)+2\right]\sdot4\right]\sdot12=\left[\left(15+2\right)\sdot4\right]\sdot12=\left(17\sdot4\right)\sdot12=68\sdot12=816}}</math>

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:<math>\scriptstyle\left(5+\frac{2}{7}\right):4=20:X</math>

:<math>\scriptstyle\left(5+\frac{2}{7}\right):4=20:X</math>

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:<math>\scriptstyle20:\left(15+\frac{5}{37}\right)=\left(5+\frac{2}{7}\right):X</math>

:<math>\scriptstyle20:\left(15+\frac{5}{37}\right)=\left(5+\frac{2}{7}\right):X</math>

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:<math>\scriptstyle\frac{2}{3}:\frac{4}{9}=\frac{4}{13}:X</math>

:<math>\scriptstyle\frac{2}{3}:\frac{4}{9}=\frac{4}{13}:X</math>

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::<math>\scriptstyle\frac{2}{3}:8=9:X</math>

::<math>\scriptstyle\frac{2}{3}:8=9:X</math>

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::<math>\scriptstyle\left(5+\frac{2}{3}\right):\left(6+\frac{3}{4}\right)=\left(8+\frac{5}{12}\right):X</math>

::<math>\scriptstyle\left(5+\frac{2}{3}\right):\left(6+\frac{3}{4}\right)=\left(8+\frac{5}{12}\right):X</math>

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::<math>\scriptstyle1:\frac{4}{5}=\left(2+\frac{1}{2}\right):X</math>

::<math>\scriptstyle1:\frac{4}{5}=\left(2+\frac{1}{2}\right):X</math>

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::<math>\scriptstyle\frac{2}{3}:\left(4+\frac{1}{2}\right)=6:X</math>

::<math>\scriptstyle\frac{2}{3}:\left(4+\frac{1}{2}\right)=6:X</math>

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:<span style=color:Green>numerator</span>: In the example of this category, we extract the numerator by that we multiply 4 by the 2 that is beneath the line next to it; the result is 8. We add to it the 1 that is above the 2; the result is 9. We multiply it by the 6, which is the integer of the third number; the result is 54 and this were the numerator, if there were no fraction in the first number. Since there is a fraction in it, we multiply the 54 that we have by the digit that is beneath the line of that fraction, which is 3; the result is 162 and this is the numerator.

:<span style=color:Green>numerator</span>: In the example of this category, we extract the numerator by that we multiply 4 by the 2 that is beneath the line next to it; the result is 8. We add to it the 1 that is above the 2; the result is 9. We multiply it by the 6, which is the integer of the third number; the result is 54 and this were the numerator, if there were no fraction in the first number. Since there is a fraction in it, we multiply the 54 that we have by the digit that is beneath the line of that fraction, which is 3; the result is 162 and this is the numerator.

:<math>\scriptstyle{\color{blue}{\left[\left[\left(4\sdot2\right)+1\right]\sdot6\right]\sdot3=\left[\left(8+1\right)\sdot6\right]\sdot3=\left(9\sdot6\right)\sdot3=54\sdot3=162}}</math>

:<math>\scriptstyle{\color{blue}{\left[\left[\left(4\sdot2\right)+1\right]\sdot6\right]\sdot3=\left[\left(8+1\right)\sdot6\right]\sdot3=\left(9\sdot6\right)\sdot3=54\sdot3=162}}</math>

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!The sixth category: one of the last numbers is an integer alone and the other is a fraction alone.

!The sixth category: one of the last numbers is an integer alone and the other is a fraction alone.

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::<math>\scriptstyle9:\frac{2}{3}=8:X</math>

::<math>\scriptstyle9:\frac{2}{3}=8:X</math>

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:<math>\scriptstyle\frac{2}{3}:\left(4+\frac{1}{2}\right)=6:X</math>

:<math>\scriptstyle\frac{2}{3}:\left(4+\frac{1}{2}\right)=6:X</math>

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:<span style=color:Green>numerator:</span> Then, we extract the numerator by that we multiply the 4 of the second number by the 2, which is beneath the 1 next to it; the result is 8. We add the 1 that is above the 2 of the second number; the result is 9. We multiply it by the 6 that is the third number; the result is 54. We multiply it also by 3 that is beneath the 2 of the first number; the result is 162 and this is the numerator.

:<span style=color:Green>numerator:</span> Then, we extract the numerator by that we multiply the 4 of the second number by the 2, which is beneath the 1 next to it; the result is 8. We add the 1 that is above the 2 of the second number; the result is 9. We multiply it by the 6 that is the third number; the result is 54. We multiply it also by 3 that is beneath the 2 of the first number; the result is 162 and this is the numerator.

:<math>\scriptstyle{\color{blue}{\left[\left[\left(4\sdot2\right)+1\right]\sdot6\right]\sdot3=\left[\left(8+1\right)\sdot6\right]\sdot3=\left(9\sdot6\right)\sdot3=54\sdot3=162}}</math>

:<math>\scriptstyle{\color{blue}{\left[\left[\left(4\sdot2\right)+1\right]\sdot6\right]\sdot3=\left[\left(8+1\right)\sdot6\right]\sdot3=\left(9\sdot6\right)\sdot3=54\sdot3=162}}</math>

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:<math>\scriptstyle6:\left(40+\frac{1}{2}\right)=\frac{2}{3}:X</math>

:<math>\scriptstyle6:\left(40+\frac{1}{2}\right)=\frac{2}{3}:X</math>

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:<math>\scriptstyle9:\frac{2}{3}=8:X</math>

:<math>\scriptstyle9:\frac{2}{3}=8:X</math>

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==== Chapter Seven: Knowing the Ratio of the Six Proportional Numbers [= the Proportional Hexad] ====

==== Chapter Seven: Knowing the Ratio of the Six Proportional Numbers [= the Proportional Hexad] ====

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|There is another way of ratios, which is that if there are six proportional numbers, such that the ratio of the first to the second is as the ratio of the third to the fourth and the ratio of the fifth to the second is as the ratio of the sixth to the fourth, then the ratio of the sum of the first and the fifth to the second is as the ratio of the sum of the third and the sixth to the fourth.

|There is another way of ratios, which is that if there are six proportional numbers, such that the ratio of the first to the second is as the ratio of the third to the fourth and the ratio of the fifth to the second is as the ratio of the sixth to the fourth, then the ratio of the sum of the first and the fifth to the second is as the ratio of the sum of the third and the sixth to the fourth.

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*Example of six proportional numbers that are:

*Example of six proportional numbers that are:

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|style="text-align:right;"|בעבור שבהרבה מקומות יש חלוף באלו הדברי' הנזכרי' צריך שנדבר מאי זה דרך כללי שבו נוכל לדעת כל חלוף מאלה שנרצה

|style="text-align:right;"|בעבור שבהרבה מקומות יש חלוף באלו הדברי' הנזכרי' צריך שנדבר מאי זה דרך כללי שבו נוכל לדעת כל חלוף מאלה שנרצה

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|style="text-align:right;"|ולדעת זה נסדר כל אלו המספרי' ככה

|style="text-align:right;"|ולדעת זה נסדר כל אלו המספרי' ככה

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|6||3||9||6||4

|6||3||9||6||4

|İskenderun||İskenderun||Bursa ||Bursa ||Constantinople

|İskenderun||İskenderun||Bursa ||Bursa ||Constantinople

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|ו||ג||ט||ו||ד

|ו||ג||ט||ו||ד

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|Since our question is how much six from Alexandria are worth in Constantinople, we should know first how much the three from Alexandria are worth in Constantinople.

|Since our question is how much six from Alexandria are worth in Constantinople, we should know first how much the three from Alexandria are worth in Constantinople.

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|But, three from Alexandria are worth nine in Bursa, so, when we know how much nine from Bursa are worth in Constantinople, then we will know how much the three from Alexandria are worth in Constantinople.

|But, three from Alexandria are worth nine in Bursa, so, when we know how much nine from Bursa are worth in Constantinople, then we will know how much the three from Alexandria are worth in Constantinople.

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|We already know that the six from Bursa are worth four from Constantinople, therefore we say: if 6 are equal 4, how much 9 are equal?

|We already know that the six from Bursa are worth four from Constantinople, therefore we say: if 6 are equal 4, how much 9 are equal?

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|<math>\scriptstyle\begin{cases}\scriptstyle b-1=a+1\\\scriptstyle2\sdot\left(a-1\right)=b+1\end{cases}</math>

|<math>\scriptstyle\begin{cases}\scriptstyle b-1=a+1\\\scriptstyle2\sdot\left(a-1\right)=b+1\end{cases}</math>

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|Therefore, it is visible that when 2, which is the mean number, is added to the smaller, it makes it half the whole; and if we subtract the 2 that we added to the smaller number and add it to the greater number, the greater becomes double the smaller.

|Therefore, it is visible that when 2, which is the mean number, is added to the smaller, it makes it half the whole; and if we subtract the 2 that we added to the smaller number and add it to the greater number, the greater becomes double the smaller.

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|Hence, it is clear that the greater is two parts of the whole and the smaller is one part, so they are three equal parts of the whole.

|Hence, it is clear that the greater is two parts of the whole and the smaller is one part, so they are three equal parts of the whole.

|So, we say that if the sum of the mean number with the smaller number is the reason that the smaller number is half the whole and their difference is the reason that it is a third of the whole, then we can deduce from this that the mean number is the difference from the third to the half. But, the difference from the third to the half is a sixth of the whole. To know that this is indeed the difference, we subtract the third from the half; a sixth remains. Hence the mean is a sixth of the whole.

|So, we say that if the sum of the mean number with the smaller number is the reason that the smaller number is half the whole and their difference is the reason that it is a third of the whole, then we can deduce from this that the mean number is the difference from the third to the half. But, the difference from the third to the half is a sixth of the whole. To know that this is indeed the difference, we subtract the third from the half; a sixth remains. Hence the mean is a sixth of the whole.

|style="text-align:right;"|א"כ נאמ' שאם חבור המספר האמצעי למספר הקטן <s>ה</s>היה סבה שיהיה המספר הקטן חצי הכל וחסרונם היה סבה שיהיה שליש הכל א"כ יראה לנו מזה שהמספר האמצעי הוא הבדל שיש מהשליש אל <s>האמצעי</s> [החצי]&#x202B;<ref>marg.</ref> אבל ההבדל מהשליש אל החצי הוא ששית הכל ולדעת שזהו ההבדל נחסר השליש מהחצי וישאר ששית וא"כ יראה שהאמצעי הוא ששית הכל

|style="text-align:right;"|א"כ נאמ' שאם חבור המספר האמצעי למספר הקטן <s>ה</s>היה סבה שיהיה המספר הקטן חצי הכל וחסרונם היה סבה שיהיה שליש הכל א"כ יראה לנו מזה שהמספר האמצעי הוא הבדל שיש מהשליש אל <s>האמצעי</s> [החצי]&#x202B;<ref>marg.</ref> אבל ההבדל מהשליש אל החצי הוא ששית הכל ולדעת שזהו ההבדל נחסר השליש מהחצי וישאר ששית וא"כ יראה שהאמצעי הוא ששית הכל

:<math>\scriptstyle{\color{OliveGreen}{C=\left(A+C\right)-A=\left[\frac{1}{2}\sdot\left(A+B+C\right)\right]-\left[\frac{1}{3}\sdot\left(A+B+C\right)\right]=\left(\frac{1}{2}-\frac{1}{3}\right)\sdot\left(A+B+C\right)=\frac{1}{6}\sdot\left(A+B+C\right)}}</math>

:<math>\scriptstyle{\color{OliveGreen}{C=\left(A+C\right)-A=\left[\frac{1}{2}\sdot\left(A+B+C\right)\right]-\left[\frac{1}{3}\sdot\left(A+B+C\right)\right]=\left(\frac{1}{2}-\frac{1}{3}\right)\sdot\left(A+B+C\right)=\frac{1}{6}\sdot\left(A+B+C\right)}}</math>

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|If the mean is a sixth of the whole, and it is 2, then the whole is 12, which is 6 times two.

|If the mean is a sixth of the whole, and it is 2, then the whole is 12, which is 6 times two.

|style="text-align:right;"|כי בזה ייוסדו כל אלה השאלות ובדרך זה נעשה בכל הדומה וזה מה שרצינו

|style="text-align:right;"|כי בזה ייוסדו כל אלה השאלות ובדרך זה נעשה בכל הדומה וזה מה שרצינו

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|So, we say that if the sum of the mean number with the smaller number makes it half the whole and its subtraction makes it one part of 21 parts of the whole, then it can be deduced from this that the mean number is [the difference from] one part of 21 parts of the whole to a half of the whole. But, the difference from one part of 21 parts of the whole to a half of the whole is 19 parts of 42 parts of the whole. So, the mean is 19 parts of 42 parts of the whole. To know that this is indeed the difference, we subtract one part of 21 parts of the whole from a half; 19 parts of 42 parts of the whole remain.

|So, we say that if the sum of the mean number with the smaller number makes it half the whole and its subtraction makes it one part of 21 parts of the whole, then it can be deduced from this that the mean number is [the difference from] one part of 21 parts of the whole to a half of the whole. But, the difference from one part of 21 parts of the whole to a half of the whole is 19 parts of 42 parts of the whole. So, the mean is 19 parts of 42 parts of the whole. To know that this is indeed the difference, we subtract one part of 21 parts of the whole from a half; 19 parts of 42 parts of the whole remain.

|style="text-align:right;"|אם כן נאמר] שאם חבור [המספר] האמצעי [למספר הקטן ע]שהו חצי הכל [וחסרנו] עשהו [אחד מכ"א חלקי הכ]ל א"כ יראה מזה [שה]מספר האמצעי הוא [...] שיש [...]לה אחד מכ"א חלקי הכל ובין חצי הכל אבל [ההבדל בין] חלק [אחד מכ]"א חלקי ה[כל] ובין [חצי] הכל י"ט חלקי' ממ"ב חלקי הכל א[ם כן ה]אמצ[עי] הוא י"ט חלקי' ממ"ב של הכל [ול]דעת שזה [ההבדל] נחסר חלק אחד מכ"א חלקי ש[לם מה]חצי וישאר י"ט חלקי' ממ"ב חלקי השלם

|style="text-align:right;"|אם כן נאמר] שאם חבור [המספר] האמצעי [למספר הקטן ע]שהו חצי הכל [וחסרנו] עשהו [אחד מכ"א חלקי הכ]ל א"כ יראה מזה [שה]מספר האמצעי הוא [...] שיש [...]לה אחד מכ"א חלקי הכל ובין חצי הכל אבל [ההבדל בין] חלק [אחד מכ]"א חלקי ה[כל] ובין [חצי] הכל י"ט חלקי' ממ"ב חלקי הכל א[ם כן ה]אמצ[עי] הוא י"ט חלקי' ממ"ב של הכל [ול]דעת שזה [ההבדל] נחסר חלק אחד מכ"א חלקי ש[לם מה]חצי וישאר י"ט חלקי' ממ"ב חלקי השלם

:<math>\scriptstyle{\color{OliveGreen}{C=\left(A+C\right)-A=\left[\frac{1}{2}\sdot\left(A+B+C\right)\right]-\left[\frac{1}{21}\sdot\left(A+B+C\right)\right]=\left(\frac{1}{2}-\frac{1}{21}\right)\sdot\left(A+B+C\right)=\frac{19}{42}\sdot\left(A+B+C\right)}}</math>

:<math>\scriptstyle{\color{OliveGreen}{C=\left(A+C\right)-A=\left[\frac{1}{2}\sdot\left(A+B+C\right)\right]-\left[\frac{1}{21}\sdot\left(A+B+C\right)\right]=\left(\frac{1}{2}-\frac{1}{21}\right)\sdot\left(A+B+C\right)=\frac{19}{42}\sdot\left(A+B+C\right)}}</math>

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|It is clear that the mean is 19 parts of 42 parts of the whole. But, if it is 19 parts of 42 parts of the whole and it is 2, then the whole is 4 integers and 8 parts of 19 parts of the whole.

|It is clear that the mean is 19 parts of 42 parts of the whole. But, if it is 19 parts of 42 parts of the whole and it is 2, then the whole is 4 integers and 8 parts of 19 parts of the whole.

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:<math>\scriptstyle\frac{1}{3}X+\frac{1}{4}X=10+\frac{1}{2}</math>

:<math>\scriptstyle\frac{1}{3}X+\frac{1}{4}X=10+\frac{1}{2}</math>

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:<math>\scriptstyle\frac{1}{4}X+\frac{1}{5}X=5</math>

:<math>\scriptstyle\frac{1}{4}X+\frac{1}{5}X=5</math>

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:<math>\scriptstyle\frac{1}{3}X+\frac{1}{5}X=10</math>

:<math>\scriptstyle\frac{1}{3}X+\frac{1}{5}X=10</math>

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:<math>\scriptstyle\frac{1}{3}X+\frac{1}{4}X=20</math>

:<math>\scriptstyle\frac{1}{3}X+\frac{1}{4}X=20</math>

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:<math>\scriptstyle\frac{1}{3}X+\frac{1}{4}X=6</math>

:<math>\scriptstyle\frac{1}{3}X+\frac{1}{4}X=6</math>

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:We divide it by the denominator, which is 7; the result of division is 6 and 2-sevenths; and this is the number, whose third and quarter summed together are equal to six.  

:We divide it by the denominator, which is 7; the result of division is 6 and 2-sevenths; and this is the number, whose third and quarter summed together are equal to six.  

:<math>\scriptstyle{\color{blue}{X=\frac{72}{7}=10+\frac{2}{7}}}</math>

:<math>\scriptstyle{\color{blue}{X=\frac{72}{7}=10+\frac{2}{7}}}</math>

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!Another teaching of proportions

!Another teaching of proportions

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:Answer: we take the known number, which is 12, then add its half to it; it is 18. We take a half of this sum; it is 9. We add another 12 to the 9; it is 21. We take its half; it is ten and a half and this is the original number that we seek for.

:Answer: we take the known number, which is 12, then add its half to it; it is 18. We take a half of this sum; it is 9. We add another 12 to the 9; it is 21. We take its half; it is ten and a half and this is the original number that we seek for.

|style="text-align:right;"|<big>התשובה</big> נקח המספר הידוע שהוא י"ב ונחבר אליו חציו ויהיו י"ח ומכל זה נקח החצי ויהיה ט' ואלו הט' נחבר עוד י"ב ויהיו כ"א ונקח חצים והם עשרה וחצי וזהו המספר שבקשנו בתחלה

|style="text-align:right;"|<big>התשובה</big> נקח המספר הידוע שהוא י"ב ונחבר אליו חציו ויהיו י"ח ומכל זה נקח החצי ויהיה ט' ואלו הט' נחבר עוד י"ב ויהיו כ"א ונקח חצים והם עשרה וחצי וזהו המספר שבקשנו בתחלה

:<math>\scriptstyle{\color{blue}{X=\frac{1}{2}\sdot\left[12+\frac{1}{2}\sdot\left[12+\left(\frac{1}{2}\sdot12\right)\right]\right]=\frac{1}{2}\sdot\left[12+\left(\frac{1}{2}\sdot18\right)\right]=\frac{1}{2}\sdot\left(12+9\right)=\frac{1}{2}\sdot21=10+\frac{1}{2}}}</math>

:<math>\scriptstyle{\color{blue}{X=\frac{1}{2}\sdot\left[12+\frac{1}{2}\sdot\left[12+\left(\frac{1}{2}\sdot12\right)\right]\right]=\frac{1}{2}\sdot\left[12+\left(\frac{1}{2}\sdot18\right)\right]=\frac{1}{2}\sdot\left(12+9\right)=\frac{1}{2}\sdot21=10+\frac{1}{2}}}</math>

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|Likewise, we can say this for any number, whether integers, or fractions, either few or many.  

|Likewise, we can say this for any number, whether integers, or fractions, either few or many.  

:Answer: we take the one that remains lastly plus one; it is 2. We double it; it is 4, and add one; it is 5. We double it; it is ten, and add one to it; it is 11. We double it; it is 22, and this is the original number that we seek for.

:Answer: we take the one that remains lastly plus one; it is 2. We double it; it is 4, and add one; it is 5. We double it; it is ten, and add one to it; it is 11. We double it; it is 22, and this is the original number that we seek for.

|style="width:45%; text-align:right;"|<big>התשובה</big> נקח האחד שנשאר באחרונה ויותר אחד ויהיו ב' ונכפול אותם ויהיו ד' ונוסיף יותר אחד ויהיו ה' ונכפול אותם ויהיו עשרה ונוסיף עליהם &#x202B;<ref>44v</ref>עוד אחד ויהיו י"א ו{{#annot:term|785,1230|DHAI}}נכפלם{{#annotend:DHAI}} ויהיו כ"ב וזהו המספר הראשון שנשאל עליו

|style="width:45%; text-align:right;"|<big>התשובה</big> נקח האחד שנשאר באחרונה ויותר אחד ויהיו ב' ונכפול אותם ויהיו ד' ונוסיף יותר אחד ויהיו ה' ונכפול אותם ויהיו עשרה ונוסיף עליהם &#x202B;<ref>44v</ref>עוד אחד ויהיו י"א ו{{#annot:term|785,1230|DHAI}}נכפלם{{#annotend:DHAI}} ויהיו כ"ב וזהו המספר הראשון שנשאל עליו

:<math>\scriptstyle{\color{blue}{X=2\sdot\left[1+\left[2\sdot\left[1+\left[2\sdot\left(1+1\right)\right]\right]\right]\right]=2\sdot\left[1+\left[2\sdot\left[1+\left(2\sdot2\right)\right]\right]\right]=2\sdot\left[1+\left(2\sdot5\right)\right]=2\sdot\left(1+10\right)=2\sdot11=22}}</math>

:<math>\scriptstyle{\color{blue}{X=2\sdot\left[1+\left[2\sdot\left[1+\left[2\sdot\left(1+1\right)\right]\right]\right]\right]=2\sdot\left[1+\left[2\sdot\left[1+\left(2\sdot2\right)\right]\right]\right]=2\sdot\left[1+\left(2\sdot5\right)\right]=2\sdot\left(1+10\right)=2\sdot11=22}}</math>

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|Likewise, you can do even if you add each time two, or three, or as much as you wish, instead of the one.

|Likewise, you can do even if you add each time two, or three, or as much as you wish, instead of the one.

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|style="text-align:right;"|ובעבור שהאורך {{#annot:term|1621,1872|Nm1C}}ישוער{{#annotend:Nm1C}} בקוים נאמ' מהו הקו

|style="text-align:right;"|ובעבור שהאורך {{#annot:term|1621,1872|Nm1C}}ישוער{{#annotend:Nm1C}} בקוים נאמ' מהו הקו

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|style="text-align:right;"|והקו הוא כמות אורך בלתי רוחב ועומק וקצותיו הם ב' נקודות{{#annotend:wrX5}}

|style="text-align:right;"|והקו הוא כמות אורך בלתי רוחב ועומק וקצותיו הם ב' נקודות{{#annotend:wrX5}}

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:*plane

:*plane

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|When you want to know the height, whichever it may be, take a pole whose size you know and place it on the surface, standing straight with no inclination

|When you want to know the height, whichever it may be, take a pole whose size you know and place it on the surface, standing straight with no inclination

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|Turn your eyes to the ground while looking at the edge of the height you want in a way that the ray of sight passes through the edge of pole's height in a straight line from your eye to the edge of the height you want.

|Turn your eyes to the ground while looking at the edge of the height you want in a way that the ray of sight passes through the edge of pole's height in a straight line from your eye to the edge of the height you want.

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|When we want to know the length of a surface, a straight pole, whose size we know, is placed on the ground where we want, standing straight with no inclination.

|When we want to know the length of a surface, a straight pole, whose size we know, is placed on the ground where we want, standing straight with no inclination.

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|style="text-align:right;"|וראשונה נדבר מגדרו

|style="text-align:right;"|וראשונה נדבר מגדרו

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|style="text-align:right;"|<big>השטח</big> הוא כמות בעל אורך ו{{#annot:term|317,1488|SLfo}}רוחב {{#annotend:SLfo}}בלתי עומק ש{{#annot:term|279,1887|tYDo}}תכליותיו{{#annotend:tYDo}} ב' קוים{{#annotend:gJ6u}}

|style="text-align:right;"|<big>השטח</big> הוא כמות בעל אורך ו{{#annot:term|317,1488|SLfo}}רוחב {{#annotend:SLfo}}בלתי עומק ש{{#annot:term|279,1887|tYDo}}תכליותיו{{#annotend:tYDo}} ב' קוים{{#annotend:gJ6u}}

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:If we want to know its area, we divide its side into two equal segments, then draw a straight line from the midpoint D to the opposite vertex, which is A.

:If we want to know its area, we divide its side into two equal segments, then draw a straight line from the midpoint D to the opposite vertex, which is A.

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==== Chapter Two: Knowing the Area of the Isosceles Triangle ====

==== Chapter Two: Knowing the Area of the Isosceles Triangle ====

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:To know its area, we divide the unequal side into two equal segments, then we draw a straight line from the midpoint D to the opposite vertex, which is A.

:To know its area, we divide the unequal side into two equal segments, then we draw a straight line from the midpoint D to the opposite vertex, which is A.

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==== Chapter Three: Knowing the Area of the Scalene Triangle ====

==== Chapter Three: Knowing the Area of the Scalene Triangle ====

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:Therefore, Archimedes and al-Fārābī said that whoever gives the circle a perimeter of three times the diameter and ten parts of 70 of it, gives it more than appropriate, and whoever gives it three times the diameter and ten parts of 71 of it, gives it less than appropriate. So, the true size of the perimeter is between 70 parts and 71 parts of the diameter.

:Therefore, Archimedes and al-Fārābī said that whoever gives the circle a perimeter of three times the diameter and ten parts of 70 of it, gives it more than appropriate, and whoever gives it three times the diameter and ten parts of 71 of it, gives it less than appropriate. So, the true size of the perimeter is between 70 parts and 71 parts of the diameter.

:<math>\scriptstyle{\color{OliveGreen}{\left(3+\frac{10}{71}\right)\sdot\mathbf{diameter}<\mathbf{perimeter}<\left(3+\frac{10}{70}\right)\sdot\mathbf{diameter}}}</math>

:<math>\scriptstyle{\color{OliveGreen}{\left(3+\frac{10}{71}\right)\sdot\mathbf{diameter}<\mathbf{perimeter}<\left(3+\frac{10}{70}\right)\sdot\mathbf{diameter}}}</math>

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|This is what we wanted concerning the plane shapes.

|This is what we wanted concerning the plane shapes.

|style="text-align:right;"|ולכן ראוי שנאמ' קודם מה הוא ה{{#annot:term|587,1850|x4VO}}גשם{{#annotend:x4VO}}

|style="text-align:right;"|ולכן ראוי שנאמ' קודם מה הוא ה{{#annot:term|587,1850|x4VO}}גשם{{#annotend:x4VO}}

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|style="text-align:right;"|הגשם הוא כמות שיש לו ג' {{#annot:term|315,1891|sNrA}}מרחקי'{{#annotend:sNrA}} שהם אורך ורוחב ועומק שתכליותיו הם ב' שטחי&#x202B;'{{#annotend:LrBm}}

|style="text-align:right;"|הגשם הוא כמות שיש לו ג' {{#annot:term|315,1891|sNrA}}מרחקי'{{#annotend:sNrA}} שהם אורך ורוחב ועומק שתכליותיו הם ב' שטחי&#x202B;'{{#annotend:LrBm}}

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|Whoever is well versed in these principles can easily study all the books of Euclid and any science that obtains its principles from arithmetic.

|Whoever is well versed in these principles can easily study all the books of Euclid and any science that obtains its principles from arithmetic.

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|We have already discussed at length everything that is necessary for our purpose.

|We have already discussed at length everything that is necessary for our purpose.

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:<math>\scriptstyle\frac{20}{3}\times\frac{19}{5}</math>

:<math>\scriptstyle\frac{20}{3}\times\frac{19}{5}</math>

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:<math>\scriptstyle\frac{3}{5}\div\frac{2}{3}</math>

:<math>\scriptstyle\frac{3}{5}\div\frac{2}{3}</math>

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:<math>\scriptstyle20:\left(15+\frac{5}{37}\right)=\left(5+\frac{2}{7}\right):X</math>

:<math>\scriptstyle20:\left(15+\frac{5}{37}\right)=\left(5+\frac{2}{7}\right):X</math>

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|style="text-align:right;"|ונשלם הספר הזה

|style="text-align:right;"|ונשלם הספר הזה

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|style="text-align:right;"|ספר המספר לעלי לידי לי אליאו גבה בכ"ר אליעזר יצ"ו

|style="text-align:right;"|ספר המספר לעלי לידי לי אליאו גבה בכ"ר אליעזר יצ"ו

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