Difference between revisions of "ספר האלזיברא"

|style="text-align:right;"|ושורש ס"ג {{#annot:term|941,1633|fd1E}}הוא המבוקש{{#annotend:fd1E}}

|

:{{#annot:definition|811|Nl0O}}This is because the ratio of a square to a square is the same as the ratio of its side to its side duplicate.

:<math>\scriptstyle{\color{OliveGreen}{a^2:b^2=\left(a:b\right)^2}}</math>

|

:According to {{#annot:reference|811|D3f5}}[[ספר_היסודות_לאקלידס#Elements_VIII_11|<span style=color:blue>'''Euclid, Elements, Book VIII, proposition 11'''</span>]]

|

:{{#annot:definition|812|eDWE}}This is because the ratio of a cube to a cube is the same as the ratio of its side to its side triplicate

:<math>\scriptstyle{\color{OliveGreen}{a^3:b^3=\left(a:b\right)^3}}</math>

|

:{{#annot:reference|812|jD6N}}According to [[ספר_היסודות_לאקלידס#Elements_VIII_12|<span style=color:blue>'''Euclid, Elements, Book VIII, proposition 12'''</span>]]

*{{#annot:√9׳√8|737|z96Z}}In order to teach you, I will give you an example of numbers that have roots and say: you wish to multiply the square root of 9, which is 3, by the cube root of 8, which is 2.

:<math>\scriptstyle{\color{blue}{\sqrt[3]{8}\sdot\sqrt{9}=2\sdot3}}</math>

:It is known that the product of 3 by 2 is 6 and this is the required.

:<math>\scriptstyle{\color{blue}{2\sdot3=6}}</math>

:According to the way that we mentioned:

:9 should be cubed; it is 729.

:<math>\scriptstyle{\color{blue}{9^3=729}}</math>

:8 should be squared; it is 64.

:<math>\scriptstyle{\color{blue}{8^2=64}}</math>

:Multiply 64 by 729; the result is 46656.

:<math>\scriptstyle{\color{blue}{64\sdot729=46656}}</math>

:The square root of the cube root of 46656 is the required.

:<math>\scriptstyle{\color{blue}{\sqrt[3]{8}\sdot\sqrt{9}=\sqrt{\sqrt[3]{46656}}}}</math>

:To add a further explanation, we reduce the required to a number; we can do this, since the numbers we took in our example have roots:

:We say that the cube root of 46656 is 36 and the square root of 36 is 6. So, 6 is the required as we said at the beginning.

:<math>\scriptstyle{\color{blue}{\sqrt[3]{8}\sdot\sqrt{9}=\sqrt{\sqrt[3]{46656}}=\sqrt{36}=6}}</math>

*{{#annot:⁴√4×⁴√7|737|XzLn}}Example: you wish to multiply the square root of the root of 4 by the square root of the root of 7.

:<math>\scriptstyle{\color{blue}{\sqrt{\sqrt{4}}\sdot\sqrt{\sqrt{7}}}}</math>

:Multiply 4 by 7; the result is 28.

:<math>\scriptstyle{\color{blue}{4\sdot7=28}}</math>

:The square root of the square root of 28 is the required.

:<math>\scriptstyle{\color{blue}{\sqrt{\sqrt{4}}\sdot\sqrt{\sqrt{7}}=\sqrt{\sqrt{28}}}}</math>

*{{#annot:2×⁴√5|737|BT9a}}Example: you wish to multiply the square root of the root of 5 by 2.

:<math>\scriptstyle{\color{blue}{2\sdot\sqrt{\sqrt{5}}}}</math>

:Square 2; it is 4. Square 4; it is 16.

:<math>\scriptstyle{\color{blue}{\left(2^2\right)^2=4^2=16}}</math>

:Multiply 5 by 16; the result is 80.

:<math>\scriptstyle{\color{blue}{5\sdot16=80}}</math>

:The square root of the root of 80 is what you want to know.  

:<math>\scriptstyle{\color{blue}{2\sdot\sqrt{\sqrt{5}}=\sqrt{\sqrt{80}}}}</math>

*{{#annot:³√3×⁴√4|737|NYWq}}Example: you wish to multiply the cube root of 3 by the square root of the square root of 4.  

:<math>\scriptstyle{\color{blue}{\sqrt[3]{3}\sdot\sqrt{\sqrt{4}}}}</math>

:Square the cube, which is 3; it is 9. Square 9; it is 81.

:<math>\scriptstyle{\color{blue}{\left(3^2\right)^2=9^2=81}}</math>

:Cube 4; it is 64.

:<math>\scriptstyle{\color{blue}{4^3=64}}</math>

:Multiply 81 by 64; the result is 5184.

:<math>\scriptstyle{\color{blue}{81\sdot64=5184}}</math>

:The square root of the square root of the cube root of 5184 is what you want.

:<math>\scriptstyle{\color{blue}{\sqrt[3]{3}\sdot\sqrt{\sqrt{4}}=\sqrt{\sqrt{\sqrt[3]{5184}}}}}</math>

:Follow this way:

:<math>\scriptstyle{\color{blue}{5^2=25}}</math>

|style="text-align:right;"|כפול ה' בעצמו יעלה כ"ה

:Multiply also a root of 6 by itself; the result is 6.

:<math>\scriptstyle{\color{blue}{\sqrt{6}^2=6}}</math>

:[The sum] is 31. Keep it.

:<math>\scriptstyle{\color{blue}{25+6=31}}</math>

:Multiply also 5 twice by a root of 6 according to this way:

:<math>\scriptstyle{\color{blue}{2\sdot\left(5\sdot\sqrt{6}\right)}}</math>

:First, multiply 5 by a root of 6: square 5; it is 25; multiply it by 6; the result is 150. A root of 150 is the result of multiplication of 5 by a root of 6.

:<math>\scriptstyle{\color{blue}{5\sdot\sqrt{6}=\sqrt{5^2\sdot6}=\sqrt{25\sdot6}=\sqrt{150}}}</math>

|

:In order for you to learn this, I describe a multiplication diagram for you:

:I draw lines from each of the numbers in the figure to the numbers by which they should be multiplied.

:{{#annot:(√32-3)×(√32-3)|737|vr1a}}10) If you wish to multiply a root of 32 minus 3 by itself.

:<math>\scriptstyle\left(\sqrt{32}-3\right)^2</math>

:First, square 3; it is 9.

:<math>\scriptstyle{\color{blue}{3^2=9}}</math>

:Now, multiply a root of 32 minus a root of 9 by itself this way:

:<math>\scriptstyle{\color{blue}{\sqrt{32}^2=32}}</math>

|style="text-align:right;"|כפול תחלה שרש ל"ב בעצמו יעלה ל"ב

:Multiply also a subtractive root of 9 by itself; the result is an additive 9.

:<math>\scriptstyle{\color{blue}{\left(-\sqrt{9}\right)^2=9}}</math>

|

:Because, you should know that the result of multiplication of a subtractive by a subtractive is an additive, as I will explain.

:[<math>\scriptstyle{\color{OliveGreen}{\left(-\right)\times\left(-\right)=\left(+\right)}}</math>]

:Therefore, add 9 to 32; the result is 41. Keep it.

:<math>\scriptstyle{\color{blue}{9+32=41}}</math>

:Multiply also a root of 32 twice by a subtractive root of 9 according to the above-mentioned way; you receive a subtractive root of 1152.

:<math>\scriptstyle{\color{blue}{2\sdot\left[\sqrt{32}\sdot\left(-\sqrt{9}\right)\right]=-\sqrt{1152}}}</math>

|

:Because, the result of multiplication of any [additive] number or measure by a subtractive is a subtractive.

:[<math>\scriptstyle{\color{OliveGreen}{\left(+\right)\times\left(-\right)=\left(-\right)}}</math>]

:Say that 41 that you kept minus a root of 1152 is the required.

:<math>\scriptstyle{\color{blue}{\left(\sqrt{32}-3\right)^2=41-\sqrt{1152}}}</math>

:{{#annot:(√48+√10)×(√48-√10)|737|V3f6}}11) If you wish to multiply a root of 48 plus a root of 10 by a root of 48 minus a root of 10.

:<math>\scriptstyle\left(\sqrt{48}+\sqrt{10}\right)\times\left(\sqrt{48}-\sqrt{10}\right)</math>

:First, multiply a root of 48 by itself; the result is 48.

:<math>\scriptstyle{\color{blue}{\sqrt{48}^2=48}}</math>

:Multiply also an additive root of 10 by a subtractive root of 10; the result is a subtractive 10.

:<math>\scriptstyle{\color{blue}{\left(+\sqrt{10}\right)\sdot\left(-\sqrt{10}\right)=-10}}</math>

:Subtract it from 48; the remainder is 38.

:<math>\scriptstyle{\color{blue}{48-10=38}}</math>

:Multiply also a root of 48 by an additive root of 10; the result is an additive root of 480.

:<math>\scriptstyle{\color{blue}{\sqrt{48}\sdot\left(+\sqrt{10}\right)=+\sqrt{480}}}</math>

:You have 38 plus an additive root of 480.

:<math>\scriptstyle{\color{blue}{38+\sqrt{480}}}</math>

:Multiply also a root of 48 by a subtractive root of 10; the result is a subtractive root of 480.

:<math>\scriptstyle{\color{blue}{\sqrt{48}\sdot\left(-\sqrt{10}\right)=-\sqrt{480}}}</math>

!<span style=color:Green>Rules of multiplication for expressions of the type (a+b) and (a-b)</span>

*The result of multiplication of any [additive] number by an additive is an additive.

:<math>\scriptstyle{\color{OliveGreen}{a\times\left(+\right)=\left(+\right)}}</math>

*The result of multiplication of any [additive] number by a subtractive is a subtractive.

:<math>\scriptstyle{\color{OliveGreen}{a\times\left(-\right)=\left(-\right)}}</math>

*The result of multiplication of an additive by an additive is an additive.

:<math>\scriptstyle{\color{OliveGreen}{\left(+\right)\times\left(+\right)=\left(+\right)}}</math>

*The result of multiplication of an additive by a subtractive is a subtractive.

:<math>\scriptstyle{\color{OliveGreen}{\left(+\right)\times\left(-\right)=\left(-\right)}}</math>

*The result of multiplication of a subtractive by a subtractive is an additive, as we have said above.

:<math>\scriptstyle{\color{OliveGreen}{\left(-\right)\times\left(-\right)=\left(+\right)}}</math>

|

:To show you a proof of it we describe a geometric illustration and present a numerical example:

*We wish to multiply the number 12 minus the number 4 by the number 8 minus the number 2.

:<math>\scriptstyle{\color{blue}{\left(12-4\right)\times\left(8-2\right)}}</math>

|style="text-align:right;"|רצינו לכפול מספר י"ב פחות מספר ד' במספר ח' פחות מספר ב&#x202B;'

:[[File:Alzibra 11.png|thumb|200px|left]]

|[[File:אלזיברא 11.png|thumb|200px]]

:Let there be a surface ABGD.

:Its side AB is 12 measures.

:<math>\scriptstyle{\color{blue}{AB=12}}</math>

:Its side AG is 8 measures.

:<math>\scriptstyle{\color{blue}{AG=8}}</math>

:We cut segment AH, which is 4 measure, from side AB.

:<math>\scriptstyle{\color{blue}{AH=4}}</math>

:We cut segment AW, which is 2 measure, from side AG.

:<math>\scriptstyle{\color{blue}{AW=2}}</math>

:We draw line HZ from point H, parallel to lines AG, BD and line WC from point W, parallel to lines AB, GD.

|style="text-align:right;"|ונעביר מ{{#annot:term|833,1606|oWim}}נקדת{{#annotend:oWim}} ה' {{#annot:term|592,1450|n2q8}}קו{{#annotend:n2q8}} ה"ז {{#annot:term|825,1821|cNKO}}נכוחי ל{{#annotend:cNKO}}קוי א"ג וב"ד<br>

ומנקדת ו' קו ו"ח נכוחי לקוי א"ב וג"ד

:These two lines intersect in surface [ABGD] at point T.

:They divide [ABGD] into four areas:

:Surface TG, surface TA, and surface TB - we call these three together the gnomon of the shape; and the fourth surface, TD, which we call the sought-after, because its area is equal to the required number that is the product of the mentioned numbers, as you can see.

:<math>\scriptstyle{\color{blue}{\left[ABGD\right]=TG+TA+TB}}</math>

|style="text-align:right;"|לשטח ט"ג ושטח ט"א ושטח ט"ב שלשתם יחד נקראם {{#annot:term|815,1823|9D4Q}}רושם התמונה{{#annotend:9D4Q}}<br>

:There is no need to further elaborate this proof for you.

|

:Now we multiply the above mentioned numbers by each other, according to the aforementioned method:

:We start by multiplying 12 by 8; the result is 96, as the area of the whole surface AD.

:<math>\scriptstyle{\color{blue}{AD=12\times8=96}}</math>

:We multiply 12 also by a subtractive 2; the result is a subtractive 24, as the area of TA plus the area of TB.

:<math>\scriptstyle{\color{blue}{-\left(TA+TB\right)=12\times\left(-2\right)=-24}}</math>

:We also multiply 8 also by a subtractive 4; the result is a subtractive 32, as the area of TA plus the area of TG.

:<math>\scriptstyle{\color{blue}{-\left(TA+TG\right)=8\times\left(-4\right)=-32}}</math>

:If we sum up 24 and 32, the result is 56, as the area of the gnomon summed with the area of TA.

:If we subtract it from the area of the whole surface AD, which is 96, the remainder is the required surface TD minus surface TA. Keep it.

:Finish multiplying the mentioned numbers: multiply a subtractive 2 by a subtractive 4; the result is 8, as the area of surface AT.

:<math>\scriptstyle{\color{blue}{AT=\left(-2\right)\times\left(-4\right)=8}}</math>

:You should add it to the reserved in order to complete the required surface TD.

:<math>\scriptstyle{\color{blue}{TD=AD-\left[\left(TA+TB\right)+\left(TA+TG\right)\right]+TA}}</math>

|style="text-align:right;"|וצריך אתה להוסיפו על {{#annot:term|960,1238|0xSg}}השמור{{#annotend:0xSg}} להשלים שטח ט"ד {{#annot:term|941,1633|wTdp}}המבקש{{#annotend:wTdp}}

|

:<span style=color:Green>The conclusion:</span> therefore it is said that the result of multiplication of a subtractive by a subtractive is an additive.

:<math>\scriptstyle{\color{OliveGreen}{\left(-\right)\times\left(-\right)=\left(+\right)}}</math>

|

:In order to bring it closer to your perception I will describe for you also a multiplication diagram in the way I drew the previous multiplication figure:

|style="text-align:right;"|ולקרבו אל ציורך אתאר לך גם תמונת הכפל <s>הקודמת</s> באופן אשר {{#annot:term|819|uW6o}}צירתי{{#annotend:uW6o}} תמונת הכפל הקודמת  

:{{#annot:√12+√48|738|PyOq}}12) If you wish to add a root of 12 to a root of 48, for example.

:<math>\scriptstyle\sqrt{12}+\sqrt{48}</math>

|style="text-align:right;"|<big>יב ואם</big> רצית לחבר שורש י"ב בשרש מ"ח דרך משל{{#annotend:PyOq}}

:Multiply 12 by 48; the result is 576.

:<math>\scriptstyle{\color{blue}{12\sdot48=576}}</math>

:The root of 576 is 24.

:<math>\scriptstyle{\color{blue}{\sqrt{576}=24}}</math>

:Take its double; the result is 48.

:<math>\scriptstyle{\color{blue}{2\sdot24=48}}</math>

:Add to it the two squares, which are 12 and 48; the result is 108.

:<math>\scriptstyle{\color{blue}{\sqrt{12}+\sqrt{48}=\sqrt{12+48+48}=\sqrt{108}}}</math>

|style="text-align:right;"|והנה שרש ק"ח הוא המבוקש

|

:<span style=color:Green>Geometric proof (no figure is given)</span>: Its proof is that we attach the side of a square of 12 with the side of a square of 48 straightly, so they are two parts of one straight line.

:<math>\scriptstyle{\color{blue}{\sqrt{12}+\sqrt{48}}}</math>

|style="text-align:right;"|<big>מופת</big> זה נחבר צלע מרבע י"ב וצלע מרבע מ"ח {{#annot:term|942,2038|6UED}}על יושר{{#annotend:6UED}} ויהיו שני חלקי קו אחד {{#annot:term|817,1847|wtKL}}ישר{{#annotend:wtKL}}

|

:{{#annot:reference|252|Eeue}}It was already clarified in [[ספר_היסודות_לאקלידס#Elements_II_4|<span style=color:blue>'''Euclid, Elements, Book II, proposition 4'''</span>]] that:

|style="text-align:right;"|וכבר נתבאר ב'''תמונת הרביעית מן המאמר השני לאקלידס'''{{#annotend:Eeue}}

|

:{{#annot:definition|252|JPnC}}When a straight line is cut randomly into two segments, the square on the whole line equals the sum of the two squares that are generated from the two segments plus twice the rectangle encompassed by the two segments.

:<math>\scriptstyle{\color{OliveGreen}{\left(a+b\right)^2=a^2+b^2+2ab}}</math>

|style="text-align:right;"|כי כאשר נחלק קו ישר לב' חלקים איך שקרה הנה מרבע הקו כלו שוה לשני המרבעים ההווים משני החלקים ולכפל השטח הנצב הזויות אשר יקיפו בו שני החלקים{{#annotend:JPnC}}

:When we multiply the two squares by each other, the resulting root, which is 576, is equal to the right-angled surface formed by the two segments.

:<math>\scriptstyle{\color{blue}{\sqrt{12\sdot48}=\sqrt{576}}}</math>

|style="text-align:right;"|ועתה הנה כאשר כפלנו שני המרובעים זה בזה הנה שרש העולה שהוא תקע"ו הוא שוה לשטח הנצב הזויות &#x202B;<ref>126r</ref>ההוה משני החלקי&#x202B;'

:When we take its double we get double the right-angled surface that is encompassed by both segments and when we add the two squares to it, we get the square of the whole line, whose root is the sought-after.

|style="text-align:right;"|וכאשר לקחנו שני דמיוניו עלה בידנו כפל {{#annot:term|591,1824|oh24}}השטח הנצב הזויות{{#annotend:oh24}} אשר יקיפו בו שני החלקים וכאשר חברנו אל זה שני המרובעים {{#annot:term|875,1240|lULj}}עלה בידינו{{#annotend:lULj}} מרבע הקו כלו ושרשו הוא המבקש

:{{#annot:√8+√19|738|mDwl}}13) If you wish to add a root of 8 to a root of 19.

:<math>\scriptstyle\sqrt{8}+\sqrt{19}</math>

:Multiply 8 by 19; the result is 152, which has no root.

:<math>\scriptstyle{\color{blue}{8\sdot19=152}}</math>

|style="text-align:right;"|כפול ח' בי"ט יעלה קנ"ב והנה אין לו שורש

:Take double a root of 152 this way: multiply a root of 152 by 2; the result is a root of 608. Keep it.

:<math>\scriptstyle{\color{blue}{2\sdot\sqrt{152}=\sqrt{4\sdot152}=\sqrt{608}}}</math>

|style="text-align:right;"|קח {{#annot:term|387|hCqO}}שני דמיוני{{#annotend:hCqO}} שרש קנ"ב על הדרך הזאת כפול שרש קנ"ב במספר ב' יעלה שרש תר"ח שמרהו

:Sum up the two squares, which are 8 and 19; the result is 27.

:<math>\scriptstyle{\color{blue}{\left(\sqrt{8}\right)^2+\left(\sqrt{19}\right)^2=8+19=27}}</math>

:Say that the root resulting from adding a root of 27 to a root of 608 is the required.

:<math>\scriptstyle{\color{blue}{\sqrt{8}+\sqrt{19}=\sqrt{27+\sqrt{608}}}}</math>

|

:The proof of this teaching is clear from the one that precedes it.

:{{#annot:³√96+³√324|738|nF3r}}14) If you wish to add a cube root of 96 to a cube root of 324.

:<math>\scriptstyle\sqrt[3]{96}+\sqrt[3]{324}</math>

:Take the greatest common divisor of these two numbers, which is 12.

:Divide 96 by it; the result of division is 8.

:<math>\scriptstyle{\color{blue}{\frac{96}{12}=8}}</math>

:Divide 324 also by it; the result is 27.

:<math>\scriptstyle{\color{blue}{\frac{324}{12}=27}}</math>

:So, 96 is 8 parts of 27 of 324.

:<math>\scriptstyle{\color{blue}{96=324\sdot\frac{8}{27}}}</math>

:Extract the cube root of 8 parts of 27; it is 2-thirds.

:<math>\scriptstyle{\color{blue}{\sqrt[3]{\frac{8}{27}}=\frac{2}{3}}}</math>

|style="text-align:right;"|{{#annot:term|881,1375|Yevf}}קח שרש מעקב{{#annotend:Yevf}} ח' חלקים מכ"ז יהיה ב' שלישי&#x202B;'

:Hence, the cube root of 96 is 2 parts of 3 of a [cube] root of 324.

:<math>\scriptstyle{\color{blue}{\sqrt[3]{96}=\frac{2}{3}\sdot\sqrt[3]{324}}}</math>

:Sum up 2 and 3; the result is 5.

:<math>\scriptstyle{\color{blue}{2+3=5}}</math>

:So, we sum the two cube roots and the result is 5.

:<math>\scriptstyle{\color{blue}{\sqrt[3]{96}+\sqrt[3]{324}=\frac{2}{3}\sdot\sqrt[3]{324}+\sqrt[3]{324}=\frac{5}{3}\sdot\sqrt[3]{324}}}</math>

:Convert 5 into a cube; it is 125.

:<math>\scriptstyle{\color{blue}{5=\sqrt[3]{125}}}</math>

|style="text-align:right;"|אחר כן תעשה מן ה' מעקב והוא קכ"ה

:We receive a cube [root] divided by a sum of two cube roots. Keep it.

:<math>\scriptstyle{\color{blue}{\sqrt[3]{96}+\sqrt[3]{324}=\frac{5}{3}\sdot\sqrt[3]{324}=\frac{\sqrt[3]{125}}{3}\sdot\sqrt[3]{324}}}</math>

:Now, to know the measure of each of the 125 by the measure of the cube 96 and the cube 324, proceed in this way:

|style="text-align:right;"|ועתה לדעת שעור כל אחד מאלו הקכ"ה במדה שבה המעקב האחד צ"ו והמעקב השני שכ"ד עשה על הדרך הזאת

:Take one part of the five mentioned parts; it is half a cube root of 96.

:<math>\scriptstyle{\color{blue}{\frac{1}{3}\sdot\sqrt[3]{324}=\frac{1}{2}\sdot\sqrt[3]{96}}}</math>

|style="text-align:right;"|קח החלק האחד מהחמשה החלקי' הנזכ' והנהו חצי שרש מעקב צ"ו

:Cube the half; you get one-eighth.

:<math>\scriptstyle{\color{blue}{\left(\frac{1}{2}\right)^3=\frac{1}{8}}}</math>

:So, the cube of one part is an eighth of 96; which is 12.

:<math>\scriptstyle{\color{blue}{\frac{1}{3}\sdot\sqrt[3]{324}=\sqrt[3]{\frac{1}{8}\sdot96}=\sqrt[3]{12}}}</math>

|style="text-align:right;"|הנה &#x202B;<ref>126v</ref>מעקב החלק האחד הוא שמיני' מספר צ"ו שהוא י"ב

:Multiply 12 by the 125 you kept; the result is 1500.

:<math>\scriptstyle{\color{blue}{125\sdot12=1500}}</math>

:The cube root of 1500 is the required.

:<math>\scriptstyle{\color{blue}{\sqrt[3]{96}+\sqrt[3]{324}=\sqrt[3]{125\sdot12}=\sqrt[3]{1500}}}</math>

|style="text-align:right;"|והנה שרש <s>ה</s>מעקב אלף ^ות"ק הוא המבקש

|

:According to my way, when "I have led you in paths of uprightness" to calculate this calculation, "I have taught you" the proof "in the way of wisdom" [Proverbs 4, 11].

|style="text-align:right;"|הנה לפי דרכי בהדריכי אותך ''{{#annot:Pr4-11|494|s5Kc}}במעגלי יושר{{#annotend:s5Kc}}'' {{#annot:term|229,1269|NOzT}}לחשוב{{#annotend:NOzT}} זה {{#annot:term|228,1200|TLWR}}החשבון{{#annotend:TLWR}} ''{{#annot:Pr4-11|494|OOqD}}בדרך חכמה הוריתיך{{#annotend:OOqD}}''&#x202B;<ref group=note>משלי ד, יא</ref> המופת

:{{#annot:√30÷√6|740|4Ayu}}15) If you wish to divide a root of 30 by a root of 6.

:<math>\scriptstyle\sqrt{30}\div\sqrt{6}</math>

|style="text-align:right;"|<big>טו ואם</big> רצית לחלק שרש ל' על שרש ו&#x202B;'{{#annotend:4Ayu}}

:Divide 30 by 6; the result is 5 and its root is the required.

:<math>\scriptstyle{\color{blue}{\frac{\sqrt{30}}{\sqrt{6}}=\sqrt{\frac{30}{6}}=\sqrt{5}}}</math>

|style="text-align:right;"|חלק ל' לו' יגיע ה' ושרשו הוא המבקש

|

:This is because {{#annot:definition|811|QcHW}}the ratio of a square to a square is as the ratio of its side to its side duplicated. [ [[ספר_היסודות_לאקלידס#Elements_VIII_11|<span style=color:blue>'''Euclid, Elements, Book VIII, proposition 11'''</span>]] ]

:<math>\scriptstyle{\color{OliveGreen}{a^2:b^2=\left(a:b\right)^2}}</math>

|style="text-align:right;"|וזה כי מפני כי יחס מרבע אל מרבע כיחס צלעו אל צלעו שנוי{{#annotend:QcHW}}

:{{#annot:20÷√10|740|QIDd}}16) If you wish to divide 20 by a root of 10.

:<math>\scriptstyle20\div\sqrt{10}</math>

:Square 20; it is 400. Divide 400 by 10; the result is 40 and the root of 40 is the sought-after.

:<math>\scriptstyle{\color{blue}{\frac{20}{\sqrt{10}}=\sqrt{\frac{20^2}{10}}=\sqrt{\frac{400}{10}}=\sqrt{40}}}</math>

|style="text-align:right;"|עשה מן כ' מרבע והוא ת' חלק ת' לי' יגיע מ' ושרש מ' הוא המבקש

|I present a teaching, if you look carefully, you will understand from it the proof of the two teachings that follow it:

|style="text-align:right;"|הנה אקדים למוד אחד אם תתבונן תבין ממנו מופתי שני הלמודים הנמשכי' אחריו  

:{{#annot:(√8+√4)×(√8-√4)|737|Gjqg}}17) If you wish to multiply a root of 8 minus a root of 4 by a root of 8 plus a root of 4, for example.

:<math>\scriptstyle\left(\sqrt{8}-\sqrt{4}\right)\times\left(\sqrt{8}+\sqrt{4}\right)</math>

|style="text-align:right;"|<big>יז אם</big> רצית לכפול שורש ^ח' פחות שרש ד' בשרש ח' ושרש ד' {{#annot:term|878|s4BT}}יותר{{#annotend:s4BT}} דרך משל{{#annotend:Gjqg}}

:Subtract 4 from 8; 4 remains and the remaining 4 is the required.

:<math>\scriptstyle{\color{blue}{\left(\sqrt{8}-\sqrt{4}\right)\times\left(\sqrt{8}+\sqrt{4}\right)=8-4=4}}</math>

|style="text-align:right;"|חסר ד' מח' ישאר ד' ומספר ד' {{#annot:term|184,1236|MC5r}}הנשאר{{#annotend:MC5r}} הוא המבקש

|

:In order that you will know that this is so, we describe a multiplication diagram, in which we multiply the numbers according to the known method:

:We start by multiplying a root of 8 by a root of 8; the result is 8.

:<math>\scriptstyle{\color{blue}{\sqrt{8}\sdot\sqrt{8}=8}}</math>

:We multiply also a root of 8 by an additive root of 4; the result is an additive root of 32.

:<math>\scriptstyle{\color{blue}{\sqrt{8}\sdot\left(+\sqrt{4}\right)=+\sqrt{32}}}</math>

:We have 8 plus a root of 32. We keep it.

:<math>\scriptstyle{\color{blue}{8+\sqrt{32}}}</math>

:We finish our calculation by multiplying an additive root of 4 by a subtractive root of 4; the result is a subtractive 4.

:<math>\scriptstyle{\color{blue}{\left(+\sqrt{4}\right)\sdot\left(-\sqrt{4}\right)=-4}}</math>

:We multiply also a root of 8 by a subtractive root of 4; the result is subtractive root of 32.

:<math>\scriptstyle{\color{blue}{\sqrt{8}\sdot\left(-\sqrt{4}\right)=-\sqrt{32}}}</math>

:18) If you wish to multiply a root of 8 minus a root of 4 by two other roots, so that the result is a root of 64, not a root of 16.

:<math>\scriptstyle\left(\sqrt{8}-\sqrt{4}\right)\times\left(\sqrt{a}+\sqrt{b}\right)=\sqrt{64}</math>

|style="text-align:right;"|<big>יח ואם</big> רצית לכפול שרש ח' פחות שרש ד' בשני שרשים אחרים יהיה העלה שרש ס"ד לא שרש י"ו

:Divide 64 by 16; the result of division is 4.

:<math>\scriptstyle{\color{blue}{\frac{64}{16}=4}}</math>

:Now, multiply 4 by 8; the result is 32.

:<math>\scriptstyle{\color{blue}{4\sdot8=32}}</math>

|style="text-align:right;"|ועתה כפול ד' בח' יעלה ל"ב

:Multiply it also by 4; the result is 16.

:<math>\scriptstyle{\color{blue}{4\sdot4=16}}</math>

|style="text-align:right;"|עוד תכפלהו בד' יעלה י"ו

:Hence, it should be multiplied by a root of 32 plus a root of 16, so the result is a root of 64 and this is self-explanatory.

:<math>\scriptstyle{\color{blue}{\sqrt{a}+\sqrt{b}=\sqrt{32}+\sqrt{16}}}</math>

|style="text-align:right;"|והנה בשרש ל"ב ושרש י"ו ראוי לכופלם ויהיה העולה שרש ס"ד וזה מובן בעצמו

:{{#annot:√64÷(√8-√4)|740|pY8l}}19) If you wish to divide a root of 64 by a root of 8 minus a root of 4.

:<math>\scriptstyle\sqrt{64}\div\left(\sqrt{8}-\sqrt{4}\right)</math>

:Subtract 4 from 8; 4 remains. Multiply the remaining 4 by itself; the result is 16.

:<math>\scriptstyle{\color{blue}{\left(8-4\right)^2=4^2=16}}</math>

:Divide 64 by 16; the result is 4.

:<math>\scriptstyle{\color{blue}{\frac{64}{16}=4}}</math>

:Now, multiply the quotient by 8; the result is 32.

:<math>\scriptstyle{\color{blue}{4\sdot8=32}}</math>

:Multiply it also by 4, whose root you want to subtract from a root of 8; the result is 16.

:<math>\scriptstyle{\color{blue}{4\sdot4=16}}</math>

|style="text-align:right;"|עוד תכפלהו <s>בל'</s> בד' אשר אמרת לפחות שרשו משרש ח' יעלה י"ו

:So, a root of 32 plus a root of 16 summed together is the sought-after.

:<math>\scriptstyle{\color{blue}{\sqrt{64}\div\left(\sqrt{8}-\sqrt{4}\right)=\sqrt{32}+\sqrt{16}}}</math>

|style="text-align:right;"|והנה שרש ל"ב ושרש י"ו {{#annot:term|178,2083|6rFP}}מחברים{{#annotend:6rFP}} הוא המבוקש

|

:This teaching follows the way of the previous teaching, because the dividend is always equal to the result of multiplication of the number resulting from division by the divisor [lit. the number by which the dividend is divided].

:<math>\scriptstyle{\color{OliveGreen}{\frac{a}{b}\sdot b=a}}</math>

|style="text-align:right;"|זה הלמוד הולך בדרך הלמוד הקודם לו מפני כי לעולם {{#annot:term|605,1563|wHYI}}המספר המתחלק{{#annotend:wHYI}} הוא שוה ל{{#annot:term|241|7pPN}}מספר העולה מכפל{{#annotend:7pPN}} {{#annot:term|783|hrnK}}המספר המגיע בחלוק{{#annotend:hrnK}} ב{{#annot:term|604|qwTn}}מספר אשר אליו יתחלק{{#annotend:qwTn}} {{#annot:term|605,1563|cnxx}}המתחלק{{#annotend:cnxx}}

:Subtract 4 from 8; 4 remains. Multiply the remaining 4 by itself; the result is 16.

:<math>\scriptstyle{\color{blue}{\left(8-4\right)^2=4^2=16}}</math>

:Divide 64 by 16; the result is 4.

:<math>\scriptstyle{\color{blue}{\frac{64}{16}=4}}</math>

:Now, multiply the quotient by 8; the result is 32.

:<math>\scriptstyle{\color{blue}{4\sdot8=32}}</math>

:Multiply it also by 4; the result is 16.

:<math>\scriptstyle{\color{blue}{4\sdot4=16}}</math>

:So, a root of 32 minus a root of 16 is the result of division.

:<math>\scriptstyle{\color{blue}{\sqrt{64}\div\left(\sqrt{8}+\sqrt{4}\right)=\sqrt{32}-\sqrt{16}}}</math>

|style="text-align:right;"|והנה שרש ל"ב פחות שרש י"ו הוא המגיע בחלק

|

:You can see that this teaching follows the technique of the previous teaching exactly.

|style="text-align:right;"|הנך רואה בעיניך כי זה הלמוד הולך בדרך הלמוד הקודם לו לא פחות ולא יתר

|

:Only that instead of your saying in the previous teaching that the sought-after is a root of 32 plus a root of 16 summed together, in the present teaching you say that it is a root of 32 minus a root of 16.

|style="text-align:right;"|רק תחת אמרך בלמוד הקודם לו שהמבוקש הוא שרש ל"ב ושרש י"ו מחברים בזה הלמוד אמרת &#x202B;<ref>127v</ref>שהוא שרש ל"ב פחות שרש י"ו

:Square 8; it is 64.

:<math>\scriptstyle{\color{blue}{8^2=64}}</math>

:Square 2; it is 4.

:<math>\scriptstyle{\color{blue}{2^2=4}}</math>

|style="text-align:right;"|וממספר ב' מרבע יהיה ד&#x202B;'

|

:Now, you return to the two teachings that precede this one. Deduce from this.

:Cube 6; it is 216.

:<math>\scriptstyle{\color{blue}{6^3=216}}</math>

|style="text-align:right;"|עשה מן ו' מעקב [יעלה]&#x202B;<ref>marg.</ref> רי"ו

:Square 10; it is 100.

:<math>\scriptstyle{\color{blue}{10^2=100}}</math>

:By this procedure you equalized [the degrees of] the roots and you made each of them a square root of a cube root.

:Now, divide 216 by 100; the result is 2 and 4 parts of 25.

:<math>\scriptstyle{\color{blue}{\frac{216}{100}=2+\frac{4}{25}}}</math>

:The square root of the cube root of 2 and 4 parts of 25 is the required.

:<math>\scriptstyle{\color{blue}{\sqrt{6}\div\sqrt[3]{10}=\sqrt{\sqrt[3]{2+\frac{4}{25}}}}}</math>

|style="text-align:right;"|ושרש מרבע מן שרש מעקב ב' וד' חלקי' מכ"ה הוא המבוקש

:Square 5 twice; it is 625.

:<math>\scriptstyle{\color{blue}{\left(5^2\right)^2=625}}</math>

|style="text-align:right;"|עשה מן ה' {{#annot:term|855|LGS1}}מרבע מרובע{{#annotend:LGS1}} ויהיה תרכ"ה

:Cube 8; it is 51[2].

:<math>\scriptstyle{\color{blue}{8^3=51{\color{red}{2}}}}</math>

:By this you equalized [the degrees of] the roots.

:Divide 625 by 51[2]; the result is one and 113 parts of 51[2].

:<math>\scriptstyle{\color{blue}{\frac{625}{51{\color{red}{2}}}=1+\frac{113}{51{\color{red}{2}}}}}</math>

|style="text-align:right;"|חלק תרכ"ה בתקי"ג ויגיע א' וקי"ג חלקים מתקי"ג

:The square root of the square root of the cube root of 1 and 113 parts of 51[2] is the required.

:<math>\scriptstyle{\color{blue}{\sqrt[3]{5}\div\sqrt{\sqrt{8}}=\sqrt{\sqrt{\sqrt[3]{1+\frac{113}{51{\color{red}{2}}}}}}}}</math>

|style="text-align:right;"|והנה שרש שרש מרבע מן שרש מעקב א' וקי"ג חלקים מתקי"ג הוא המבוקש

:Multiply 8 by 18; the result is 144.

:<math>\scriptstyle{\color{blue}{8\sdot18=144}}</math>

:Exatract its root; it is 12.

:<math>\scriptstyle{\color{blue}{\sqrt{144}=12}}</math>

:Take its double; it is 24.

:<math>\scriptstyle{\color{blue}{2\sdot12=24}}</math>

:Sum up 8 and 18; it is 26.

:<math>\scriptstyle{\color{blue}{8+18=26}}</math>

:Subtract 24 from 26; 2 remains and the root of 2 is what you want.

:<math>\scriptstyle{\color{blue}{\sqrt{18}-\sqrt{8}=\sqrt{26-24}=\sqrt{2}}}</math>

|style="text-align:right;"|חסר כ"ד מכ"ו ישאר ב' ושורש ב' הוא מה שרצית

|

:To show you the proof for this it I should teach you that:

|

:{{#annot:definition|255|cvaS}}When a straight line is cut randomly into two segments, the sum of the squares on both segments equals twice the rectangle encompassed by both segments plus the square that is generated from the excess of the larger segment over the smaller segment. [ [[ספר_היסודות_לאקלידס#Elements_II_7|<span style=color:blue>'''Euclid, Elements, Book II, proposition 7'''</span>]] ]

|style="text-align:right;"|כי כאשר נחלק קו ישר &#x202B;<ref>128r</ref>לשני חלקים איך שקרה הנה מרבעי שני החלקים שוים לכפל השטח הנצב הזויות אשר יקיפו בו שני החלקים ולמרבע ההוה ממותר החלק הגדול על הקטן{{#annotend:cvaS}}

!<span style=color:Green>Geometric illustration</span>

:[[File:Alzibra 24.png|thumb|150px|left]]

|[[File:אלזיברא 24.png|thumb|150px]]

:Let line AB be cut randomly at point G.

|style="text-align:right;"|ויהיה {{#annot:term|817,1847|4AfE}}קו ישר{{#annotend:4AfE}} עליו <s>על</s> א"ב ו{{#annot:to be cut randomly at point|820|S8Q5}}יחולק איך שקרה על נקדת{{#annotend:S8Q5}} ג&#x202B;'

:We also cut segment AZ from line AG, so that it equals the smaller segment GB.

:<math>\scriptstyle{\color{blue}{AZ=GB}}</math>

|style="text-align:right;"|עוד נחלק מן קו א"ג חלק א"ז ממנו שוה לקו <s>א"ב</s> ג"ב שהוא החלק הקטן

:Line ZG remains, which is the excess of the larger segment over the smaller segment.

|

:<span style=color:Green>Supposition:</span> I say that double the right-angled surface encompassed by lines AG and GB, summed with the square that is formed by ZG, equals the two squares that are formed by AG and GB, when they are summed together.

:<math>\scriptstyle{\color{blue}{\left[2\sdot\left(AG\times GB\right)\right]+ZG^2=AG^2+GB^2}}</math>

|style="text-align:right;"|הנה אומר כי כפל השטח הנצב הזויות {{#annot:term|2526,1855|noZK}}אשר יקיפו בו קוי{{#annotend:noZK}} א"ג ג"ב עם המרבע ההוה מן ז"ג מחברים {{#annot:term|429,1247|X8k9}}יהיו שוים ל{{#annotend:X8k9}}שני המרבעים ההוים מן א"ג וג"ב כאשר {{#annot:term|178,1346|NbAk}}יחברו{{#annotend:NbAk}}

|

:<span style=color:Green>Proof:</span>

:We construct square AGDH from line AG, and square GBKW from line GB.

:<math>\scriptstyle{\color{blue}{AGDH=AG^2\quad GBKW=GB^2}}</math>

|style="text-align:right;"|ו{{#annot:term|1015|Mz5i}}נעשה מן קו א"ג מרבע{{#annotend:Mz5i}} אגד"ה ומן קו ג"ב מרבע גבח"ו

:We draw line ZI from point Z, parallel to AD and GH.

:<math>\scriptstyle{\color{blue}{ZI\parallel AD,\;GH}}</math>

|style="text-align:right;"|ומנקדת ז' {{#annot:term|822,2061|67xx}}נמשיך קו{{#annotend:67xx}} ז"י נכחי לשני קוי א"ד ג"ה

:We extend line WK straight until it meets line ZI at point C.

|style="text-align:right;"|ונמשיך קו ו"ח על יושר עד אשר {{#annot:term|824|UM04}}יפגוש קו{{#annotend:UM04}} ז"י על נקדת כ&#x202B;'

:Now, since GB is equal to line AZ, line ZB is equal to line AG, which is the larger segment, and line BW to line GB, which is the smaller segment.

|style="text-align:right;"|ועתה מפני כי ג"ב שוה לקו א"ז יהיה קו ז"ב שוה לקו א"ג שהוא החלק הגדול וקו ב"ו לקו ג"ב שהוא החלק הקטן

:Therefore, surface BC is equal to the right-angled surface encompassed by the two lines AG and GB that are the two parts of the whole line.

:<math>\scriptstyle{\color{blue}{BC=\left[ZB\times BW\right]=\left(AG\times GB\right)}}</math>

|style="text-align:right;"|אם כן שטח ב"כ שוה ל{{#annot:term|591,1824|vcTz}}שטח נצב הזויות{{#annotend:vcTz}} אשר יקיפו בו שני קוי א"ג ג"ב אשר הם שני חלקי הקו כלו

:Since line AD is equal to line AG and line AZ is equal to line GB, surface ZD is also equal to the right-angled surface encompassed by the two lines AG and GB.

:<math>\scriptstyle{\color{blue}{AD=AG,\;AZ=GB\longrightarrow ZD=\left[AD\times AZ\right]=\left(AG\times GB\right)}}</math>

|style="text-align:right;"|וגם כן מפני כי קו א"ד שוה לקו א"ג וקו א"ז שוה לקו ג"ב יהיה שטח ז"ד גם כן שוה לשטח הנצב הזויות אשר יקיפו בו שני קוי א"ג ג"ב

:So, the two surfaces CB and ZD are equal to double the right-angled surface encompassed by the two lines AG and GB.

:<math>\scriptstyle{\color{blue}{CB+ZD=2\sdot\left(AG\times GB\right)}}</math>

|style="text-align:right;"|אם כן שני שטחי כ"ב וז"ד שוים לכפל השטח הנצב הזויו' אשר יקיפו בו שני קוי א"ג ג"ב

:Surface CH remaining from [the subtraction of] the squares of the two segments is a square that equals the square formed by ZG, which is the excess of the larger segment over the smaller.

:<math>\scriptstyle{\color{blue}{CH=2\sdot\left(AG\times GB\right)-\left(AG^2+GB^2\right)=ZG^2}}</math>

|style="text-align:right;"|ושטח כ"ה {{#annot:term|184,1236|Xt9P}}הנשאר מן{{#annotend:Xt9P}} שני מרבעי שני החלקים הוא מרבע שוה למרבע ההוה מן ז"ג שהוא מותר החלק הגדול על הקטן

:Because, line CK is equal to line ZG; and line HK, which is its other side is equal to line ZG.

:<math>\scriptstyle{\color{blue}{CK=ZG,\;HK=ZG}}</math>

|style="text-align:right;"|מפני &#x202B;<ref>128v</ref>כי קו כ"ח שוה לקו ז"ג<br>

וקו ה"ח שהוא צלעו השני שוה לקו ז"ג

:Also, because it is the excess of line GH, which is equal to the larger segment, over line GK, which is equal to the smaller segment.

:<math>\scriptstyle{\color{blue}{CH=GH-GK}}</math>

|style="text-align:right;"|גם כן מפני כי הוא מותר קו ג"ה שהוא שוה לחלק הגדול על קו ג"ח שהוא שוה לחלק הקטן

:Therefore, the two squares formed by AG and GB summed together are equal to the two surfaces BC and ZD, each of which equals the surface encompassed by lines AG and GB that are the two segments of the line, summed with CH that is equal to the square formed by ZG, which is the excess of the larger segment over the smaller segment.

|style="text-align:right;"|הנה שני המרבעים ההוים מן א"ג וג"ב מחברים שוים לשני שטחי ב"כ וז"ד אשר כל אחד מהם שוה לשטח אשר יקיפו בו קוי א"ג ג"ב שהם שני חלקי הקו ול{{#annot:term|305,1263|D62w}}מרבע{{#annotend:D62w}} כ"ה שהוא שוה למרבע ההוה מן ז"ג שהוא מותר החלק הגדול על הקטן כאשר יחברו

|colspan=2|

:<math>\scriptstyle{\color{blue}{AG^2+GB^2=BK+ZD+KH=\left(AG\times GB\right)+\left(AG\times GB\right)+ZG^2}}</math>

:Q.E.D.

|style="text-align:right;"|{{#annot:term|780,2080|pMsQ}}וזה מה שרצינו לבאר{{#annotend:pMsQ}}

:We give a numerical example:

|style="text-align:right;"|ו{{#annot:term|898,1712|wBZD}}נעשה דמיון במספר{{#annotend:wBZD}}

:Let line A[B] the side of a square whose area is 18.

:<math>\scriptstyle{\color{blue}{AB^2=18}}</math>.

|style="text-align:right;"|ויהיה קו א"ג {{#annot:term|325,1464|I58U}}צלע מרבע{{#annotend:I58U}} {{#annot:term|816,1243|Empf}}שבריו{{#annotend:Empf}} י"ח

:Its segment AG is the side of a square whose area is 8.

:Which is the square AH.

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

|style="text-align:right;"|והוא {{#annot:term|305,1263|mzGp}}מרבע{{#annotend:mzGp}} א"ה

|style="text-align:right;"|והנה כאשר כפלנו ח' בי"ח הנה שרש העולה שוה לשטח נצב הזויו' אשר יקיפו בו שני החלקים

|style="text-align:right;"|וכאשר לקחנו שני דמיוניו עלה בידינו כפל השטח הנצב הזויות {{#annot:term|2526,1855|hrMO}}אשר יקיפו בו שני החלקים{{#annotend:hrMO}}

|style="text-align:right;"|וכאשר חסרנו מכ"ו שהוא שברי שני המרובעים נשאר בידינו מרבע מותר החלק הגדול על הקטן

|style="text-align:right;"|וטרם החלי אציע הצעה מבוארה

|style="text-align:right;"|<big>ואומר</big> ראוי שתשכיל ותדע כי יחס מרבע המרבע אל המעקב כיחס המעקב אל המרבע וכיחס המרבע אל הדבר וכיחס הדבר אל האחד

|style="text-align:right;"|וזה מפני כי מספר {{#annot:term|287,2309|3eXG}}האחדים{{#annotend:3eXG}} אשר בדבר כמספר הדברים אשר במרבע וכמספר המרובעים אשר במעקב וכמספר המעקבים אשר במרבע המרבע

|style="text-align:right;"|וזאת ההצעה שמרה כי תצטרך אליה במופתי הלמודים הבאים אחריה

|style="text-align:right;"|&#x202B;<ref>129r</ref>וזה החלי