How prove the $\sqrt{n^2+1}$ is irrational

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I want to know how can we prove $\sqrt{n^2+1}$ is irrational. I know that when I use to find the square root of something by the long division method as suggested in the below image( it about some other expression and does not demonstrate my question but I attached the image to show the method of long division to find square roots.) and we get a remainder them it is irrational but I am not able to know why getting a remainder means irrational.

I want it to be opened because the duplicate suggested what I have already answered but not what I want to know, I want to know how can prove that $\sqrt{n^2+1}$ is irrational by the long division method used to find square roots.

I am talking about this me thod. Here it is used to find $\sqrt{4x^4-28x^3+37x^2+42x+9}$ This is just an example to show the method. My question is above the picture.

Best Answer

Suppose $\sqrt{n^2 + 1}$ is rational (in lowest terms). $$ \sqrt{n^2 + 1} = \frac{a}{b} \text{,} $$ where we may require $a,b \in \Bbb{Z}$, $b > 0$, and $\gcd(a,b) = 1$. We have $$ n^2 + 1 = \frac{a^2}{b^2} \text{.} $$ The left-hand side is an integer, so the right-hand side is an integer. Since $\gcd(a,b) = 1$, this requires $b = 1$. Therefore, $$ n^2 + 1 = a^2 \text{.} $$ But then \begin{align*} 1 &= a^2 - n^2 \\ &= (a+n)(a-n) \text{.} \end{align*} If $n = 0$, this is a factorization of $1=a^2$ and so $a =-1$ or $a=1$ is a solution. If $n \neq 0$, this is a factorization of $1$ into two distinct integers, which is impossible. (The only divisors of $1$ are $-1$ and $1$ and we have already seen how those factors produce $1$.)

Therefore, either $n = 0$ or $\sqrt{n^2+1}$ is irrational.

This is part one of the answer , which covers just the Galley method for division and for square roots, the Tartaglia way. The second part will cover a procedure for nth roots.

Introduction

The Galley method of division , according to Prof. Lam Lay Yong[1], originated in the work Sunzi Suanjing, written in about $400$AD. It was also used by Al-Khwarizmi (for example, in a work completed in $825$AD), and by the likes of Galileo and Tartaglia. It was usurped by long division in ~ $1700$, but remained the primary method of division until then.

Consult [2] for some more details on the history of this method.

We will first see what Galley division is , in general. Then, we'll see how it can be used for square, and then general $n$th root-type calculations.

The idea behind galley division

Take a look at this picture :

This is galley division, as performed by hand. It doesn't look like a galley, does it? I say it does, and you must trust me. Otherwise, do your best to create a boat-like structure around the galley division I perform, and you'll see where the name comes from. (Do it in the comments!)

I picked up this image from [3], which is another excellent well-written resource. However, we should take a look at the idea behind this division now.

Indeed, the idea behind galley division is the same as with long division : begin with the most significant digit, reduce the dividend by a digit at each turn, and simplify.

The key insight, however, is the following : imagine that one is not doing calculations on paper, but on sand instead (or think about writing on paper, but with a pencil whose marks are erasable). In that case, one can "cancel" a particular expression by simply erasing it, and then replace it with another term. This allows us to forget about initial terms and keep our field of view very small, constantly.

That seems to be the basic idea behind galley division : a lot of cancellations, but fewer numbers in total than long division. However, because cancellation marks were difficult to print (thankfully, they're not too difficult to MathJax!) , this method went out of sight, although it's not completely out of use : it's still taught in some parts of North Africa and the Middle East.

Note that in the figure above, figures are "cancelled" not by a strikethrough, but rather by having some other number written above them. That is one of three or four variants of this method, and the one that cancels properly (i.e. like writing on sand) is likely to give a better image of a boat.

A BABY example

Let's start with a baby example : take $858$ divided by $7$.

Step 1 : To begin, write the dividend below the divisor, in such a way that the leftmost digits of the dividend and divisor match up. Draw a vertical bar that is tall enough to capture both the dividend and divisor. This will be extended linearly if other rows are created, but it will be ensured that the quotient forms to the right of this bar, and the remainder will be what is left on the left-hand side, following cancellations.

$$ 858 | \\ 7\ \ \ \ | $$

Step 2 : The "temporary" dividend is formed as follows : as written, we find the position of the rightmost digit of the divisor. Then, everything to the left of that, in the dividend, forms the "temporary" dividend. In this case, that is only $8$. We will divide $8$ by $7$, and need only the quotient for now, which is $1$. Write a $1$ as the first digit of the quotient.

$$ 858 |1 \\ 7 \ \ \ \ | \ \ $$

Step 3 : Formally, we find the remainder of this division by using a cancellation. Note that $8 - 7 \times 1 = 1$, so we cancel the $8$ and write a $1$ above it, giving us a new dividend of $158$. Cancel out the divisor $7$ as well.

$$ \require{cancel} 1 \ \ \ \ |\\ \cancel{8} 58 |1 \\ \ \cancel{7} \ \ \ | \ \ $$

Step 4 : Shift the divisor one step to the right : this can be done by cancelling it and moving to a new line (which is what we do), or can be done by erasure. Now the $7$ will lie underneath the $5$. $$ 1 \ \ \ \ |\\ \cancel{8} 58 |1 \\ \cancel{7} \ \ \ \ | \ \ \\ \ \ 7 \ \ | $$

Step 5 : We are back to step 2. The position of the $7$ is now below the $5$, so taking everything to the right of $4$ gives us a temporary dividend of $15$. We know that $15$ divided by $7$ has a quotient of $2$. So write the $2$ now, after the $1$. $$ 1 \ \ \ \ |\\ \ \ \cancel{8} 58 |1 2 \\ \cancel{7} \ \ \ \ | \ \ \\ \ \ 7 \ \ | $$

Step 6 : The cancellation step gives $15 - (7 \times 2) = 1$. This is represented by cancelling the top $1$ entirely, and then crossing out the $5$ and writing a $1$ above it, so that effectively, $15$ has been rewritten as $1$. Cancel out the divisor $7$ in this process.

$$ \cancel{1} \ 1 \ \ \ | \ \ \\ \ \ \cancel{8} \cancel{5}8 |1 2 \\ \cancel{7} \ \ \ \ \ \ | \ \ \\ \ \ \cancel{7} \ \ | $$

Step 7 : Shift the divisor $7$ one step to the right, like in step 4.

$$ \cancel{1} \ 1 \ \ \ | \ \ \\ \ \ \cancel{8} \cancel{5}8 |1 2 \\ \cancel{7} \ \ \ \ \ \ | \ \ \\ \ \ \cancel{7} \ \ | \\ \ \ \ \ \ \ 7 | $$

Step 8 : The temporary dividend is given by everything to the right of $7$ as located, which leads to $18$ now. Clearly, the quotient is $2$, so we write that $2$ over there.

$$ \cancel{1} \ 1 \ \ \ | \ \ \\ \ \ \ \ \cancel{8} \cancel{5}8 |1 2 2 \\ \cancel{7} \ \ \ \ \ \ | \ \ \\ \ \ \cancel{7} \ \ | \\ \ \ \ \ \ \ 7 | $$

Step 9 : The cancellation step is performed, and $18 - (7 \times 2) = 4$. This is illustrated by cancelling the $1$, and writing a $4$ above the $8$. In effect, we have replaced the $18$ by a $4$. Cancel out the divisor $7$ again.

$$ \ \ \cancel{1} \ \cancel{1} \ 4\ | \ \ \ \ \ \ \ \\ \ \ \cancel{8} \cancel{5}\cancel{8} |1 2 2 \\ \cancel{7} \ \ \ \ \ \ \ \ | \ \ \ \ \\ \cancel{7} \ \ \ \ | \\ \ \ \ \ \cancel{7} | $$

Step 10 : Everything on the right hand side is the quotient in that order. Everything that's not struck out on the left hand side is the remainder in that order.$$ \frac{858}{7} \quad \rightarrow \quad \text{Quotient : } 122, \text{ Remainder : } 4 $$

That's just the baby example. Let's now do a bigger, more glamorous example that involves more cancellation (and as a corollary, a bigger MathJax alignment headache).

A large, super-illustrative example

Let's do, well, I don't know. My MSE ID is $316409$ (it's a semiprime!) , let's use it as a dividend. This is my $1825$th answer so the divisor will be $1825$.

Actually, nothing much changes. The only thing that does change is the cancellation step : because we don't have a small divisor anymore, we will have to deal in way more cancellations. However, that will be taken care of.

Step 1 : Write the dividend and divisor so that the leftmost digit of the divisor is directly underneath that of the dividend. Draw a bar to the left of the divisor. $$ \begin{matrix} 3&1&6&4&0&9&| \\ 1&8&2&5& & &| \end{matrix} $$

Step 2 : Form the temporary divisor. The rightmost digit of $1825$ is the $5$, and everything to the left of that is $3164$. So we get $3164$ as the temporary divisor. Clearly, $3164 / 1825$ has a quotient of $1$. Write the $1$. $$ \begin{matrix} 3&1&6&4&0&9&|&1 \\ 1&8&2&5& & &|& \end{matrix} $$

Step 3a : We have , now a very elaborate cancellation step. It begins as follows. The quotient digit is $1$. Multiply this with the first digit of the divisor i.e. $1$, and subtract it from the number above i.e. a $3$. This gives $3 - (1 \times 1) = 2$. Write that $2$ above the $3$ with a cancellation. Cancel the $1$ from the divisor, which has now been used in the subtraction.

$$ \begin{matrix} 2&&&&&&|&\\ \cancel{3}&1&6&4&0&9&|&1 \\ \cancel{1}&8&2&5& & &|& \end{matrix} $$

Step 3b : In the second part of the cancellation step, we now multiply the quotient digit with the next uncancelled digit of the divisor, and subtract it from the corresponding digit of the dividend. In this case, we have $1 - (1 \times 8)$. That's negative, though!

To counteract this, we borrow from the $2$ on the top right column, making it $21 - (1 \times 8) = 13$ instead. To show that the $2$ has been borrowed, we strike it out and write a $1$ above it. Write a $3$ above the $1$ , so that $21$ has been effectively replaced with $13$. Strike out the $8$.

$$ \begin{matrix} 1&&&&&&|& \\ \cancel{2}&3&&&&&|&\\ \cancel{3}&\cancel{1}&6&4&0&9&|&1 \\ \cancel{1}&\cancel{8}&2&5& & &|& \end{matrix} $$

Step 3c : The next divisor digit is $2$. Multiply that with the quotient digit $1$, and try to subtract it from the digit above the $2$, which is $6$. This leads to $6-(2 \times 1) = 4$. Cancel the $6$ and put a $4$ above it. Cancel the $2$.

$$ \begin{matrix} 1&&&&&&|& \\ \cancel{2}&3&4&&&&|&\\ \cancel{3}&\cancel{1}&\cancel{6}&4&0&9&|&1 \\ \cancel{1}&\cancel{8}&\cancel{2}&5& & &|& \end{matrix} $$

Step 3d : In the last of these cancellation steps, the remaining digit of the divisor $5$ is multiplied with the quotient digit $1$, and we attempt to subtract it from the number above the $5$, which is a $4$. However, $4 - (5 \times 1)$ is negative.

To counter this, we borrow the $4$ that's to the left and above this $4$, giving the number $44$. Now, subtraction gives $44 - (5 \times 1) = 39$. Cut the upper $4$ out and write a $3$ above it. Cut the lower $4$ out and write a $9$ above it. Cut the $5$ out.

$$ \begin{matrix} 1&&3&&&&|& \\ \cancel{2}&3&\cancel{4}&9&&&|&\\ \cancel{3}&\cancel{1}&\cancel{6}&\cancel{4}&0&9&|&1 \\ \cancel{1}&\cancel{8}&\cancel{2}&\cancel{5}& & &|& \end{matrix} $$

Step 4 : The divisor $1825$ is now written on the last line, but it's now shifted one place to the right compared to the previous $1825$. That gives $$ \begin{matrix} 1&&3&&&&|& \\ \cancel{2}&3&\cancel{4}&9&&&|&\\ \cancel{3}&\cancel{1}&\cancel{6}&\cancel{4}&0&9&|&1 \\ \cancel{1}&\cancel{8}&\cancel{2}&\cancel{5}& & &|& \\ &1&8&2&5&&|& \end{matrix} $$

We just need to repeat all these steps again, with $316409 \to 133909$ and a $1$ on the quotient. Let's do that!

Step 5 : Form the temporary divisor like so. The last digit of the divisor is a $5$, and everything to the above-left of that in the dividend results in $13390$ in this case.

We now need to divide $13390$ by $1825$. A rough grasp at a quotient may be done by seeing that $18 \times 8 = 144$ and $18 \times 7 = 126$, so we expect that $1825 \sim 1800$ will also yield a quotient of $7$. This is true, and we write the quotient $7$ after the $1$ now.

$$ \begin{matrix} 1&&3&&&&|& \\ \cancel{2}&3&\cancel{4}&9&&&|&&\\ \cancel{3}&\cancel{1}&\cancel{6}&\cancel{4}&0&9&|&1&7 \\ \cancel{1}&\cancel{8}&\cancel{2}&\cancel{5}& & &|&& \\ &1&8&2&5&&|&& \end{matrix} $$

Step 6a : Begin the cancellation step. The quotient digit is a $7$, and we multiply it with the divisor's first digit, which is a $1$, and then try to remove that from the digit above a $1$ (in this case, a $3$). However, $3 - (1 \times 7)$ is negative. Borrowing the top $1$, however, makes this $13 - (1 \times 7) = 6$. The $1$ is cancelled (no point of writing a $0$ above it, although one should do so for completeness), we cancel $3$ and write a $6$ above it, and cancel the divisor digit $1$.

$$ \begin{matrix} \cancel{1}&6&3&&&&|& \\ \cancel{2}&\cancel{3}&\cancel{4}&9&&&|&&\\ \cancel{3}&\cancel{1}&\cancel{6}&\cancel{4}&0&9&|&1&7 \\ \cancel{1}&\cancel{8}&\cancel{2}&\cancel{5}& & &|&& \\ &\cancel{1}&8&2&5&&|&& \end{matrix} $$

Step 6b : You've seen it many times now, so let's be a little snappy. $7 \times 8 =56$ which is bigger than the $3$, so borrow the $6$ to give $63 - (7 \times 8) = 7$. Cancel the $6$, replace the $3$ by a $7$, cancel the $8$.

$$ \begin{matrix} &&7&&&&|&\\ \cancel{1}&\cancel{6}&\cancel{3}&&&&|& \\ \cancel{2}&\cancel{3}&\cancel{4}&9&&&|&&\\ \cancel{3}&\cancel{1}&\cancel{6}&\cancel{4}&0&9&|&1&7 \\ \cancel{1}&\cancel{8}&\cancel{2}&\cancel{5}& & &|&& \\ &\cancel{1}&\cancel{8}&2&5&&|&& \end{matrix} $$

Step 6c : $7 \times 2 =14$ which is bigger than the $9$, so borrow the $7$ to give $79- (7 \times 2) = 65$. Replace the $79$ by a $65$, cancel the $2$.

$$ \begin{matrix} &&6&&&&|&\\ &&\cancel{7}&&&&|&\\ \cancel{1}&\cancel{6}&\cancel{3}&5&&&|& \\ \cancel{2}&\cancel{3}&\cancel{4}&\cancel{9}&&&|&&\\ \cancel{3}&\cancel{1}&\cancel{6}&\cancel{4}&0&9&|&1&7 \\ \cancel{1}&\cancel{8}&\cancel{2}&\cancel{5}& & &|&& \\ &\cancel{1}&\cancel{8}&\cancel{2}&5&&|&& \end{matrix} $$

Step 6d : $7\times 5 = 35>0$ (well, I mean, obviously, eh?) , so borrowing the $5$ gives $50 - (7 \times 5) = 15$, replace $50$ by $15$ and cancel the $5$.

$$ \begin{matrix} &&6&&&&|&\\ &&\cancel{7}&1&&&|&\\ \cancel{1}&\cancel{6}&\cancel{3}&\cancel{5}&&&|& \\ \cancel{2}&\cancel{3}&\cancel{4}&\cancel{9}&5&&|&&\\ \cancel{3}&\cancel{1}&\cancel{6}&\cancel{4}&\cancel{0}&9&|&1&7 \\ \cancel{1}&\cancel{8}&\cancel{2}&\cancel{5}& & &|&& \\ &\cancel{1}&\cancel{8}&\cancel{2}&\cancel{5}&&|&& \end{matrix} $$

Step 7 : Shift the divisor one place to the right and write it below the cancelled divisor. $$ \begin{matrix} &&6&&&&|&\\ &&\cancel{7}&1&&&|&\\ \cancel{1}&\cancel{6}&\cancel{3}&\cancel{5}&&&|& \\ \cancel{2}&\cancel{3}&\cancel{4}&\cancel{9}&5&&|&&\\ \cancel{3}&\cancel{1}&\cancel{6}&\cancel{4}&\cancel{0}&9&|&1&7 \\ \cancel{1}&\cancel{8}&\cancel{2}&\cancel{5}& & &|&& \\ &\cancel{1}&\cancel{8}&\cancel{2}&\cancel{5}&&|&& \\ &&1&8&2&5&|&& \end{matrix} $$

Step 8: The temporary dividend is $6159$, and upon division by $1825$ it gives a quotient of $3$. Write the $3$ to the right of the $7$ : $$ \begin{matrix} &&6&&&&|&\\ &&\cancel{7}&1&&&|&\\ \cancel{1}&\cancel{6}&\cancel{3}&\cancel{5}&&&|& \\ \cancel{2}&\cancel{3}&\cancel{4}&\cancel{9}&5&&|&&\\ \cancel{3}&\cancel{1}&\cancel{6}&\cancel{4}&\cancel{0}&9&|&1&7&3 \\ \cancel{1}&\cancel{8}&\cancel{2}&\cancel{5}& & &|&& \\ &\cancel{1}&\cancel{8}&\cancel{2}&\cancel{5}&&|&& \\ &&1&8&2&5&|&& \end{matrix} $$

The last cancellation step is left as an exercise, with only the diagrams shown.

Step 9a : $$ \begin{matrix} &&3&&&&|& \\ &&\cancel{6}&&&&|&\\ &&\cancel{7}&1&&&|&\\ \cancel{1}&\cancel{6}&\cancel{3}&\cancel{5}&&&|& \\ \cancel{2}&\cancel{3}&\cancel{4}&\cancel{9}&5&&|&&\\ \cancel{3}&\cancel{1}&\cancel{6}&\cancel{4}&\cancel{0}&9&|&1&7&3 \\ \cancel{1}&\cancel{8}&\cancel{2}&\cancel{5}& & &|&& \\ &\cancel{1}&\cancel{8}&\cancel{2}&\cancel{5}&&|&& \\ &&\cancel{1}&8&2&5&|&& \end{matrix} $$

Step 9b : $$ \begin{matrix} &&\cancel{3}&&&&|& \\ &&\cancel{6}&7&&&|&\\ &&\cancel{7}&\cancel{1}&&&|&\\ \cancel{1}&\cancel{6}&\cancel{3}&\cancel{5}&&&|& \\ \cancel{2}&\cancel{3}&\cancel{4}&\cancel{9}&5&&|&&\\ \cancel{3}&\cancel{1}&\cancel{6}&\cancel{4}&\cancel{0}&9&|&1&7&3 \\ \cancel{1}&\cancel{8}&\cancel{2}&\cancel{5}& & &|&& \\ &\cancel{1}&\cancel{8}&\cancel{2}&\cancel{5}&&|&& \\ &&\cancel{1}&\cancel{8}&2&5&|&& \end{matrix} $$

Step 9c :

$$ \begin{matrix} &&\cancel{3}&6&&&|& \\ &&\cancel{6}&\cancel{7}&&&|&\\ &&\cancel{7}&\cancel{1}&&&|&\\ \cancel{1}&\cancel{6}&\cancel{3}&\cancel{5}&9&&|& \\ \cancel{2}&\cancel{3}&\cancel{4}&\cancel{9}&\cancel{5}&&|&&\\ \cancel{3}&\cancel{1}&\cancel{6}&\cancel{4}&\cancel{0}&9&|&1&7&3 \\ \cancel{1}&\cancel{8}&\cancel{2}&\cancel{5}& & &|&& \\ &\cancel{1}&\cancel{8}&\cancel{2}&\cancel{5}&&|&& \\ &&\cancel{1}&\cancel{8}&\cancel{2}&5&|&& \end{matrix} $$

Step 9d :

$$ \begin{matrix} &&\cancel{3}&6&&&|& \\ &&\cancel{6}&\cancel{7}&&&|&\\ &&\cancel{7}&\cancel{1}&8&&|&\\ \cancel{1}&\cancel{6}&\cancel{3}&\cancel{5}&\cancel{9}&&|& \\ \cancel{2}&\cancel{3}&\cancel{4}&\cancel{9}&\cancel{5}&4&|&&\\ \cancel{3}&\cancel{1}&\cancel{6}&\cancel{4}&\cancel{0}&\cancel{9}&|&1&7&3 \\ \cancel{1}&\cancel{8}&\cancel{2}&\cancel{5}& & &|&& \\ &\cancel{1}&\cancel{8}&\cancel{2}&\cancel{5}&&|&& \\ &&\cancel{1}&\cancel{8}&\cancel{2}&\cancel{5}&|&& \end{matrix} $$

Step 10 : The quotient is everything to the right, in that order. The remainder is everything not cancelled to the left, in that order. $$ \frac{316409}{1825} \quad\rightarrow \quad \text{Quotient : } 173 , \text{ Remainder : } 684 $$

While you're at it, take a look at [4] for another example, and some more methods of division. [5] looks at variants of this method (including the "erasure" method which is convenient for sand-based calculations, and the printer method where cancellation is just avoided as a convention).

What about square roots?

We know that long division, or a variant of it, can be used for square roots. It is not surprising that the long division method can be used for $n$th roots as well. It's just a variant of the binomial theorem, a brief discussion can be found at [6].

It turns out that the square root Galley method of Tartaglia is the same as that of modified long-division, but suited to cancellation. We describe only square roots below : first, an easy version, then a hard one.

Square roots, the galley method

Let's start with a baby example, as usual. Say I want the square root of $7056$.

Step 1 : Split $7056$ into blocks of $2$ from the right : in that case, it would be $(70)(56)$. If the number were $361$, it'd be grouped like $(3)(61)$ instead.

Step 2 : Following grouping, the notation is to write the number (remembering the grouping mentally, we'll forget the brackets) , followed by a bar, after which the answer will appear. $$ \begin{matrix} 7&0&5&6&| \end{matrix} $$

Step 3 : Find the single digit number $x$, whose square is the largest one below the first group ( from the left) of numbers. In this case, the first group is $70$, so that number is $8$. Write $8$ on the right side of the bar. Then, write $8$ below the rightmost digit of the group (in this case, below the $0$). $$ \begin{matrix} 7&0&5&6&|&8 \\ &8&&&& \end{matrix} $$

Step $4$ : we subtract the product of $8$ with itself from the first group, in this case $70$. By the choice of $8$, this will always be negative, and we'll get $70 - 8^2 = 6$. The $7$ and $0$ will be cancelled out to reflect this. $$ \begin{matrix} &6&&&|&\\ \cancel{7}&\cancel{0}&5&6&|&8 \\ &8&&&|& \end{matrix} $$

Step 5 : Double the $8$ in the last row to get $16$. Write the $1$ below the $8$ (by cancellation) and write the $6$ next to the $8$. You've basically shifted the divisor one to the right by doing this, and that's exactly what's done in galley division as well.

$$ \begin{matrix} &6&&&|&\\ \cancel{7}&\cancel{0}&5&6&|&8 \\ &\cancel{8}&6&&|& \\ &1&&&|& \end{matrix} $$

Step 6 : This is the "doubling" step of the algorithm. We must now find the next digit of the quotient. However, how do we do that? The answer is to look back to the long method. We must find $x$ such that $16x \times x$ (where $16x$ is treated as a three digit number) is smaller than $656$. Suppose we find the biggest such value of $x$. Then we place $x$ in the positions shown below $$ \begin{matrix} &6&&&|&&\\ \cancel{7}&\cancel{0}&5&6&|&8&x \\ &\cancel{8}&6&x&|&& \\ &1&&&|&& \end{matrix} $$

In our case, $x=4$ works out. We write $$ \begin{matrix} &6&&&|&&\\ \cancel{7}&\cancel{0}&5&6&|&8&4 \\ &\cancel{8}&6&4&|&& \\ &1&&&|&& \end{matrix} $$

Step 7 : Now, we do the cancellation step of the galley method, but with divisor $164$ , dividend $656$ and quotient digit $4$. However, there is one key difference : we DO NOT cancel out the divisor digits (in this case, $164$), we ONLY cancel digits from the dividend (in this case, $656$). The divisor cancellation will be done following a "doubling step" in the next stage.

That leads to the following sequence of steps. $$ \begin{matrix} &2&&&|&& \\ &\cancel{6}&&&|&&\\ \cancel{7}&\cancel{0}&5&6&|&8&4 \\ &\cancel{8}&6&4&|&& \\ &1&&&|&& \end{matrix} $$

Then

$$ \begin{matrix} &\cancel{2}&&&|&& \\ &\cancel{6}&1&&|&&\\ \cancel{7}&\cancel{0}&\cancel{5}&6&|&8&4 \\ &\cancel{8}&6&4&|&& \\ &1&&&|&& \end{matrix} $$

Finally $$ \begin{matrix} &\cancel{2}&&&|&& \\ &\cancel{6}&\cancel{1}&&|&&\\ \cancel{7}&\cancel{0}&\cancel{5}&\cancel{6}&|&8&4 \\ &\cancel{8}&6&4&|&& \\ &1&&&|&& \end{matrix} $$

Step $8$ : We have exhausted all the digits of the original number. Now, we strike out the entire divisor as written (which is $164$) and look at what is left over $$ \begin{matrix} &\cancel{2}&&&|&& \\ &\cancel{6}&\cancel{1}&&|&&\\ \cancel{7}&\cancel{0}&\cancel{5}&\cancel{6}&|&8&4 \\ &\cancel{8}&\cancel{6}&\cancel{4}&|&& \\ &\cancel{1}&&&|&& \end{matrix} $$

In Galley division, this leads to a "quotient" of $84$ (the square root of the biggest perfect square smaller than $7056$) and a "remainder" (the discrepancy between those two numbers) of $0$. This basically means that $7056 = 84^2+0 = 84^2$.

We'll do another example for good measure.

A bigger, bolder example

Let's do one of the examples that Tartaglia himself uses, in his reference "General Tratto di Numeri", whose untranslated version can be found at [7].

Tartaglia is dealing with the number $968372$.

Step 1 : The grouping , in groups of $2$, leads to $(96)(83)(72)$. Write down the number with a bar at the end.

$$ \begin{matrix} 9&6&8&3&7&2&| \end{matrix} $$

Step 2 : Deal with the first group i.e. $96$. The number $9$ is the largest one-digit number such that $9^2 \leq 96$, so we write $9$ down both in the quotient and in the divisor position.

$$ \begin{matrix} 9&6&8&3&7&2&|&9 \\ &9&&&&&|& \end{matrix} $$

Step 3 : The square of $9$ is $9^2 = 81$. This can't be subtracted from $6$ so we borrow the $9$ to get $96-9^2 = 15$. Cancellations ensue, as usual.

$$ \begin{matrix} 1&5&&&&&|& \\ \cancel{9}&\cancel{6}&8&3&7&2&|&9 \\ &9&&&&&|& \end{matrix} $$

Step 4 : Double the bottommost $9$ to get $18$. Cancel the $9$ and write $18$ below the $9$, so that it is shifted one to the right (just like how the divisor is shifted one space to the right in the usual galley division procedure).

$$ \begin{matrix} 1&5&&&&&|& \\ \cancel{9}&\cancel{6}&8&3&7&2&|&9 \\ &\cancel{9}&8&&&&|& \\ &1&&&&&|& \end{matrix} $$

Step 5 : Now, we will search for the next quotient digit. Mimicking the earlier example, we need the largest single digit $x$ so that $18x \times x \leq 1583$. That $x$ would be inserted at the position $$ \begin{matrix} 1&5&&&&&|& \\ \cancel{9}&\cancel{6}&8&3&7&2&|&9&x \\ &\cancel{9}&8&x&&&|& \\ &1&&&&&|& \end{matrix} $$

That $x$ can be found. Indeed, $x=8$ has $188 \times 8 = 1504$, while $x=9$ gives $189 \times 9 = 1701$, so $x=8$ fits the bill. Write the $8$ in place of $x$.

$$ \begin{matrix} 1&5&&&&&|& \\ \cancel{9}&\cancel{6}&8&3&7&2&|&9&8 \\ &\cancel{9}&8&8&&&|& \\ &1&&&&&|& \end{matrix} $$

Step 6 : Now, just like the cancellation step in the original galley division, we will use $8$ as the last quotient digit, $188$ as the divisor, and $1583$ as the dividend. However, recall the key difference : we DO NOT cancel out the divisor digits (in this case, $188$), we ONLY cancel digits from the dividend (in this case, $1583$). The divisor cancellation will be done following a change to a new divisor in the next stage.

For the digit $1$ : $$ \begin{matrix} &7&&&&&|&& \\ \cancel{1}&\cancel{5}&&&&&|& \\ \cancel{9}&\cancel{6}&8&3&7&2&|&9&8 \\ &\cancel{9}&8&8&&&|& \\ &1&&&&&|& \end{matrix} $$ For the first $8$ : $$ \begin{matrix} &1&&&&&|&& \\ &\cancel{7}&&&&&|&& \\ \cancel{1}&\cancel{5}&4&&&&|& \\ \cancel{9}&\cancel{6}&\cancel{8}&3&7&2&|&9&8 \\ &\cancel{9}&8&8&&&|& \\ &1&&&&&|& \end{matrix} $$

For the second $8$ : here, even if we borrow the digit $4$ which comes before the $3$, we still end up with $43 < 8 \times 8$, so we must borrow the $1$ before the $4$ as well. We get $143-64 = 79$, so $$ \begin{matrix} &\cancel{1}&&&&&|&& \\ &\cancel{7}&7&&&&|&& \\ \cancel{1}&\cancel{5}&\cancel{4}&9&&&|& \\ \cancel{9}&\cancel{6}&\cancel{8}&\cancel{3}&7&2&|&9&8 \\ &\cancel{9}&8&8&&&|& \\ &1&&&&&|& \end{matrix} $$

Step 7 : We will now create a new divisor as follows : the existing divisor is $188$. The last digit of the current quotient is an $8$. Their sum is $196$ : and that's the new divisor. The $196$ will be created by shifting $188$ one step to the right, as usual. NOW, we cancel the $188$.

$$ \begin{matrix} &\cancel{1}&&&&&|&& \\ &\cancel{7}&7&&&&|&& \\ \cancel{1}&\cancel{5}&\cancel{4}&9&&&|& \\ \cancel{9}&\cancel{6}&\cancel{8}&\cancel{3}&7&2&|&9&8 \\ &\cancel{9}&\cancel{8}&\cancel{8}&6&&|& \\ &\cancel{1}&1&9&&&|& \end{matrix} $$

Step 8 : Once again, we seek a largest possible single digit $y$ , this time such that $196y \times y \leq 7972$. That value can be found using the heuristic that $20 \times 4 = 80$, so $y=4$ is tested and we find that $1964 \times 4 = 7856$ is very close to $7972$. Hence, $y=4$ is used, and when we substitute it we get

$$ \begin{matrix} &\cancel{1}&&&&&|&& \\ &\cancel{7}&7&&&&|&& \\ \cancel{1}&\cancel{5}&\cancel{4}&9&&&|& \\ \cancel{9}&\cancel{6}&\cancel{8}&\cancel{3}&7&2&|&9&8&4 \\ &\cancel{9}&\cancel{8}&\cancel{8}&6&4&|& \\ &\cancel{1}&1&9&&&|& \end{matrix} $$

Step 9 : Cancellation , with the divisor as $1964$, the dividend as $7972$ and the last digit of the quotient as $4$. As a reminder that it isn't done in the middle, we don't cancel the divisor digits, only the dividend digits.

The $1$ :

$$ \begin{matrix} &\cancel{1}&3&&&&|&& \\ &\cancel{7}&\cancel{7}&&&&|&& \\ \cancel{1}&\cancel{5}&\cancel{4}&9&&&|& \\ \cancel{9}&\cancel{6}&\cancel{8}&\cancel{3}&7&2&|&9&8&4 \\ &\cancel{9}&\cancel{8}&\cancel{8}&6&4&|& \\ &\cancel{1}&1&9&&&|& \end{matrix} $$

The $9$ :$$ \begin{matrix} &\cancel{1}&\cancel{3}&&&&|&& \\ &\cancel{7}&\cancel{7}&3&&&|&& \\ \cancel{1}&\cancel{5}&\cancel{4}&\cancel{9}&&&|& \\ \cancel{9}&\cancel{6}&\cancel{8}&\cancel{3}&7&2&|&9&8&4 \\ &\cancel{9}&\cancel{8}&\cancel{8}&6&4&|& \\ &\cancel{1}&1&9&&&|& \end{matrix} $$

The $6$ :$$ \begin{matrix} &\cancel{1}&\cancel{3}&1&&&|&& \\ &\cancel{7}&\cancel{7}&\cancel{3}&&&|&& \\ \cancel{1}&\cancel{5}&\cancel{4}&\cancel{9}&3&&|& \\ \cancel{9}&\cancel{6}&\cancel{8}&\cancel{3}&\cancel{7}&2&|&9&8&4 \\ &\cancel{9}&\cancel{8}&\cancel{8}&6&4&|& \\ &\cancel{1}&1&9&&&|& \end{matrix} $$

The $4$ : $$ \begin{matrix} &\cancel{1}&\cancel{3}&1&&&|&& \\ &\cancel{7}&\cancel{7}&\cancel{3}&1&&|&& \\ \cancel{1}&\cancel{5}&\cancel{4}&\cancel{9}&\cancel{3}&6&|& \\ \cancel{9}&\cancel{6}&\cancel{8}&\cancel{3}&\cancel{7}&\cancel{2}&|&9&8&4 \\ &\cancel{9}&\cancel{8}&\cancel{8}&6&4&|& \\ &\cancel{1}&1&9&&&|& \end{matrix} $$

Step 10 : All the dividend digits have been struck out. At this point, we strike out all the digits of the divisor first :

$$ \begin{matrix} &\cancel{1}&\cancel{3}&1&&&|&& \\ &\cancel{7}&\cancel{7}&\cancel{3}&1&&|&& \\ \cancel{1}&\cancel{5}&\cancel{4}&\cancel{9}&\cancel{3}&6&|& \\ \cancel{9}&\cancel{6}&\cancel{8}&\cancel{3}&\cancel{7}&\cancel{2}&|&9&8&4 \\ &\cancel{9}&\cancel{8}&\cancel{8}&\cancel{6}&\cancel{4}&|& \\ &\cancel{1}&\cancel{1}&\cancel{9}&&&|& \end{matrix} $$

To be left with $$ 968372 = 984^2 + 116 $$

Check also that this matches directly with Tartaglia's working.