How can you prove that the square root of two is irrational?

I have read a few proofs that $\sqrt{2}$ is irrational.

I have never, however, been able to really grasp what they were talking about.

Is there a simplified proof that $\sqrt{2}$ is irrational?

• mathoverflow.net/questions/32011/direct-proof-of-irrationality/… – user977 Aug 10 '10 at 19:17
• $\sqrt{2}$ isn't integer (it's strictly in between 1 and 2). So if it's rational, it's equal to an irreducible fraction $p/q$. Then the fraction $p^2 / q^2$ is also irreducible, but it is equal to 2, which is an integer! – Alexei Averchenko Apr 3 '11 at 1:16
• Plato said I can't call myself human unless I can prove this. – Mike Jones May 7 '11 at 19:48
• math.stackexchange.com/questions/699002/… – Sawarnik Mar 26 '14 at 8:24
• The first irrational known (diagonal of square of side 1), contrary to the reasoning of people knowing just integers; hence "irrational". – Piquito May 29 '15 at 15:40

You use a proof by contradiction. Basically, you suppose that $$\sqrt{2}$$ can be written as $$p/q$$. Then you know that $$2q^2 = p^2$$. As squares of integers, both $$q^2$$ and $$p^2$$ have an even number of factors of two. $$2q^2$$ has an odd number of factors of 2, which means it can't be equal to $$p^2$$.

• This is by far my favorite proof of $\sqrt{2}$ irrational. I believe it is due to Chaitin (at least I think it's his -- it's in his book Meta Math! (p. 98), and he does not attribute it to anyone else). Of course, this depends on unique prime factorization, but it's still quite elementary. The descent method in the standard proof is, of course, hidden in the prime factorization proof, but that's a fine place for it. Note that the original poster couldn't grasp the popular proof, and I bet the descent with contradiction is the obstacle -- I've seen that with many students. – David Lewis Apr 3 '11 at 10:53
• This proof easily generalizes to any exponent k and ratio b >= 2 which is not a perfect power of k, as follows (not in Chaitin's book, but it ain't so hard)... Assume $m^k = b n^k$ Then the unique prime factorizations of $m^k$ and $n^k$ must have all exponents that are multiples of k, and that must also therefore be true of b. But that means b is a perfect k-th power, $b = c^k$ for some integer c. The case k = b = 2 is the classical theorem, with 2 not a perfect square. – David Lewis Apr 3 '11 at 10:54
• It is interesting that the Greeks (and I guess most everybody since) missed this proof, because they had unique prime factorization (cf. Euclid's algorithm), and this proof makes clear that that the irrationality of non-perfect roots is intimately related to it. – David Lewis Apr 3 '11 at 10:55
• @David This proof is not due to Chaitin - it is ancient. It does not depend on unique factorization. Rather it uses only $\:2\ |\ n^2\:$ $\Rightarrow$ $\:2\ |\ n\:$ which is provable by brute-force checking of the multiplication table (mod $2).\:$ Ditto for $\:p\ |\ n^2\:$ $\Rightarrow$ $\:p\ |\ n\:$ for a fixed prime $\:p.\:$ But uniqueness of prime factorizations is equivalent to the much stronger form: $\:\!$ for all primes $\:p,\$ $\:p\ |\ nk\:$ $\Rightarrow$ $\:p\ |\ n\:$ or $\:p\ |\ k.$ – Bill Dubuque Mar 14 '12 at 13:36
• @Bill -- thanks. Makes sense that this proof is not due to Chaitin. It's just that I had never seen it, and it is so attractive (IMHO) that I assumed it must be recent, or it would be better known. He does not attribute it, but I guess that is normal with ancient, folklore proofs. I wonder if anyone does know where or from whom it did originate -- did the Greeks know it? As for not requiring unique prime factorization, you are correct, mathematically. But I was thinking more pedagogically -- it's feasible to introduce this proof in school when prime factorization has been taught. – David Lewis Mar 14 '12 at 17:27

Another method is to use continued fractions (which was used in one of the first proofs irrationality of $\displaystyle \pi$).

Instead of $\displaystyle \sqrt{2}$, we will consider $\displaystyle 1 + \sqrt{2}$.

Now $\displaystyle v = 1 + \sqrt{2}$ satisfies

$$v^2 - 2v - 1 = 0$$

i.e

$$v = 2 + \frac{1}{v}$$

This leads us to the following continued fraction representation

$$1 + \sqrt{2} = 2 + \cfrac{1}{2 + \cfrac{1}{2 + \dots}}$$

Any number with an infinite simple continued fraction is irrational and any number with a finite simple continued fraction is rational and has at most two such simple continued fraction representations.

Thus it follows that $\displaystyle 1 + \sqrt{2}$ is irrational, and so $\displaystyle \sqrt{2}$ is irrational.

Exercise: Show that the Golden Ratio is irrational.

• The task was to write a proof that's easy for almost anyone to understand. Sophie Alpert's answer already does that and this one doesn't. Also, almost anyone would fine Sophie Alpert's answer satisfactory so this answer is not needed at all so I downvoted it. – Timothy Apr 21 '18 at 3:15

If $\sqrt 2$ were rational, we could write it as a fraction $a/b$ in lowest terms. Then $$a^2 = 2 b^2.$$ Look at the last digit of $a^2$. It has to be $0$, $1$, $4$, $5$, $6$ or $9$. Now look at the last digit of $2b^2$. It has to be $0$, $2$ or $8$. As $a^2$ and $2b^2$ are the same number, its last digit must be $0$. But that's only possible if $a$ ends in $0$ and $b$ ends in $0$ or $5$. Either way both $a$ and $b$ are multiples of $5$ contradicting $a/b$ being in lowest terms.

• Can you explain how you got the numbers, 0, 1, 4, 5, 6, or 9 – Tyler Hilton Aug 10 '10 at 19:59
• The last digit of the square of a number depends only on the last digit of the number. To see this, just think about how you usually multiply two numbers (by hand) and focus on what can contribute to the 1's column. From here, you just compute 0^2, 1^2, 2^2,..., 9^2 and record the last digits to get 0,1,4,9,6,5,6,9,4,1, which, not counting multiples, is 0,1,4,5,6, or 9. – Jason DeVito Aug 16 '10 at 2:20
• @TylerHilton $0^2$ ends in $0$, $1^2$ ends in $1$, $2^2$ ends in $4$, $3^2$ ends in $9$, $4^2$ ends in $6$, $5^2$ ends in $5$, $6^2$ ends in $6$, $7^2$ ends in $9$, $8^3$ ends in $4$, and $9^2$ ends in $1$. After that it just repeats itself. – Akiva Weinberger May 29 '15 at 15:50
• Everyone who finds this answer good enough would also find Sophie Alpert's answer good enough so this answer doesn't add anything to it so I'm not sure this answer is needed at all so I downvoted it. – Timothy Apr 21 '18 at 3:21

Assume that $\sqrt{2}$ is rational. Then there exists some rational $R=\sqrt{2}=\frac{Q}{D}$, where $Q$ and $D$ are positive integers and relatively prime (since $R$ can be expressed in simplified form).

Now consider $R^2 = 2 = \frac{Q^2}{D^2}$. Since $Q$ and $D$ are relatively prime, this means that only $Q^2$ can have $2$ in its prime decomposition, and the exponent must be one. Thus, $Q^2 = 2^1 x$, for some odd integer $x$. But $Q^2$ is a square, and thus the exponents for all of its prime factors must be even. Here we have a contradiction.

Thus, $\sqrt{2}$ must be irrational.

• If you are implicitly using uniqueness of prime factorizations then you need to explicitly state that, and state how it applies to yields your deduction. This is essential for proofs at this level. – Bill Dubuque Apr 14 '12 at 0:19

You can also use the rational root test on the polynomial equation $x^2-2=0$ (whose solutions are $\pm \sqrt{2}$). If this equation were to have a rational solution $\frac{a}{b}$, then $a \vert 2$ and $b \vert 1$, hence $\frac{a}{b}\in \{\pm 1, \pm 2\}$. However, it's straightforward to check that none of $1,-1,2,-2$ satisfy the equation $x^2-2=0$. Therefore the equation has no rational roots and $\sqrt{2}$ is irrational.

The continued fraction proof in Aryabhata's answer can be recast into an elementary form that requires no knowledge of continued fractions. Below is a variant of such that John Conway (JHC) often mentions, followed by my (WGD) reformulation that highlights the key role played by the principality of (denominator) ideals in $$\:\mathbb Z\:$$ (which I call unique fractionization) in order to emhpasize its very close relationship unique factorization).

Theorem  (JHC) $$\quad \rm r = \sqrt{n}\ \:$$ is integral if rational,$$\:$$ for $$\:\rm n\in\mathbb{N}$$

Proof $$\ \ \$$ Put $$\ \ \displaystyle\rm r = \frac{A}B ,\;$$ least $$\rm\; B>0\:.\;$$ $$\ \displaystyle\rm\sqrt{n}\; = \frac{n}{\sqrt{n}} \ \Rightarrow\ \frac{A}B = \frac{nB}A.\ \:$$ Taking fractional parts yields $$\rm\displaystyle\ \frac{b}B = \frac{a}A\$$ for $$\rm\ 0 \le b < B\:.\$$ But $$\rm\displaystyle\ B\nmid A\ \Rightarrow\:\ b\ne 0\ \:\Rightarrow\ \frac{A}B = \frac{a}b\$$ contra $$\rm B$$ least. $$\:$$ QED

Abstracting out the Euclidean descent at the heart of the above proof yields the following

Theorem  (WGD) $$\quad \rm r = \sqrt{n}\ \:$$ is integral if rational,$$\:$$ for $$\:\rm n\in\mathbb{N}$$

Proof $$\ \$$ Put $$\ \ \displaystyle\rm r = \frac{A}B ,\;$$ least $$\rm\; B>0\:.\;$$ $$\ \displaystyle\rm\sqrt{n}\; = \frac{n}{\sqrt{n}} \ \Rightarrow\ \frac{A}B = \frac{nB}A\ \Rightarrow\ B\:|\:A\$$ by this key result:

Unique Fractionization $$\$$ The least denominator $$\rm\:B\:$$ of a fraction divides every denominator.

Proof $$\rm\displaystyle\ \ \frac{A}B = \frac{C}D\ \Rightarrow\ \frac{D}B = \frac{C}A \:.\$$ Taking fractional parts $$\rm\displaystyle\ \frac{b}B = \frac{a}A\$$ where $$\rm\ 0 \le b < B\:.\$$ But

$$\rm\displaystyle\ \:B\nmid D\ \Rightarrow\ b\ne 0\ \Rightarrow\ \frac{A}B = \frac{a}b\ \$$ contra leastness of $$\rm\:B.\,$$ Thus $$\rm\,B\mid D\,$$ as claimed $$\quad$$ QED

Thus JHC's proof essentially "inlines" the above proof - which can be more conceptually viewed as the principality of (denominator) ideals in $$\mathbb Z,\,$$ cf. my post here. See also this sci.math discussion between John Conway and me (click "plain text" to get correct format).

There is also a proof of this theorem that uses the well ordering property of the set of positive integers, that is in a non empty set of positive integers there is always a least element. The proof follows the approach of proof by contradiction but uses the well ordering principle to find the contradiction :) -

Let us assume $\sqrt{2}$ is rational, hence it can be written down in the form $\sqrt{2}=a/b$ assuming that both $a$ and $b$ are positive integers in that case if we look at the set $S = \{k\sqrt{2} \mid k, k\sqrt{2}\text{ are integers}\}$ we find that it's a non empty set of positive integers, it's non empty because $a = b\sqrt{2}$ is in the above set. Now using the Well ordering principle we know that every set of positive integers which is non-empty has a minimum element, we assume that smallest element to be $s$ and let it equal to $s =t\sqrt{2}$. Now an interesting thing happens if we take the difference between the following quantities $s\sqrt{2} - s = (s-t)\sqrt{2} = s(\sqrt{2} - 1)$ which is a smaller element than $s$ itself, hence contradicting the very existence of $s$ being the smallest element. Hence we find that $\sqrt{2}$ is irrational.

I know the proof but I am still amazed at how the author came up with the set assumption. Sometimes such assumptions make you feel kinda dumb :). If anyone has some insight regarding how to come up with such assumptions kindly post your answer in the comment, otherwise I would just assume that it was a workaround.

Another one that is understandable by high schoolers and below.

We will use the following lemma:

If $n$ is an integer, $n^2$ is even (resp. odd) iff $n$ is even (resp. odd).

For the high-schoolers, the proof is about writing $(2k)^2 = 2(2k^2)$ and $(2k+1)^2=2(2k^2+2k)+1$ ...

Now, assume $\sqrt 2 = \frac{a}{b}$ with $a$ and $b$ strictly positive integers.

Then $a^2=2b^2$, $\implies a^2$ is even ($=2b^2$), $\implies a$ is even (from the lemma), $\implies a=2a_1$ with $a_1 \in \mathbb N^*$, $\implies b^2=2a_1^2$.

Repeat with $b$ to find that $b=2b_1$ with $b_1 \in \mathbb N^*$ and $(a_1,b_1)$ verifies $a_1^2=2b_1^2$.

By repeating these two steps, we build two sequences $(a_n)_{n\in \mathbb N}$ and $(b_n)_{n\in \mathbb N}$ with values in $\mathbb N^*$ and strictly decreasing, which is impossible, ergo $\sqrt{2}$ is irrational.

(Here of course we use the well-ordering principle which most high schoolders would not know about, but the intuition that the sequence would hit $0$ after at most $a_0=a$ steps is easy to get).

• You could stop at $b^2=2a_1^2$ by noting that both $a^2$ and $b^2$ cannot be even. – ahorn Oct 22 '15 at 7:07

Here's a short algebraic proof. It nowhere uses rules about primes or even numbers.

You need to first show that $$1<\sqrt{2}<2,$$ but that is obvious.

We first assume that $$\sqrt{2}$$ is rational. Then pick the smallest positive $$q$$ so that $$p=q\sqrt{2}$$ is an integer. Then $$q

Now compute:

$$\left(\frac{2q-p}{p-q}\right)^2 = \frac{4q^2-4pq+p^2}{p^2-2pq+q^2}=\frac{6q^2-4pq}{3q^2-2pq}=2$$

But $$q means $$0, and $$\frac{2q-p}{p-q}=\sqrt{2},$$ contradicting the assumption that $$q$$ was the least possible positive denominator.

More generally

We can prove, more generally, that if $$n$$ is an integer and $$n^2 then $$\sqrt{D}$$ is irrational. In the case $$D=2$$ we have $$n=1$$.

If $$\sqrt{D}$$ is rational, find the least positive $$q$$ such that there is a $$p$$ such that $$\frac{p}{q}=\sqrt{D}$$. So $$p^2=Dq^2$$ means $$n^2q^2 and hence $$nq.

Therefore $$0.

But then:

$$\left(\frac{Dq-pn}{p-qn}\right)^2=\frac{D^2q^2-2Dpqn + p^2n^2}{p^2-2pqn+q^2n^2}=\frac{D^2q^2-2Dpqn + Dq^2n^2}{Dq^2-2pqn+q^2n^2}=D\tag{*}$$

contradicting the fact that $$q$$ was the smallest positive denominator for $$\sqrt{D}$$.

You can prove if $$D\geq 0$$ is an integer, then there is exactly one non-negative integer $$n$$ such that $$n^2\leq D<(n+1)^2$$. We first prove $$n$$ exists:

Since $$(1+D)^2=D+(1+D+D^2)$$, we know that $$D<(1+D)^2$$ and hence there is a least positive $$m$$ such that $$D We know $$m\neq 0$$ because $$D\geq 0^2$$, so $$m\geq 1$$. Let $$n=m-1$$. Then $$n^2\leq D<(n+1)^2$$.

Uniqueness follows from:

If $$0\leq m then $$1\leq m+1\leq n$$ and thus $$(m+1)^2\leq n^2$$.

So if $$n^2\leq D< (n+1)^2$$ and $$m^2\leq D< (m+1)^2$$, then we can't have $$m or we'd have $$D<(m+1)^2\leq n^2\leq D$$. Similarly, we can't have $$m>n$$. So we must have $$m=n$$.

Together, the above say that if $$D\geq 0$$ then $$\sqrt{D}$$ is rational if and only if $$\sqrt{D}$$ is an integer.

(*) The magic trick in the above computation is that If $$\frac{np}{nq}=\sqrt{D}$$ then $$\frac{Dq}{p}=\sqrt{D}.$$ And if $$\frac{a}{b}=\frac{c}{d}$$ then $$d\neq b$$ then $$\frac {a-c}{b-d}=\frac{a}{b}.$$

The expression is arrived by computing (using that $$p=q\sqrt{D}):$$

$$0<\left(p+q\sqrt{D}\right)\left(\sqrt{D}-n\right)=(qD-np)+(p-nq)\sqrt{D}$$

From this we see $$(qD-np)^2-D(p-nq)^2=0.$$

And $$p-nq=q(\sqrt{D}-n) since $$0<\sqrt{D}-n<1.$$

• Note  The proofs above are essentially special cases of the simple proof posted in this answer $6$ years earlier. – Bill Dubuque Dec 18 '20 at 14:22

Here are some of my favorite (sketches) of proofs for the irrationality of $\sqrt{2}$.

• Using Newton's method to approximate roots of the polynomial $f(x) = x^2 - 2$, then showing that the sequence does not converge to a rational number.
• Proof by contradiction, assume that $\sqrt{2} = \frac{n}{m}$ for some $n,m \in \mathbb{Z}$ with $m \neq 0$, then $2m^{2} = n^2$, hence $n$ must be even and we can let $n = 2k$ for some $k \in \mathbb{Z}$, but then $m^2 = 2k^2$ will also be even, which is impossible if $\frac{n}{m}$ is reduced. Therefore, $\sqrt{2}$ cannot be expressed as a ratio of integers.
• Since $f(x) = x^2 -2$ is irreducible over $\mathbb{Q}[x]$, its roots must lie in some finite extension field $\mathbb{Q}(\sqrt{2})$ over the rationals.

[Reposted from closed topicProve the square root of 2 is irrational

• It would help to give further details about the first and third methods. I know a lot about these topics yet I cannot be sure precisely what you have in mind. – Bill Dubuque Nov 7 '12 at 20:15

Let $x^2-2=0$ be the polynomial equations this have a possibles rational solutions $\pm1,\pm2$ and no one of this is a solution then the solution is irrational and we now that this are $\pm \sqrt2$

The irrationality of the square root of 2 follows from our knowledge of how Pythagorean triples behave, specifically, that for positive integers x, y, and z, if x^2 + y^2 = z^2, then x is not equal to y. But if the square root of 2 were rational, then there would exist positive integers a and b such that a/b = the square root of 2. Then a^2/b^2 = 2. Then a^2 = 2b^2. Then b^2 + b^2 = a^2, and so we would have a Pythagorean triple with x = y, a contradiction.

• How do you know that $x \ne y$? Seems to assume the conclusion. – marty cohen May 15 '18 at 20:43

Proposition: In base $$2$$ any square must end in an even number of trailing zeros.

The proposition comes directly from, for example, multiplying a binary number with itself using the standard algorithm or simply by squaring

$$(\sum_{k=t}^{N} b_k 2^k)^2=B2^{2t+1}+b_t2^{2t}$$

If we can represent $$\sqrt{2}=\frac{p}{q}$$ then

$$2q^2=p^2$$

Multiplying by $$2$$ is shifting all bits of a binary number to the left, so if the number was ending in $$m$$ trailing zeros after multiplication by $$2$$ it will end in $$m+1$$ zeros.

This means that $$2q^2$$ is ending in odd while $$p^2$$ is ending in an even number of zeros. Thus, these two cannot be equal.

*Notice that this proof does not care about the relative primality of $$p$$ and $$q$$.

• I find this the most intuitive of the proofs posted so far. (Disclaimer: I studied electronic engineering.) – timtfj Dec 15 '18 at 1:30

Let me give a proof based on (partly) Newton's method and (mainly) Pell's equation. The proof technique is also similar to that of continued fractions.

Consider the iterations: $$x_{n+1} = \frac{x_n}{2}+\frac{1}{x_n}$$. This is the formula given by Newton's method for the function $$f(x) = x^2 -2$$. Let's use a special initial value: $$x_0 = \frac{3}{2}$$.

Denote $$x_n = \frac{p_n}{q_n}$$. We immediately get the formula: $$$$p_{n+1} = p_n^2 + 2q_n^2, q_{n+1} = 2p_n q_n.$$$$

Furthermore, you might notice that $$p_n, q_n$$ are all the solutions of the Pell's equation $$x^2-2y^2 = 1$$. But we only need the following calculation: $$$$p_{n+1}^2-2q_{n+1}^2 = (p_n^2 -2q_n^2)^2,$$$$ $$$$p_0^2-2q_0^2 = 3^2-2\cdot 2^2 = 1.$$$$

From the above formula, we have the relation: $$\frac{2}{x_n}< \sqrt{2} < x_n$$. Now assume $$\sqrt{2} = \frac{s}{t}$$. We can get the estimate: $$$$|\frac{p_n}{q_n}-\frac{s}{t}| = |x_n - \sqrt{2}| < |x_n - \frac{2}{x_n}| = \frac{1}{p_n q_n}.$$$$

So $$|tp_n-sq_n| < \frac{t}{p_n}$$. Because $$p_n$$ grows to infinity, we have $$|tp_n - sq_n|<1$$ for some $$n$$. Then $$tp_n = sq_n$$ and $$x_n = 2$$. Contradiction!

If you know more about Pell's equation ($$x^2-ny^2 = 1$$ has solution for all positive nonsquare integer $$n$$), you can prove that $$\sqrt{n}$$ is irrational.

The irrationality of $$\sqrt{2}$$ is equivalent to that of $$1+\sqrt{2}$$, which is equivalent to there being no length $$\ell$$ such that the hypotenuse and perimeter of an isosceles right triangle are both whole-number multiples of $$\ell$$. Suppose there were such a length. We say that the hypotenuse (indicated schematically by the black path in the triangle on the left) and the perimeter (indicated by the brown path in the triangle on the left) are measured by $$\ell$$. To find $$\ell$$, we apply the Euclidean algorithm, measuring off the perimeter in units of the hypotenuse, and observing that any remainder is measured by $$\ell$$. In any right triangle, the perimeter contains at least two hypotenuse-lengths (since the legs together are greater than the hypotenuse) but never three (since the hypotenuse is greater than each leg separately). (The isosceles right triangle, in fact maximizes the ratio of perimeter to hypotenuse, this ratio being $$1+\sqrt{2}\approx 2.4$$.) The two hypotenuse-lengths are indicated schematically by the black diagonal path and the green L-shaped path in the middle figure. The remainder is the short black vertical path. We claim that the brown triangular path in the middle figure is also a hypotenuse-length path, and therefore also measured by $$\ell$$. But then this small isosceles right triangle, which is less than half as big (in linear dimensions) as the original isosceles right triangle, has hypotenuse and perimeter that are measured by $$\ell$$. Which is a contradiction: the Euclidean algorithm can be iterated, always producing a new isosceles right triangle of less than half the size of the previous one whose hypotenuse and perimeter are measured by $$\ell$$. Such repeated reductions to less than half must eventually result in a figure whose size is less than any conceivable $$\ell$$.

To establish the claim that the brown triangular path is hypotenuse-length, note that if three congruent isosceles right triangles are arranged like the gray triangles in the middle figure, and the rectangular gap in the middle filled in, the result is an isosceles right triangle. Then the black, green, and brown paths all have length $$2s+d$$, using the notation of the figure on the right.

For a more organic derivation, let $$\triangle ABC$$ be an isosceles right triangle. Draw a circle with center $$C$$ and radius $$\overline{CB}$$ intersecting $$\overline{AC}$$ at $$G$$. The length of $$\overline{AG}$$ is the amount by which the length of the hypotenuse exceeds that of a leg, which is the length that the short segment of the green L in the middle figure needs to have. Now construct a segment perpendicular to $$\overline{AC}$$ at $$G$$, intersecting $$\overline{AB}$$ at $$D$$. Triangle $$AGD$$ is an isosceles right triangle, so $$\overline{DG}\cong\overline{AG}$$. But also $$\overline{DB}\cong\overline{DG}$$ as required, which can be seen by observing that the right triangles $$CGD$$ and $$CBD$$ share a hypotenuse and have corresponding legs $$\overline{CG}$$ and $$\overline{CB}$$ congruent, and are therefore congruent. The points $$F$$ and $$E$$ can be constructed by a symmetric procedure.

One final image: if the hypotenuse and perimeter of the largest triangle in the fractal below are both measured by $$\ell$$, then the hypotenuse and perimeter of every similar triangle in the fractal, and the perimeter of every similar rectangle in the fractal, are measured by $$\ell$$ as well.

Remarks: This proof is essentially the same as the continued fraction proof of Aryabhata and the algebraic proof of Thomas Andrews. The only added feature is the geometric interpretation and the pictures.

To relate it to continued fractions, let the hypotenuse be $$1$$, so that the perimeter is $$1+\sqrt{2}$$. Taking out two hypotenuse lengths gives $$1+\sqrt{2}=2+(\sqrt{2}-1).$$ The quantity in parentheses is the length of the short black vertical segment in the middle diagram, which is the hypotenuse of the original triangle scaled down by a factor of $$1+\sqrt{2}$$. So $$1+\sqrt{2}=2+\frac{1}{1+\sqrt{2}}.$$ Iterating gives $$1+\sqrt{2}=2+\frac{1}{2+\frac{1}{1+\sqrt{2}}}=2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\ldots}}}.$$

To relate it to the algebraic proof of Thomas Andrews, refer to the rightmost diagram and let $$\sqrt{2}=\frac{p}{q}=\frac{2s+d}{s+d},$$ where $$p$$ has been interpreted as the length of the hypotenuse and $$q$$ as the length of a leg. But the small right triangles in the diagram make it clear that also $$\sqrt{2}=\frac{d}{s}=\frac{2q-p}{p-q}.$$ If $$\frac{p}{q}$$ was in lowest terms, this is a contradiction, which is another way of expressing the infinite descent idea of the geometric proof.

(Since p and q are coprime integers it is supposed to be in lowest form. But we demonstrated the existence of a fraction having lower positive integral denominator.) It is proof by Contradiction

• Please typeset your post; do not post text or equations as images. Also, this is much the same as Thomas Andrews' answer. – Will Orrick Dec 3 '20 at 9:44