Visual walkthrough — Irrational numbers — √2, π, e — proof that √2 is irrational

We will keep meeting a small set of words. Let us pin them down before we use them.

The whole proof is a machine with one job: assume √2 is a fraction, then break that fraction. The tool that breaks it is even-versus-odd — because squaring interacts with even/odd in a rigid, predictable way. That is why we choose parity and not, say, size or decimals: parity is the property that survives squaring cleanly.

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Step 1 — Assume √2 is a tidy fraction

WHAT. Suppose, just to see what happens, that √2 is rational. Then by the definition above we can write $\underset{\text{the number}}{\underbrace{\sqrt{2}}}=\underset{\text{a fraction}}{\underbrace{\frac{p}{q}}},\underset{\text{integers}}{\underbrace{p,q}},\underset{\text{no divide by zero}}{\underbrace{q\ne 0}},\underset{\text{lowest terms}}{\underbrace{\gcd (p,q)=1}}.$

WHY. This is Proof by Contradiction: we grant the enemy claim and hunt for nonsense. We demand lowest terms because any fraction can be reduced first — $\frac{6}{4}$ becomes $\frac{3}{2}$ — so assuming it costs us nothing but hands us a weapon later.

PICTURE. Every rational sits somewhere on the Real Number Line. We are betting √2 lands on one of the fraction tick-marks. The picture stamps a flag saying "lowest terms — no shared factors" onto that bet.

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Step 2 — Square both sides to kill the root

WHAT. Multiply each side by itself: $(\sqrt{2}{)}^2=(\frac{p}{q}{)}^2⟹\underset{\text{root gone}}{\underbrace{2}}=\frac{p^2}{q^2}.$

WHY. The square root is awkward — it is not an integer operation. Squaring is its exact inverse (that is why we pick squaring and not, say, adding): $(\sqrt{2}{)}^2$ un-does the root and leaves the clean integer $2$. Now everything in sight is an integer or a ratio of integers.

PICTURE. Think of √2 as the side of a square whose area is 2. Squaring the equation is literally reading off that area. On the right, $\frac{p}{q}$ squared is the area of a square of side $\frac{p}{q}$ — and it must also equal $2$.

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Step 3 — Clear the fraction into a whole-number equation

WHAT. Multiply both sides by $q^2$: $2q^2=p^2.$ Here $\underset{\text{even: it’s }2\times \text{something}}{\underbrace{2q^2}}$ equals $\underset{\text{a perfect square}}{\underbrace{p^2}}$.

WHY. Fractions are hard to reason about with parity; whole numbers are easy. Removing the denominator gives a pure integer equation where the even/odd argument can bite.

PICTURE. Picture two identical $q\times q$ tiles laid side by side (total area $2q^2$) exactly filling one $p\times p$ tile. The two-of-them is the key: the left side is visibly double something.

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Step 4 — So $p^2$ is even

WHAT. The left side $2q^2$ is $2\times (\text{integer})$, so it is even by definition. It equals $p^2$. Therefore $p^2\text{ is even.}$

WHY. "Even" just means "a multiple of $2$", and $2q^2$ wears its factor of $2$ on its sleeve. Whatever $p^2$ equals, it inherits that factor.

PICTURE. Split the big $p\times p$ tile into two equal halves — you can, because its area is $2q^2$, a clean double. Evenness = "splits into two equal whole piles".

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Step 5 — Therefore $p$ itself is even

WHAT. From "$p^2$ even" we get "$p$ even". Check the only other option and watch it fail: if $p$ were odd, write $p=2k+1$, then $p^2=(2k+1{)}^2=4k^2+4k+1=2\underset{\text{integer}}{\underbrace{(2k^2+2k)}}+1=\text{odd}.$ An odd $p$ forces an odd $p^2$ — but we just proved $p^2$ is even. So $p$ cannot be odd; $p$ is even. Write $p=2m$.

WHY. This is the heart of the machine: squaring preserves parity. even² is even, odd² is odd — never crossed. That rigidity is exactly why parity, not some fuzzy property, cracks the proof.

PICTURE. Two side-by-side columns: the "odd" column always ends with one leftover square after pairing (odd² stays odd); the "even" column pairs perfectly. Only the even column matches our even $p^2$.

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Step 6 — Feed $p=2m$ back in

WHAT. Substitute $p=2m$ into $2q^2=p^2$: $2q^2=(2m{)}^2=4m^2⟹\underset{\text{after }÷2}{\underbrace{q^2}}=2m^2.$

WHY. We learned something concrete about $p$ (it's even), so we push that fact back into the equation to see what it forces onto $q$. Dividing both sides by $2$ isolates $q^2$.

PICTURE. Replace the $p\times p$ tile with a $2m\times 2m$ tile ($=4m^2$). Two $q$-tiles fill it, so one $q$-tile equals $2m^2$ — again a clean double.

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Step 7 — So $q$ is even too

WHAT. $q^2=2m^2$ says $q^2$ is even. By the exact same parity logic as Step 5, $q$ is even. Write $q=2n$.

WHY. The same "squaring preserves parity" rule runs a second time. We deliberately reuse it — one rule, applied twice, catches both $p$ and $q$.

PICTURE. Mirror image of Step 4/5: the $q\times q$ tile splits cleanly into two, so $q$ pairs up perfectly — $q$ is even.

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Step 8 — The contradiction you can see

WHAT. We now know $p=2m$ and $q=2n$. Both are even, so both are divisible by $2$: $\gcd (p,q)\ge 2.$ But Step 1 flew the flag $\gcd (p,q)=1$. A number cannot be both $\ge 2$ and $=1$. Contradiction.

WHY. The only thing we assumed was "√2 is a lowest-terms fraction." Everything after was forced. Since the assumption produced nonsense, the assumption is false. $\boxed{\sqrt{2}\text{is irrational.}}\square$

PICTURE. The lowest-terms flag from Step 1 gets torn: both $p$ and $q$ carry a shared factor $2$, exactly what "lowest terms" forbade.

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Edge case — why the same machine does not break √4

WHAT. Run the identical machine on $\sqrt{4}$. Assume $\sqrt{4}=\frac{p}{q}$ lowest terms $\Rightarrow 4=\frac{p^2}{q^2}\Rightarrow p^2=4q^2$. So $p$ is even, $p=2m$, giving $4m^2=4q^2\Rightarrow m^2=q^2\Rightarrow m=q$. Then $p=2q$, so $\frac{p}{q}=2$. No contradiction — $\sqrt{4}=2$ really is rational.

WHY. The √2 proof needs the number under the root to be not a perfect square. When you square a fraction, every prime in the answer appears an even number of times (see Prime Factorization). But $2=\frac{p^2}{q^2}$ would need the prime $2$ to appear an odd number of times — impossible. For $4$ (already a perfect square) no such clash exists. This is why √2, √3, √5, √6, … are irrational but √4, √9 are not.

PICTURE. Side-by-side: for $2$ the parity chain hits a wall (red stop); for $4$ it loops back to a consistent answer (green tick).

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The one-picture summary

The whole argument is a loop that eats itself: assume lowest terms → force $p$ even → force $q$ even → not lowest terms after all.

Recall Feynman retelling — the walkthrough in plain words

I claim √2 is a fraction, and I even tidy it up so top and bottom share nothing — that's my promise. I square the equation to lose the root: now $2q^2=p^2$. The left side is obviously double something, so $p^2$ is even, which drags $p$ itself to be even (because odd numbers square to odd, always). Even $p$ means $p=2m$; I plug that back and — same trick — $q$ turns out even too. But wait: if both $p$ and $q$ are even, they share the factor $2$. That breaks my tidy promise that they shared nothing. I can't keep both. So my very first claim — "√2 is a fraction" — must have been a lie. √2 slips between every fraction on the number line. Done.

Recall Quick self-check

Why must we assume lowest terms at the start? ::: So that "both even" becomes a genuine contradiction — if we didn't demand it, both being even would be allowed and no nonsense would appear. Which single fact makes the parity chain work? ::: Squaring preserves parity: even²=even, odd²=odd. So evenness of $p^2$ forces evenness of $p$. Why does the same proof fail for √4? ::: $4$ is a perfect square, so $2=p^2/q^2$ has no odd-power-of-a-prime clash; the chain closes to the valid answer $p/q=2$ with no contradiction.