Visual walkthrough — Real number system — ℕ ⊂ ℤ ⊂ ℚ ⊂ ℝ

Before line one, let us agree on the words and the notation the whole proof rests on. We build them from scratch — nothing assumed.

Keep that last italic phrase in your pocket. The entire proof is a trap that forces top and bottom to both be even, which lowest terms forbids.

---

Step 1 — What number are we even talking about?

WHAT. We meet $\sqrt{2}$ as a length, not a symbol. Draw a square whose sides are length $1$, and cut it corner-to-corner: that cut splits the square into two right-angled triangles. The Pythagoras rule (pictured below) says that in any right-angled triangle, the square built on the long slanted side has exactly the same area as the two squares built on the short sides combined. Here both short sides are $1$, so the long side $d$ obeys $d^2=1^2+1^2=2$, hence $d=\sqrt{2}$.

WHY a picture of a square? Because "$\sqrt{2}$" by itself is just ink. Anchoring it to the red diagonal below makes it a real, drawable thing — a length you could measure with a ruler. The proof will show no ruler marked in equal fractions can ever land exactly on that red tip.

PICTURE. (The three attached squares show why $d^2=1^2+1^2$: the red area equals the two black areas.)

$\underset{2}{\underbrace{d}}=\underset{2}{\underbrace{1}}+\underset{2}{\underbrace{1}}=2⟹d=\sqrt{2}$

Each term: the two $1$'s are the square's sides; $d$ is the red diagonal; "${}^2$" means "times itself" (an area), and Pythagoras says the red area matches the two black areas.

---

Step 2 — Make the enemy's assumption

WHAT. We pretend the opposite of what we want. Suppose $\sqrt{2}$ is a fraction, already cancelled to lowest terms: $\sqrt{2}=\frac{p}{q},\underset{\text{integers}}{\underbrace{p,q\in \mathbb{Z}}},q\ne 0,\underset{\text{nothing left to cancel}}{\underbrace{\gcd (p,q)=1}}.$

WHY assume the thing we want to disprove? This is proof by contradiction: the tool that answers "how do I prove something is impossible?" You cannot check infinitely many fractions one by one. Instead you grant the enemy their fraction and show it leads to nonsense. If the assumption breaks reality, the assumption was false.

WHY "lowest terms"? Because every fraction can be reduced to lowest terms (keep cancelling common factors — it must stop, you cannot cancel forever from finite numbers). So assuming lowest terms costs us nothing but hands us a weapon: $p$ and $q$ are not both even.

PICTURE.

The $\gcd (p,q)=1$ term ($\gcd$ = "greatest common divisor", the biggest number dividing both) is the trap door. Watch it.

---

Step 3 — Clear the square root

WHAT. Square both sides of $\sqrt{2}=\frac{p}{q}$ (recall ${}^2$ means "times itself", and squaring the root gives back $2$), then multiply out: $2=\frac{p^2}{q^2}⟹\underset{\text{even: it’s }2\times \text{something}}{\underbrace{2q^2}}=\underset{\text{so this is even}}{\underbrace{p^2}}.$

WHY square? The root sign  ​ is awkward — it hides whether things are whole. Squaring is the tool that undoes a square root (that was its whole definition), converting a statement about an irrational-looking length into a clean statement about integers only ($p^2$, $q^2$). Integers are where "even/odd" lives, and even/odd is the crowbar we'll use.

Term by term: multiplying $2=\frac{p^2}{q^2}$ by $q^2$ moves $q^2$ to the left. The left side, $2q^2$, is literally "$2\times (\text{integer})$" — the definition of even. So the right side $p^2$ is even.

PICTURE.

---

Step 4 — Even square forces an even root

WHAT. We just learned $p^2$ is even. Claim: then $p$ itself is even. Check the only other option — suppose $p$ is odd, i.e. $p=2k+1$: $p^2=(2k+1{)}^2=4k^2+4k+1=2\underset{\text{integer}}{\underbrace{(2k^2+2k)}}+1=\text{odd}.$

WHY bother with this little sub-argument? Because "even" is a fact about $p^2$, but our trap needs a fact about $p$. This step is the bridge. The tool is case-splitting: every integer is either even or odd, no third option. An odd number squared is odd (shown above), so if $p^2$ is even, $p$ could not have been odd — $p$ must be even.

Term by term: $2k+1$ is the general odd number; expanding the square leaves a lonely $+1$ that no factoring can absorb, so the result is odd. That rules odd out.

PICTURE.

---

Step 5 — Feed "p is even" back in

WHAT. Since $p$ is even, write $p=2m$ (some integer $m$). Substitute into $2q^2=p^2$ and simplify, showing every move: $2q^2=(2m{)}^2=4m^2\stackrel{÷2}{\to }\underset{\text{now this is even}}{\underbrace{q^2}}=2m^2.$

Why may we divide by $2$? Dividing both sides of a true equation by the same nonzero number keeps it true (it just rescales both sides equally). Here $2\ne 0$, so from $2q^2=4m^2$ we divide each side by $2$: the left becomes $q^2$, the right becomes $2m^2$. Written fully: $2q^2=4m^2⟹\frac{2q^2}{2}=\frac{4m^2}{2}⟹q^2=2m^2.$

WHY substitute? We learned new information ($p=2m$); the honest move is to plug it back into the equation we already trust ($2q^2=p^2$) and see what it now forces. This is just bookkeeping — but watch what pops out.

Term by term: the right side $2m^2$ is "$2\times (\text{integer})$" — so $q^2$ is even, and by the very same Step-4 argument, $q$ is even too.

PICTURE.

---

Step 6 — Spring the trap

WHAT. Collect what we proved: $p\text{ is even}\text{AND}q\text{ is even}.$ But Step 2 assumed $\gcd (p,q)=1$ — they share no common factor. Both being even means they share the factor 2. Two facts that cannot both be true: a contradiction.

WHY is this the finish? A contradiction means one of our assumptions was rotten. Everything after Step 2 was airtight algebra. The only choice we made was Step 2's "$\sqrt{2}$ is a fraction." So that is the false statement.

PICTURE.

$\therefore \sqrt{2}\notin \mathbb{Q}.■$

---

Step 7 — The degenerate cases (nothing sneaks through)

WHAT. Cover the corners the proof might seem to dodge:

• $q=0$? Forbidden from the start — division by zero is undefined, so $\frac{p}{0}$ was never a candidate fraction.
• $p=0$? Then $\frac{p}{q}=0\ne \sqrt{2}$ (since $\sqrt{2}\approx 1.414>0$). So $p\ne 0$; both are genuine nonzero integers and the even/odd machinery applies cleanly.
• Negative $p$ or $q$? We allowed these from the definition ($p,q\in \mathbb{Z}$), and they cause no trouble: "even/odd" and $\gcd$ care only about size, not sign. $-4$ is just as even as $4$. And $\sqrt{2}>0$, so we could even take $p,q>0$ without losing anything — but the argument never needed that restriction.
• What about $\sqrt{3},\sqrt{5},\dots$? The very same trap works with the prime $3$, $5$, ... in place of $2$. It fails only for $\sqrt{4},\sqrt{9},\dots$ — perfect squares — because there the "root" is already an integer and no contradiction ever appears. That is exactly why $\sqrt{9}=3$ is rational.

WHY list these? A proof with an unchecked hole in it proves nothing. Here we confirm every input — zero, negative, perfect-square — is either impossible or handled.

PICTURE.

---

The one-picture summary

The whole argument is a loop that eats itself: assume fraction in lowest terms → force $p$ even → force $q$ even → but lowest terms forbids both even → boom. The red arrows show information flowing until it collides with the assumption it started from.

Recall Feynman retelling — say it to a friend with no notation

Imagine someone swears the diagonal of a $1\times 1$ square is a perfect fraction — top integer over bottom integer, already cancelled as small as it goes. I don't argue; I just do their algebra. Squaring gives $2q^2=p^2$, which says $p\times p$ is even. But odd-times-odd is always odd, so $p$ has to be even. Even means $p=2m$; I plug that in and out pops $q^2=2m^2$ — so $q$ is even too. Now both the top and bottom are even, meaning I could cancel another 2. But they told me it was already fully cancelled! You can't have both. Their story is impossible. So no fraction on Earth equals that diagonal — $\sqrt{2}$ is irrational, and the number line truly needs numbers beyond the fractions.

Recall Quick self-test

Why do we insist the starting fraction is in lowest terms? ::: So that discovering a shared factor of 2 is an immediate contradiction — there's nothing left to cancel. What tool answers "how do I prove something is impossible?" ::: Proof by contradiction: assume it's possible and derive nonsense. Why square both sides in Step 3? ::: Squaring undoes the root (its very definition), turning a claim about an ugly length into a clean statement about integers, where even/odd works. Where does the same proof fail, and why is that correct? ::: For perfect squares like $\sqrt{9}=3$: the root is already an integer, so no even/even contradiction ever arises — and indeed those are rational.

Where this sits: this walkthrough is the geometric heart of the real number system. The existence of even one such hole is why we needed non-terminating decimals and, ultimately, the Completeness Axiom to plug every gap. It also foreshadows algebraic irrationals — $\sqrt{2}$ solves $x^2-2=0$, so it is algebraic, unlike $\pi$.