Before the quick-fire drill, lock in the five images the questions keep returning to. Each figure IS the reasoning — the words only point at it.
1 · What "terminating" vs "repeating" actually looks like. A rational number's decimal either stops (terminating) or locks into a loop forever (repeating). Irrationals do neither — they wander with no cycle. The repeat is not a coincidence: long division of qp has only q possible remainders, so a remainder must eventually repeat, and once it does, the digits loop.
Look at the green block: the moment the same remainder returns, the whole tail must copy itself — that forced loop is why every fraction repeats.
2 · Why we demand lowest terms (gcd(p,q)=1). A fraction like 86 hides a shared factor of 2. "Lowest terms" means we have already cancelled every common factor, so p and q share nothing. That is the tripwire: if the algebra later forces a shared factor of 2, we have contradicted a cancellation we already performed.
The red crossings show common factors being removed; the final coprime pair on the right is what the proof starts from — so "both even" at the end is a factor that should not exist.
3 · Why squaring turns a root into an integer relation.2=qp has a stubborn root sign. Squaring both sides sends 2↦2 and qp↦q2p2, then clearing the denominator lands on the clean integer equation p2=2q2 — the doorway to parity arguments.
4 · Parity as pairing squares. "Even" means the dots pair up with none left over. A square of dots for an odd number always leaves a lone diagonal dot; a square for an even number pairs perfectly. This is the visual meaning of "p2 even ⇒p even" — and the Pythagorean unit-square diagonal is where 2 comes from in the first place, via Pythagoras.
5 · Dense vs complete on the number line. Rationals crowd arbitrarily close to 2 from both sides (density), yet the exact point 2 is a hole in the rationals. Filling every such hole is what "complete" means.
The blue rationals swarm the yellow target but never land on it — that gap is exactly the irrational number.
A rational number always has a terminating decimal
False — it may repeat forever (like 31=0.3); the rule (figure 1) is that a rational has a terminating OR repeatingdecimal, while irrationals do neither.
Every non-terminating decimal is irrational
False — 0.333… never terminates yet equals 31; only a non-terminating and non-repeating decimal is irrational.
2 is irrational because a calculator shows no repeating pattern
False — a calculator only shows finitely many digits, which can never prove non-repetition; the contradiction proof is what actually establishes it.
The proof would still work if we forgot to reduce qp to lowest terms
False — the whole contradiction is "p,q both even yetgcd(p,q)=1" (figure 2); without assuming lowest terms there is no GCD fact to contradict.
The proof shows √2 equals no fraction of the specific form even/even
False — it starts from an arbitrary lowest-terms fraction and derives that both must be even, so it rules out all fractions, not a special subclass.
2 is transcendental
False — it satisfies x2−2=0, a polynomial with integer coefficients, so it is algebraic (irrational but not transcendental).
Between 2 and 1.4142 there are no rational numbers
False — rationals are dense on the real line (figure 5); infinitely many fractions lie in that tiny gap, but none of them equals2.
If p2 is even then p is even
True — squaring preserves parity (figure 4): an odd p=2k+1 gives p2=2(2k2+2k)+1, which is odd, so an even square forces an even root.
If p2 is divisible by 3 then p is divisible by 3
True — this is the same prime argument as parity but with the prime 3; it is exactly why the identical proof shows 3 is irrational.
Because rationals can approximate √2 to any accuracy, √2 is "basically rational"
False — approximation is not equality; (10001414)2=1.999396=2, and no fraction squares to exactly 2.
"Assume 2=qp. Then p2=2q2, so p is even, done — contradiction."
Incomplete, not wrong: showing p even alone is no contradiction; you must push on to show q is also even, which then breaks gcd(p,q)=1.
"p2=2q2 means p2 is even, therefore 2q2 is odd."
Nonsense — 2q2 is a multiple of 2, hence even; both sides of an equation share the same parity, so they cannot disagree.
"Since p is even, write p=2m; then 2q2=2m2, so q2=m2."
Algebra error: p=2m gives 2q2=(2m)2=4m2, so q2=2m2 (not m2); dropping the factor of 4 destroys the step that proves q even.
"The same proof shows 4 is irrational, since p2=4q2 makes p even."
Error: p=2m leads to m2=q2, i.e. m=q and qp=2 — a valid integer, no contradiction; the argument only bites when the number is not a perfect square.
"√2 is irrational, and irrationals aren't real numbers."
Error: irrationals are real numbers; they fill the gaps between rationals on the number line (figure 5) — "irrational" means "not a ratio," not "not real."
"2⋅2=2, and 2 is rational, so √2 must be rational too."
Error: a product of two irrationals can be rational; that says nothing about whether each factor is rational, just as 2 times itself cancels the irrationality.
"The proof uses Pythagoras, so it only works for right triangles, not for numbers."
Error: Pythagoras is just one motivation (the diagonal in figure 4); the algebraic proof from p2=2q2 needs no geometry at all.
So that "p and q are both even" becomes a genuine contradiction (figure 2); without the coprime assumption there is nothing for the "both even" conclusion to violate.
Why does the proof need 2 to be prime?
Primeness is what lets "2∣p2⇒2∣p"; for a non-prime like 4, a square can be divisible by 4 without its root being — so the cascade breaks.
Why does squaring both sides (2=p2/q2) actually help?
It removes the square root and turns the problem into the pure integer equation p2=2q2 (figure 3), where parity/factorization arguments become available.
Why is "√2 has a non-repeating decimal" equivalent to "√2 is irrational"?
A repeating (or terminating) decimal can always be rewritten as a fraction of integers (figure 1); so non-repeating forever is exactly the failure to be any such ratio.
Why does the same argument fail for 9?
9=32 is a perfect square, so 9=3=13 satisfies the equation with no contradiction; every prime under the root appears an even number of times.
Why can we phrase the contradiction as "infinite descent"?
If qp worked, both being even lets us halve to a smaller positive fraction q/2p/2, and repeat forever — but positive integers cannot shrink endlessly, which is impossible.
Why is being dense (rationals filling every gap) not the same as being complete?
Density means fractions get arbitrarily close to 2 (figure 5), but the exact point 2 is still a hole in the rationals — completeness is what filling those holes with irrationals achieves.
Is 2 negative-square-root a separate case the proof missed?
No — −2=−qp is rational iff 2 is, so the positive-root proof covers both signs at once.
Does the proof handle q=1 (i.e. √2 an integer)?
Yes — q=1 is a lowest-terms fraction; the derivation still forces q even, but 1 is odd, an immediate contradiction, so no integer equals 2.
Is 0=0 a counterexample to "roots are irrational"?
No — 0=10 is rational; the claim is only that roots of non-perfect-squares are irrational, and 0 is a perfect square.
Does 2=qp with q=0 but p=0 slip past the proof?
No — p=0 gives q0=0=2, so that case is excluded before the argument even begins.
Family B — structure of the number
Is 2+(−2)=0 evidence that irrationals sum to rationals in general?
It shows irrationals can sum to a rational, but only in special cancelling cases; 2+2=22 stays irrational, so no general rule.
Is π irrational for the same reason as 2?
No — π is not the root of any integer polynomial (it is transcendental), so the parity/algebraic trick fails; its irrationality needs Lambert's calculus-based proof.
Family C — representation
What if someone uses base-3 instead of base-10 decimals — does √2 become rational?
No — rationality is base-independent; changing base changes the digits but not whether an exact fraction qp exists.
Every trap above lives somewhere on this flow. The green path is the honest proof; the red branches are where the misconceptions try to escape and get blocked.
Recall One-line summary of every trap
The proof works because 2 is prime (figure 4) and qp is coprime (figure 2); it fails for perfect squares because the prime under the root appears an even number of times; and approximation, density (figure 5), or a partial "both even" is never the same as an actual contradiction.