False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-24

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FALSE: some rational number squares to 2

Statement

False claim: there exists $x \in Q$ with $x^{2} = 2$ .

Facts & Assumptions

Given: The rationals as quotients of integer pairs (The rationals as equivalence classes of pairs of integers).

[A1]

Every rational can be written $a / b$ in lowest terms, with $a, b$ not both even (cancel common factors of $2$ ; the process terminates by well-ordering of $N$ ).

[A2]

Parity in $Z$ : if $k^{2}$ is even then $k$ is even (an odd $k = 2 j + 1$ has odd square $4 j^{2} + 4 j + 1$ ).

Refutation

technique · contradiction

1.1

Assume, for contradiction, that some $x \in Q$ has $x^{2} = 2$ ; write $x = a / b$ in lowest terms, $a, b$ not both even.

assume-contraA1

2.1

Clearing denominators: $a^{2} = 2 b^{2}$ .

step 1.1algebra

3.1

$a^{2}$ is even, hence $a$ is even: $a = 2 c$ .

step 2.1A2

4.1

Substituting: $4 c^{2} = 2 b^{2}$ , so $b^{2} = 2 c^{2}$ is even, hence $b$ is even.

step 3.1A2algebra

5.1

Both $a$ and $b$ are even, contradicting the lowest-terms choice of step 1.1; the assumption fails: no rational squares to $2$ .

step 4.1step 1.1discharge-contradiction∎

Depends on

Used by

Dependency tree · two levels

7 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

T. Tao, Analysis I, 3rd ed., §4.4 (standard reference, not scraped)
Square root of 2 — proofs of irrationality (Wikipedia) (standard reference, not scraped)