Definition D1
s/t denotes the x satisfying t × x = s.
Proof.
If t ≠ 0, such an x is unique. With this notation, equations can be manipulated without division.convention
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Definition D2
That n is a perfect square means there is a natural number m with n = m².
Proof.
Take n to be a natural number at least 1.convention
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Definition D3
vₚ(a) is the exponent of the prime p in the prime factorization of a.
Proof.
(1)a = ∏ p^vₚ(a), and A1 determines this value uniquely.A1
(2)For primes not dividing a, vₚ(a) = 0.(1)
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Lemma L1
For every prime p, vₚ(ab) = vₚ(a) + vₚ(b)
Proof.
(1)a = ∏ p^vₚ(a) and b = ∏ p^vₚ(b)D3
(2)ab = ∏ p^(vₚ(a)+vₚ(b))(1), A2, A5, A6
(3)The prime factorization is unique, so that is the exponent of ab.(2), A1, D3
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Lemma L2
For every prime p, vₚ(m²) is even.
Proof.
(1)vₚ(m²) = vₚ(m) + vₚ(m)L1
(2)2x means x + x.convention
(3)Therefore vₚ(m²) = 2vₚ(m), which is even.(1), (2)
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Lemma L3
If vₚ(n) is even for every prime, then n is a perfect square.
Proof.
(1)There is a natural number fₚ with vₚ(n) = 2fₚ.assumption
(2)n = ∏ p^(2fₚ)(1), D3
(3)Setting m = ∏ p^fₚ gives m² = ∏ p^(fₚ+fₚ)A2, A5, A6
(4)fₚ + fₚ = 2fₚ, so m² = n(2), (3)
(5)Therefore n is a perfect square.(4), D2
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Theorem T
If n is not a perfect square, no rational x satisfies x² = n.
Proof.
(1)Suppose such a rational x exists.assumption
(2)(−x)² = x², so it suffices to consider x > 0.(1), A7, convention
(3)There are natural numbers s, t with x = s/t.(2), A4
(4)t × x = s(3), D1
(5)(tx)(tx) = s × s, that is, t²x² = s²(4), E, A5, A6
(6)x² = n, so n t² = s²(1), (5), A5
(7)For an arbitrary prime p, vₚ(n) + vₚ(t²) = vₚ(s²)(6), L1
(8)vₚ(t²) and vₚ(s²) are even.L2
(9)Therefore vₚ(n) is also even.(7), (8), A3
(10)Even at every prime, so n is a perfect square.(9), L3
(11)This contradicts the assumption that n is not a perfect square.(10), assumption
(12)No such x exists.(11)
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Corollary C
If n is a perfect square, there is a rational x with x² = n.
Proof.
(1)There is a natural number m with n = m².assumption, D2
(2)m = m/1, so m is rational.D1
(3)x = m satisfies x² = n.(1), (2)
(4)If n = 4, then m = 2, and 2² = 4.(1), (3)
(5)4 is a perfect square, so it fails the theorem's hypothesis. This is why √4 is the exception.(4), D2
(6)Therefore the condition 'not a perfect square' cannot be dropped from the theorem.(3), (5)
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