Definition D1
The solution set of an equation is the set of all x that make it true.
Proof.
(1)Let S be the solution set of A(x) = B(x).convention
(2)Let S′ be the solution set of A(x)² = B(x)².convention
(3)Let S⁻ be the solution set of A(x) = −B(x).convention
(4)The three solution sets are counted within the x at which both A and B are defined.convention
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Lemma L1
If A(x) = B(x), then A(x)² = B(x)²
Proof.
(1)Suppose A(x) = B(x).assumption
(2)Multiplying both sides by the same number gives A(x)² = B(x)².(1), E
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Lemma L2
S′ ⊆ S ∪ S⁻
Proof.
(1)Suppose x ∈ S′. Then A(x)² = B(x)².assumption, D1
(2)Either A(x) = B(x) or A(x) = −B(x).(1), A2
(3)Hence x ∈ S or x ∈ S⁻.(2), D1
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Lemma L3
S ∪ S⁻ ⊆ S′
Proof.
(1)If x ∈ S, then A(x)² = B(x)², so x ∈ S′.L1, D1
(2)Suppose x ∈ S⁻. Then A(x) = −B(x).assumption, D1
(3)A(x)² = (−B(x))²(2), L1
(4)(−B(x))² = B(x)²A1
(5)A(x)² = B(x)², so x ∈ S′.(3), (4), D1
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Theorem T1
S ⊆ S′. Squaring loses no solutions.
Proof.
(1)Suppose x ∈ S. Then A(x) = B(x).assumption, D1
(2)A(x)² = B(x)², so x ∈ S′.(1), L1, D1
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Theorem T2
S′ = S ∪ S⁻. The added solutions are the solutions of A = −B.
Proof.
(1)S′ ⊆ S ∪ S⁻L2
(2)S ∪ S⁻ ⊆ S′L3
(3)The two sets contain each other, hence they are equal.(1), (2), A9
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Theorem T3
There is an example with S ⊊ S′.
Proof.
(1)Consider √(x + 2) = x. Both sides are defined for x with x + 2 ≥ 0.A3, D1
(2)Squaring both sides gives the equation x + 2 = x².(1), A3, D1
(3)When x = −1, we have x + 2 = 1.N2
(4)(−1)² = 1² = 1A1, N3
(5)Hence −1 is a solution of the squared equation.(3), (4), D1
(6)When x = −1, we have √(x + 2) = √1 = 1 and 1 ≠ −1.(3), A3, N3
(7)Hence −1 is not a solution of the original equation.(6), D1
(8)When x = 2, we have x + 2 = 4 and √4 = 2, so 2 is a solution of the original equation.N1, A3, D1
(9)S is nonempty, and −1 is in S′ but not in S.(5), (7), (8)
(10)Hence S ⊊ S′.(9), T1
(11)√(−1 + 2) = 1 = −(−1), so −1 is in S⁻. This is the place L2 pointed to.(6), N2, D1, L2
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Corollary C
If A(x) ≥ 0 and B(x) ≥ 0 for every x, then S = S′
Proof.
(1)Suppose 0 ≤ A(x) and 0 ≤ B(x).assumption
(2)Suppose x ∈ S⁻. Then A(x) = −B(x).assumption, D1
(3)If 0 < B(x), then 0 + (−B(x)) < B(x) + (−B(x)).A5
(4)The left side is −B(x) and the right side is 0, so −B(x) < 0.(3), A8, A7, A6
(5)Then A(x) < 0, which contradicts 0 ≤ A(x) and trichotomy.(1), (2), (4), A4
(6)Hence B(x) = 0.(1), (5), A4
(7)Since −0 = 0, we have A(x) = 0 = B(x).(2), (6), A6, A7, A8
(8)Hence x ∈ S, and S⁻ ⊆ S.(7), D1
(9)S′ ⊆ S ∪ S⁻ ⊆ SL2, (8)
(10)Since S ⊆ S′, the two sets are equal.(9), T1, A9
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