Page 128 - HOW TO PROVE IT: A Structured Approach, Second Edition

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114 Proofs
The steps we’ve used to solve for y should not appear in the final proof. In the
√
final proof we will simply say “Let y = (−1 + 1 + 4x)/2” and then prove
that y(y + 1) = x. In other words, the final proof will have this form:

Let x be an arbitrary real number.
Suppose x > 0.
√
Let y = (−1 + 1 + 4x)/2.
[Proof of y(y + 1) = x goes here.]
Thus, ∃y[y(y + 1) = x].
Therefore x > 0 →∃y[y(y + 1) = x].
Since x was arbitrary, we can conclude that ∀x(x > 0 →∃y[y(y + 1)
= x]).

To see what must be done to fill in the remaining gap in the proof, we add
√
y = (−1 + 1 + 4x)/2 to the givens list and make y(y + 1) = x the goal:
Givens Goal
x > 0 y(y + 1) = x
√
−1 + 1 + 4x
y =
2
We can now prove that the equation y(y + 1) = x is true by simply sub-
√
stituting (−1 + 1 + 4x)/2 for y and verifying that the resulting equation is
true.

Solution

Theorem. For every real number x, if x > 0 then there is a real number y such
that y(y + 1) = x.
Proof. Let x be an arbitrary real number, and suppose x > 0. Let
√
−1 + 1 + 4x
y =
2
which is defined since x > 0. Then,
√ √
−1 + 1 + 4x −1 + 1 + 4x
y(y + 1) = · + 1
2 2
√ √
1 + 4x − 1 1 + 4x + 1
= ·
2 2
1 + 4x − 1 4x
= = = x.
4 4