Prove that for every real number $x$, if $x>2$ then there is a real number $y$ such that $y+1/y = x$.

Question

I'm trying to learn about proofs and I'm stuck in Velleman's book "How to Prove it".

This is the question (ex.6 p.118):

Prove that for every real number $x$, if $x > 2$ then there is a real number $y$ such that $y+\frac{1}{y}= x$.

I couldn't do it so I went to see the answer, and it is written that we can assume that $y=(x+\sqrt{x^2-4})/2$.

How did the author get to this value for $y$?

Answer 1

Suppose that

$$y+\frac{1}{y}=x.$$

Then

$$y^2+1=xy,$$

and

$$y^2-xy+1=0.$$

It follows that

$$y=\frac{x\pm\sqrt{x^2-4}}{2}$$

either by completing the square, or by using the quadratic formula.

Answer 2

The author just pulled it out of nowhere - where it comes from isn't important to the proof, only that it works. For how they knew what to pull out of nowhere, try rearranging $y + \frac{1}{y} = x$ (which I suspect is what the formula was supposed to read?) to make $x$ the subject.

Answer 3

$y + \frac 1y = x \implies$

$y(y+\frac 1y) = yx \implies$

$y^2 - yx + 1 = 0$

Now we use quadratic equation or complete the square to solve for $y$.

$y = \frac {x \pm \sqrt{x^2 - 4}}2$

The only issue is we must have $x^2 -4 > 0$. Which is okay as $x \ge 2 \implies x^2 -4 \ge 0$