• Support PF! Buy your school textbooks, materials and every day products Here!

Real Analysis Proof.

  • Thread starter playboy
  • Start date
  • #1
playboy
Prove: For every real number x>1, there exists two distinct positive real numbers y and z such that

x = (y^2 +9)/(6y) = (z^2 +9)/(6z)

Okay.. this has real got me beat. Firstly (this sounds stupid and obvious), when they give us a proof, is it really true? Do we just naturally believe that its true and with this in mind, prove it? Or could it be wrong and we eventually find that out if we do the proof?

I first tried to isolate y and z in terms of x and set them equal to each other, but that led me to a dead end.

Then i tried the contrapositive, but that didn't make it any easier.

So.. I tried contradiction.

For every real number x>1, there DOES NOT exists two distinct positive real numbers y and z such that

x = (y^2 +9)/(6y) = (z^2 +9)/(6z)

and now im out of ideas?

Can somebody please help me out?

Thanks
 

Answers and Replies

  • #2
benorin
Homework Helper
1,085
14
Consider solving the equation

[tex]x=\frac{w^2+9}{6w},[/tex] where x>1 is a constant.

If [itex]w\neq 0[/itex], then

[tex]w^2-6xw+9=0[/tex]

so by the quadratic equation, we have

[tex]w=\frac{6x\pm\sqrt{36x^2-36}}{2}=3x\pm 3\sqrt{x^2-1}[/tex]

which are both real (since x>1) and thus we note that:

[tex]y=3x+3\sqrt{x^2-1},z=3x-3\sqrt{x^2-1}[/tex]

are certianly distinct real numbers satisfing said conditions.
 
  • #3
matt grime
Science Advisor
Homework Helper
9,395
3
There is of course another way of showing existence without actually finding the answers y aand z in terms of x.

Consider f(w)=(w^2+9)/6w - x

when w tends to zero this is positive, when w tends to infinity this is positive. If there exists any point where it is negative then the solution follows. We can finid the minimum: f'(w)= 1/6 - 3/(4w^2), and the minimal value there will be negative (for x>1) and we are done.
 

Related Threads for: Real Analysis Proof.

  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
6
Views
530
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
7
Views
788
  • Last Post
Replies
3
Views
957
  • Last Post
Replies
4
Views
1K
Top