# simple but puzzling proofs

• May 5th 2009, 05:27 PM
pberardi
simple but puzzling proofs
I am a bit puzzled on this. It is a bit embarrassing.

Is the following statement true or false?
For all y there exists an x s.t. x > y^2 + 1

The problem says nothing else about x or y and this is confusing. I have a feeling it is true because I can always find a number that is bigger than another number. Am I thinking correctly? Or is it that I choose y to be zero and x to be zero and thus it is false by counterexample. I think that this is incorrect because it is for some x not for all x.
• May 5th 2009, 05:44 PM
BoboStrategy
Quote:

Originally Posted by pberardi
I am a bit puzzled on this. It is a bit embarrassing.

Is the following statement true or false?
For all y there exists an x s.t. x > y^2 + 1

Let y be any real number. Define x so that \$\displaystyle x = y^2 + 2 \$.
Then show that \$\displaystyle x > y^2 + 1 \$

You might find the part I chapters 1-3 of my book Topology and the Language of Mathematics useful. There's a free download available through here: Bobo Strategy - Topology

Hope that helps...
• May 5th 2009, 05:53 PM
pberardi
Thanks but could you please explain what gives me the right to define x = y^2 + 1 ?
• May 5th 2009, 06:21 PM
BoboStrategy
Re:
Quote:

Originally Posted by pberardi
Thanks but could you please explain what gives me the right to define x = y^2 + 1 ?

You are supposed to show that for any y, you can come up with an x so that x is bigger than \$\displaystyle y^2 + 1 \$. That's what we've done.

For any y, we have shown there is an x that is bigger than \$\displaystyle y^2 + 1 \$. And we've defined it as a function of the arbitrary y.

Note you could also define x = 323 - it just wouldn't be useful as it wouldn't always be bigger than \$\displaystyle y^2 + 1 \$ (for example when y = 20).