Thread: Have I done this 'show that' problem correctly?

The question:

Show that if , then

I did:




Therefore:


My textbook doesn't have answers for these types of questions, so I can't check.

Originally Posted by Glitch

The question:

Show that if , then

I did:




Therefore:


My textbook doesn't have answers for these types of questions, so I can't check.

Well, I think the proof is circular because it relies on being able to take the square root of both sides of an inequality, in other words: for and for non-negative reals and we have , which is what you're trying to prove (and which is not true for general ).

All we're really saying here is that is increasing. The "usual" way is to take the derivative and show that it's greater than equal to zero on that interval. This question is posted in the Pre-Calculus subforum; does it makes sense to take the derivative given where you are in the textbook?

Originally Posted by undefined

This question is posted in the Pre-Calculus subforum; does it makes sense to take the derivative given where you are in the textbook?

Nope, this chapter is pre-calculus only, and hasn't mentioned derivatives yet.

Hint:

Originally Posted by Glitch

The question:

Show that if , then

I did:




Therefore:


My textbook doesn't have answers for these types of questions, so I can't check.

I'm not sure if this approach is necessary any more, but nevertheless here it is:

If

Substitute and you'll get:



Finally you have which is true because the square-root of a positive number is positive too.

It's not "unecessary", it is wrong. You want to prove that , yet the first line of your "proof" begins " which is exactly the same as what you want to prove. And, you are also assuming that which is a much harder theorem that was was asked here.