- Thread starter nad081
- Start date

- Joined
- Nov 12, 2017

- Messages
- 6,663

If \(\displaystyle c^2 = 1 - \left(\frac{a+b}{a-b}\right)^2\), can \(\displaystyle c^2\) ever be greater than 1? That part should be easy. (By the way, they told you that \(\displaystyle c \ne 1\), so the "inclusive" part was not really necessary.)View attachment 17365

I cannot figure out how to work out the last part … I figured out that one real solution means real and equal roots (discriminant = 0) but I cannot understand how it is possible to prove that c^2 is between 0 and 1 inclusive.

And it should be even easier to see that \(\displaystyle c^2 > 0\).

If you don't see this, please show your work in attempting it, so we can see if you are going in the wrong direction.

Many thanks for your reply.If \(\displaystyle c^2 = 1 - \left(\frac{a+b}{a-b}\right)^2\), can \(\displaystyle c^2\) ever be greater than 1? That part should be easy. (By the way, they told you that \(\displaystyle c \ne 1\), so the "inclusive" part was not really necessary.)

And it should be even easier to see that \(\displaystyle c^2 > 0\).

If you don't see this, please show your work in attempting it, so we can see if you are going in the wrong direction.

If you are subtracting from 1, the value should be less than 1. Also, yes anything square is positive, so the answer can never be less than 0. But does this involve any working? How can I represent my thoughts mathematically? This is a question from a test paper.

Thanks in advance!

- Joined
- Nov 12, 2017

- Messages
- 6,663

But you can express this using symbols together with words. Since any square is non-negative (not necessarily positive!), you can say first that \(\displaystyle c^2 \ge 0\); you still need to show that it is non-zero, which is clear from the initial form they have to find for \(\displaystyle c^2\), because you are told that a and b are non-zero.

For the other part, you can similarly state that \(\displaystyle \left(\frac{a+b}{a-b}\right)^2 \ge 0\), and derive the desired inequality from this one.

Many thanks! Appreciated

But you can express this using symbols together with words. Since any square is non-negative (not necessarily positive!), you can say first that \(\displaystyle c^2 \ge 0\); you still need to show that it is non-zero, which is clear from the initial form they have to find for \(\displaystyle c^2\), because you are told that a and b are non-zero.

For the other part, you can similarly state that \(\displaystyle \left(\frac{a+b}{a-b}\right)^2 \ge 0\), and derive the desired inequality from this one.

- Joined
- Apr 22, 2015

- Messages
- 2,223

How can c… you can say first that \(\displaystyle c^2 \ge 0\) …

- Joined
- Dec 30, 2014

- Messages
- 6,147

You have to be careful with what you say. You made two statements and they are both false.Many thanks for your reply.

If you are subtracting from 1, the value should be less than 1. Also, yes anything square is positive, so the answer can never be less than 0. But does this involve any working? How can I represent my thoughts mathematically? This is a question from a test paper.

Thanks in advance!

1)

2)

This is mathematics and what you say must be sound. Please be more careful.

Now you are going to argue that in your situation your statements follow. My problem is that you made this statements in general.

For example if had said that if x is not 0 then x^2 is never negative and that x^2 is always positive that would have been fine.

- Joined
- Nov 12, 2017

- Messages
- 6,663

You missed my point. I said "you can sayHow can c^{2}equal zero, when we have c^{2}=-4ab/(a-b)^{2}with a≠0 and b≠0?

One doesn't have to see an entire proof at once.

- Joined
- Apr 22, 2015

- Messages
- 2,223

I didn't have to, but that's what happened.… One doesn't have to see an entire proof at once …

Seriously, though, I did miss your point. I had forgotten the call for a proof, so I was thinking loosely (the fact that a and b must have opposite signs foremost in my mind). Thank you.

\(\;\)