# Thread: prove inequality

1. ## prove inequality

im in a proof writing class and am having particular trouble proving that when a and b are both either positive or negative, a<b iff (1/a) > (1/b)

2. Originally Posted by alr621 im in a proof writing class and am having particular trouble proving that when a and b are both either positive or negative, a<b iff (1/a) > (1/b)
It is almost impossible to help unless we know the set of axioms you have to work with. How basic are your axioms?

3. very basic. i have the field axioms and the order axioms

4. Say that a & b are both positive or both negative.
By order axoms: $\displaystyle x > y \Leftrightarrow \left( {x - y} \right) > 0$.
So one way is:
$\displaystyle \frac{1}{a} > \frac{1}{b} \Rightarrow \frac{1}{a} - \frac{1}{b} > 0 \Rightarrow \frac{{b - a}}{{ab}} > 0 \Rightarrow b > a$.( note that it must be true that $\displaystyle ab > 0$).

Now you can do the other way.

5. thanks! i have another too. if (a^2)<(b^2) and a, b>0, then a<b

6. Proof by contradiction
$\displaystyle a \ge b\quad \Rightarrow \quad a^2 \ge ab$.
Then what?

7. can you clarify a little more? i have to prove that a<b if (a^2)<(b^2)

8. he is saying assume a is greater than or equal to b and see where that takes you.. you will reach an inherent contradiction, finding no way to prove that a>= b and thus a must be < b

9. okay. last one. if a<b and c<d, then a+c < b+d

10. Originally Posted by alr621 can you clarify a little more? i have to prove that a<b if (a^2)<(b^2)
There are responders here who will supply posters with ready to turn in solutions. I must tell you that I am not one of those. I must also tell you that I doubt that any of those have enough ability to help you with this question. What I am telling you is this: I want you to learn to do these problems on your own. I am willing to give you substantial hints and direction, but you must show effort on your own. So dont ask me to do more unless you have done more yourself.

11. Originally Posted by Plato There are responders here who will supply posters with ready to turn in solutions. I must tell you that I am not one of those. I must also tell you that I doubt that any of those have enough ability to help you with this question. What I am telling you is this: I want you to learn to do these problems on your own. I am willing to give you substantial hints and direction, but you must show effort on your own. So dont ask me to do more unless you have done more yourself.
I'm not asking for direct answers to all of my problems... I was simply asking for clarification on that specific situation because I didn't understand it fully.

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