1. ## 'Show that' help

Suppose that ab > 0. Show that if a < b, then 1/b < 1/a.

I know this is correct, but I don't know how to show it. I did this:

aa^-2 < ba^-2

1/a < ba^-2

(b^-2)/a < (a^-2)/b

Hoping that I could find a way to swap the inequality sign around whilst creating the fractions required. But I can't seem to do it.

Any assistance would be great!

2. Originally Posted by Glitch

Suppose that ab > 0. Show that if a < b, then 1/b < 1/a.

I know this is correct, but I don't know how to show it. I did this:

aa^-2 < ba^-2

1/a < ba^-2

(b^-2)/a < (a^-2)/b

Hoping that I could find a way to swap the inequality sign around whilst creating the fractions required. But I can't seem to do it.

Any assistance would be great!
You can consider two cases: (1) a and b are both positive, or (2) a and b are both negative.

3. Ok. But am I on the right track? Is there a method to proving this?

4. Originally Posted by undefined
You can consider two cases: (1) a and b are both positive, or (2) a and b are both negative.
We can do it without "consider two cases"...

1/b<1/a

1/b-1/a<0

{a-b}/ab<0

since ab>0, and a<b, it's obvious!

5. Thanks! Looks like I was way off! :P

If I were to write that in an exam, would I have to explain why it's obvious?

6. Yes, you explain it in the next form:

{a-b}<0 since a<b(given)
ab>0 {given}

so, {a-b}/ab={-}/{+}={-}<0

ok?

7. Yup. Thanks. I really do need more practise with this stuff.

8. Thanks Also sprach Zarathustra, my way was pretty tedious compared with yours.

Originally Posted by Glitch
Thanks! Looks like I was way off! :P

If I were to write that in an exam, would I have to explain why it's obvious?
You should explain it on an exam, yes, but it's simply that the numerator is negative while the denominator is positive...

Edit: Too slow.

9. One more thing worth mentioning since "show that" means we're dealing with formal proofs here.

Originally Posted by Also sprach Zarathustra
1/b<1/a

1/b-1/a<0

{a-b}/ab<0
These lines are connected implicitly by "if and only if" as in

1/b<1/a

$\displaystyle \iff$ 1/b-1/a<0

$\displaystyle \iff$ {a-b}/ab<0

For the last line, we don't need to think about whether it's "if and only if" because we only need to go in one direction

1/b<1/a

$\displaystyle \iff$ 1/b-1/a<0

$\displaystyle \iff$ {a-b}/ab<0

$\displaystyle \Longleftarrow ab > 0 \land a < b$

I write this mainly so that nobody gets confused, thinking the proof is not valid because we assumed what we wanted to prove.

10. I've been meaning to ask, what does that upside-down 'V' symbol mean?

11. Originally Posted by Glitch
I've been meaning to ask, what does that upside-down 'V' symbol mean?
AND

12. The simplest way is to multiply 1/ab in to the both sides! and then you wont have to say that this is obvious!!
Just note that 1/ab>0 since ab>0.
a<b => a/ab<b/ab => 1/b<1/a .

13. Given $ab > 0$ and $a < b$.

Case 1: $a > 0$ and $b > 0$.

$a < b$

$\frac{a}{a} < \frac{b}{a}$

$1 < \frac{b}{a}$

$\frac{1}{b} < \frac{b}{ab}$

$\frac{1}{b} < \frac{1}{a}$.

Case 2: $a < 0$ and $b < 0$.

$a < b$

$\frac{a}{a} > \frac{b}{a}$

$1 > \frac{b}{a}$

$\frac{1}{b} < \frac{b}{ab}$

$\frac{1}{b} < \frac{1}{a}$.

Q.E.D.

14. Originally Posted by Glitch
Thanks! Looks like I was way off! :P

If I were to write that in an exam, would I have to explain why it's obvious?
The best thing to do would be to multiply both sids of $\frac{a- b}{ab}> 0$ by the positive value ab to get a- b> 0, then add b to both sides.