'Show that' help

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!

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MHF Hall of Honor

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.

Glitch

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

Also sprach Zarathustra

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!

Glitch

Thanks! Looks like I was way off!

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

Also sprach Zarathustra

Yes, you explain it in the next form:

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

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

ok?

Glitch

Glitch

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

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MHF Hall of Honor
Thanks Also sprach Zarathustra, my way was pretty tedious compared with yours.

Thanks! Looks like I was way off!

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.

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MHF Hall of Honor
One more thing worth mentioning since "show that" means we're dealing with formal proofs here.

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

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

$$\displaystyle \displaystyle \iff$$ {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

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

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

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

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

Glitch

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