'Show that' help

Apr 2010
487
9
The question asks:

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
Mar 2010
2,340
821
Chicago
The question asks:

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.
 
Apr 2010
487
9
Ok. But am I on the right track? Is there a method to proving this?
 
Apr 2010
487
9
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?
 
Dec 2009
1,506
434
Russia
Yes, you explain it in the next form:

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

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

ok?
 
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Apr 2010
487
9
Yup. Thanks. I really do need more practise with this stuff.
 

undefined

MHF Hall of Honor
Mar 2010
2,340
821
Chicago
Thanks Also sprach Zarathustra, my way was pretty tedious compared with yours.

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.
 

undefined

MHF Hall of Honor
Mar 2010
2,340
821
Chicago
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.
 
Apr 2010
487
9
I've been meaning to ask, what does that upside-down 'V' symbol mean?