Math proof question

[GoldenBear]

Wondering if what I'm doing is legal...

I have

1 > n/(x+y) in simplified terms..it's more complicated but that's basically it

so if I take each side to the -1 I'd get

(x+y)/n

Could I then say that n/(x+y) = (x+y)/n?

This is the problem.

I must show that -1 < r < 1

 

[jmcoreymv]

Well if n/(x+y) = 1 then (x+y)/n is also 1.

 

[jpsj82]

edit: if you want to learn some math that is incorrect look here, if not look below at the other posts.
i don't think so.

1 > n/(x+y)

then to the -1

1 < (x+y)/n

so

n/(x+y) < 1 < (x+y)/n

which means

n/(x+y) < (x+y)/n

so they can't be equal

 

[GoldenBear]

Oh that makes sense. I was thinking you couldn't do that, but that cleared it up.

Although that does help me with this..

 

[jpsj82]

wait, can you do this?

I am trying to think if you can do ^-1 with an inequalty. because it is like doing cross multipling. So you might only be able to do this if n/(x+y) is greater then 0.

now I am confused.

 

[GoldenBear]

Doing ^-1 would be like ^2 I would think..

 

[jpsj82]

what is the rule for ^-1 with inequalities?
look at this

5 > 3 to the ^-1 gives you

0.2 < 0.333, so we need to switch the > sign to <

now look at this
5 > -3, again to the ^-1

0.2 > -0.333, here we don't switch the > sign.

so are we allowed to even use ^-1 with the inequality signs?

 

[Whitecloak]

you cannot use "power" with inequalties ie ^-1 or ^2. multiplying both sides of the inequality with the same quantity is fine as it doesnt change the inequality

ie if a<b, then ax<bx for any x (x <> 0)

if you use powers , say ^2 then we get aa<bb which might or might not be true because both sides of the inequality have been mulitiplied witg different numbers.

 

[jpsj82]

thank you whitecloak

 

[Whitecloak]

dude
why did you give me a rating of 1? :|

 

[jpsj82]

me?, I didn't give you a rating

 

[Whitecloak]

okie dokie

 

[Shiggs]

I gave you a 1