Identity proof

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Jan 14th 2006, 02:01 PM
DenMac21

Identity proof

I have to prove that if $|x| <> 1, y <> 0, y<>-x$ then identity is valid:

$\frac{x^2-1}{y^2+xy}\cdot(\frac{1}{1-\frac{1}{y}}-1)\cdot\frac{1-xy^3-y^4+y}{1-x^2}=\frac{y^2+y+1}{y}$

I have try to prove it.This what I have done:

$\frac{x^2-1}{y^2+xy}\cdot(\frac{1}{y-1})\cdot\frac{1-xy^3-y^4+y}{1-x^2}=\frac{y^2+y+1}{y}$

$\frac{(1-x^2)(-1+xy^3+y^4-y)}{(y^2+xy)(y-1)(1-x^2)}=\frac{y^2+y+1}{y}$

$\frac{(-1+xy^3+y^4-y)}{(y^2+xy)(y-1)}=\frac{y^2+y+1}{y}$

$\frac{-1+xy^3+y^4-y}{y^3-y^2+xy^2-xy}=\frac{y^2+y+1}{y}$

$\frac{y(-1+xy^3+y^4-y)}{y(y^2-y+xy-x)}=y^2+y+1$

$\frac{-1+xy^3+y^4-y}{y^2-y+xy-x}=y^2+y+1$

$-1+xy^3+y^4-y=(y^2-y+xy-x)(y^2+y+1)$

$-1+xy^3+y^4-y=-x+xy^3+y^4-y$

$x=1$

So identity is not valid.
Am I right?
Jan 14th 2006, 02:23 PM
ThePerfectHacker

Did you make a mistake when you multiplied the 3 terms by the 4 terms?
And I also do not understand what all that stuff in the beginning is?
Jan 14th 2006, 02:29 PM
DenMac21

Quote:

Originally Posted by ThePerfectHacker

Did you make a mistake when you multiplied the 3 terms by the 4 terms?

I don't think so, I checked.

Quote:

Originally Posted by ThePerfectHacker

And I also do not understand what all that stuff in the beginning is?

What do you mean exactly?
Jan 14th 2006, 02:39 PM
ThePerfectHacker

I also checked, confirmed.

That means that if the identity is true then $x=1$ which means that if $x$ is anything else from 1 then it is false. Which means it cannot be an identity.
Jan 14th 2006, 03:47 PM
ticbol

Quote:

Originally Posted by DenMac21

I have to prove that if $|x| <> 1, y <> 0, y<>-x$ then identity is valid:

$\frac{x^2-1}{y^2+xy}\cdot(\frac{1}{1-\frac{1}{y}}-1)\cdot\frac{1-xy^3-y^4+y}{1-x^2}=\frac{y^2+y+1}{y}$

I have try to prove it.This what I have done:

[...skip...]

$x=1$

So identity is not valid.
Am I right?

As in proving trig identities, here it is better to check first if the probvlem identity is true for a numerical substitution.
Try x=0 and y=2.
You get the LHS to be 13/4, while the RHS to be 7/2.
LHS <> RHS, hence, not an identity.

So no need to crack cranium for those x's and y's.
Jan 14th 2006, 11:13 PM
CaptainBlack

Quote:

Originally Posted by ThePerfectHacker

I also checked, confirmed.

That means that if the identity is true then $x=1$ which means that if $x$ is anything else from 1 then it is false. Which means it cannot be an identity.

The " $|x| <> 1, y <> 0, y<>-x$ " is meant to be the set of conditions:

$x^2 \ne 1, y \ne 0, y \ne -x$

RonL