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**estelle j****Guest**

HI I am given a function g(x)= x^2 - 1, where x takes any real value.

I am asked to **show that this function is not a one-one function.**

Apart from the method of horizontal line test(graphical method of drawing horizontal line and show that cuts graph more than once)

what other method can use to show its not a one-one function?

thanks! (:

**ZHero****Real Member**- Registered: 2008-06-08
- Posts: 1,889

g(x)=x[sup]2[/sup]-1

is defined for all R-->R

To show that g(x) is not one-one we need to show that g(x[sub]1[/sub]) = g(x[sub]2[/sub]) for some x[sub]1[/sub] ≠ x[sub]2[/sub]

Let g(x[sub]1[/sub]) = g(x[sub]2[/sub])

x[sub]1[/sub][sup]2[/sup] - 1 = x[sub]2[/sub][sup]2[/sup] - 1

x[sub]1[/sub][sup]2[/sup] = x[sub]2[/sub][sup]2[/sup]

x[sub]1[/sub][sup]2[/sup] - x[sub]2[/sub][sup]2[/sup] = 0

(x[sub]1[/sub] - x[sub]2[/sub])(x[sub]1[/sub] + x[sub]2[/sub]) = 0

x[sub]1[/sub] - x[sub]2[/sub] = 0 or x[sub]1[/sub] + x[sub]2[/sub] = 0

x[sub]1[/sub] = x[sub]2[/sub] or x[sub]1[/sub] = - x[sub]2[/sub]

i. e. x[sub]1[/sub] = x[sub]2[/sub] or x[sub]1[/sub] ≠ x[sub]2[/sub]

However.. this may not be the only proof.

*Last edited by ZHero (2010-05-11 02:39:01)*

If two or more thoughts intersect, there has to be a point!

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi estelle j;

A function for which every element of the range of the function corresponds to exactly one element of the domain. One-to-one is often written 1-1.

This function is so simple you can do it like this.

To prove that it is not 1 - 1 you just need 1 counterexample. In this case (-4)^2 - 1 = 15 and 4^2 - 1 = 15. 2 x's yielding the same y. Your knowledge that a negative times a negative = a positive is enough to find a counterexample.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

The only comment I have is that it should have been "an interesting comment" in the hidden text...

Honestly, I have no idea what it is you're looking for.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi Rick;

I need to fill in here. What I meant was If it passed by you and her then that was sort of like peer review. Except, in this case we are not peers. The old guy is lagging 2 miles behind and out of breath.

I have been meaning to tell you this:

Recently, a question about whether a function was defined or not came up. This function had a simple pole, I think. Although you and mathsy had already explained it to me, I still made a mistake. Jane protested but I just waved her off. Armed with some new info about removable singularities, I calmy continued. Finally, a guy, called "Dr. Rick," (Ask Dr. Math) pointed it out so bluntly that it actually sunk in.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Ah, I see. Your post was spot on, bobby.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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