Math Is Fun Forum

Toast

Real Member

Registered: 2006-10-08

Posts: 1,321

Variable

We have a problem where we have to prove that

is true no matter what the value of x.

Actually, to be precise, we were given the following information:

Think of a number...Add one...Double it...Take away 3...Add the number you first thought of...Add 7... Divide by 3... Take away the number you first thought of... You should be left with 2. Explain why this works.

So I try to solve for x:

This proves that the value of x does not matter since it will be eliminated anyway.

Some of my friends, however, think that I must leave the right side equaled to 2 for it to be correct:

Does it really matter?

Last edited by Toast (2007-02-14 19:53:47)

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: Variable

Those both have a presumption that you need to determin x. And this assumption is not good.
Typically you need only evaluate the left blindly, without knowing it equating to 2, then naturally derive 2 as the result. Through this process you only assumed x  any value.

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X'(y-Xβ)=0

Toast

Real Member

Registered: 2006-10-08

Posts: 1,321

Re: Variable

Sorry I don't quite understand =.=

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: Variable

Sorry I made a mistake. The 3 ways (yours, your friend's and mine) are all correct.

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X'(y-Xβ)=0

Toast

Real Member

Registered: 2006-10-08

Posts: 1,321

Re: Variable

I've received several similar problems where the variable eliminates itself. Some of these problems contain 2 variables, x and y. If they are both eliminated that means either of them could take any value, right?

MathsIsFun

Administrator

Registered: 2005-01-21

Posts: 7,713

Re: Variable

Isn't it neater just to show that x gets cancelled?

Isolate "x": (2x+x)/3 - x = 3x/3 - x = x-x

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"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

Toast

Real Member

Registered: 2006-10-08

Posts: 1,321

Re: Variable

Hmm well I supposed so

Thanks George and mathsisfun

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Variable

MathsIsFun's method isn't enough, is it?
Surely all that would do is show that the expression is always some constant, no matter what x is. It doesn't show that that constant is 2.

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Why did the vector cross the road?
It wanted to be normal.

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: Variable

His method is enough, actually his method is what I proposed in Post 4. This method has the assumption that x is arbitary when dealing it algebraically, so when the result shows 2, it means whatever x is, the result is (always) 2.

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X'(y-Xβ)=0

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: Variable

And the method of Toast and his friend is generally the solving equation approach. And every manupulation of the equation, every changing is called equivalent transformation, until 2=2. Equivalent literally means if one holds, the other too.

Then when x satisfies what 2=2 holds, the answer is x=any. So x=any => 2=2 is true => the original equation holds. That means regardless the value of x, the original equation is true. And that means always true.

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X'(y-Xβ)=0

Toast

Real Member

Registered: 2006-10-08

Posts: 1,321

Re: Variable

Also, does

?

Last edited by Toast (2007-02-15 22:22:07)

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: Variable

Logically yes. 2*3=2*3, 6/3=6/3

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X'(y-Xβ)=0