Math question

# 1 05-08-2009 , 06:38 AM

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It's not for school, just something I don't understand and it really bothers me -

When I have this expression:
'y = (x^2)^0.5'
Solving it carefully will result in 'y = |x|'
'y = (x^0.5)^2'
Will result in 'y = x, x>=0'

However, we've been taught in school that it's possible to multiply 2 by 0.5 so the result in this case should be
'y = x^1' or simply 'y = x'

What am I missing? Hope I made my question clear...
Benny

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# 2 05-08-2009 , 07:39 AM

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it's been a while but at a guess, can you comfirm my interpretation of your notation?

|x| is the absolute value of x yes?

then i say...

assuming |x| is the absolute value of x (that is, x will always be positive)
then in that case if x >= 0 then x = |x|
therefore
y = |x| is the same as y = x, x >= 0
because you can't have a negative number if it's absolute

therefore
y = |x| is the same as y = x

if you're asking why it will always be positive then you could probably solve it by proof by induction... or proof by contradiction

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# 3 05-08-2009 , 08:28 AM

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Originally posted by Chirone
it's been a while but at a guess, can you comfirm my interpretation of your notation?

|x| is the absolute value of x yes?

then i say...

assuming |x| is the absolute value of x (that is, x will always be positive)
then in that case if x >= 0 then x = |x|
therefore
y = |x| is the same as y = x, x >= 0
because you can't have a negative number if it's absolute

therefore
y = |x| is the same as y = x

if you're asking why it will always be positive then you could probably solve it by proof by induction... or proof by contradiction

Originally Posted by

I concur.

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# 4 05-08-2009 , 08:47 AM

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Yup thats right.

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# 5 05-08-2009 , 09:07 AM

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ah, no :\

y=x and y=|x| are not the same. that's the whole problem...
Let me simplify:
Here is 'y = x'
user added image

'y = x^2'

'y = (x^2)^0.5'
user added image

One more option -
'y = x^0.5'
user added image

'y = (x^0.5)^2
[IMG]https://img14.imageshack.us/img14/444...eaterthen0.png[/IMG]

non of these is actually the same as 'y=x' which you can get to by multiplying the 0.5 and 2...
(y=(x^2)^0.5 -> y= x^(2*0.5) -> y=x^1 -> y=x)

When in doubt, delete history and freeze transformations.

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# 6 05-08-2009 , 09:10 AM

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Originally posted by Chirone
...
then in that case if x >= 0 then x = |x|
therefore
y = |x| is the same as y = x, x >= 0
...

Originally Posted by

But it's not - that's your mistake...

When in doubt, delete history and freeze transformations.

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# 7 05-08-2009 , 04:06 PM

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I think the answer comes from the fact that "in school, you've been taught that you can simply multiply exponents..." which works in 99.99% of cases, but obviously you've come up with a special case where it doesn't quite work out.

I think it's a bit like canceling out variables (though I could be mistaken here) where you actually lose a solution, so it's not really allowed.

Hope that sheds light on your conundrum... user added image

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# 8 05-08-2009 , 09:50 PM

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I thought it's something like that...
Thanks, that made that clear user added image

So the formula (a^n)^m = a^n*m isn't always correct?
can you point me to some reading material?

Thanks,
Benny

When in doubt, delete history and freeze transformations.

My latest snake game
(Requires DirectX9 SDK and Framework)

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# 9 05-08-2009 , 09:51 PM

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aaah, now i remember what it means to do a ^0.5 on something... it's the square root

when you take the square root of something you get two of the same numbers and they are either both negative or positive

so... that means that what i said is correct, and also y = x when x < 0 as well because x could have been negative or positive

eg
y = ((-2)^2)^0.5 = 2 or -2

not exactly a proof but i think the idea is right

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# 10 05-08-2009 , 10:37 PM

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Chirone, no offense, but I think you should stay out of this one user added image

((-2)^2)^0.5 can only be worth 2 (not -2) (see second post with drawings)...

Never mind that now... stwert has already answered my question.

Thanks anyway user added image

Go join Mayaniac and his pipe

Benny

When in doubt, delete history and freeze transformations.

My latest snake game
(Requires DirectX9 SDK and Framework)

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# 11 05-08-2009 , 11:19 PM

As Zbrushiac sounds stupid!
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Originally posted by BennyK

Thanks anyway Go join Mayaniac and his pipe

Originally Posted by

Hey!.... I ain't sharing!

... user added image

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# 12 05-08-2009 , 11:27 PM

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Actually Chirone does sort of have a point in that when you square root something, (or ^0.5) there are technically two answers. Thus, if you have x^2 = 4, x could be either +2 or - 2...

That's why your angular graph extends into the negative x's... and therefore
y=(x^2)^0.5 does not necessarily equal |x|... and so I take back my earlier statement. Basically I'm just confusing myself and probably you too... but I think the conclusion I've reached is that you can't simplify those exponents to an absolute value, despite how it works in practice (e.g. on a calculator), because those won't give you both answers, though the graph does.

Further... my point still stands that many times the "rules" you learn in school often only apply to the majority of cases, but the multiplying exponents rule may be correct after all.

I guess my question for you, Benny, would be, what does your solution look like?

"Solving it carefully will result in 'y = |x|'"