Does someone have a way to check this?

Kareir · 18-05-2009 07:24PM #1

Hey all,

I was kinda bored in the car on the way home today, so I thought up this very easy, very unoriginal (i think

) equation.

X^n = X/2X [2(X)^n]

Easy, but how do i check whether it works for everything or not?

Thanks,

_Kar.

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gra26 · 18-05-2009 07:53PM

I guess you could try proof by induction, but its fairly clear it works.

LeixlipRed · 18-05-2009 08:10PM

The reason it's true is that the X's and 2's cancel giving you the same thing on left and right. Similar to saying 2=(2/2)2

silverside · 18-05-2009 08:14PM

I don't think it works for x=0

LeixlipRed · 18-05-2009 08:21PM

silverside wrote: »

I don't think it works for x=0

Haha, true.

Thoie · 18-05-2009 08:26PM

http://www79.wolframalpha.com/input/?i=X^n+=+X/2X+(2(X)^n)

Thomas_S_Hunterson · 20-05-2009 10:13PM

Reminds me of the formula for the length of a piece of string.

Take half it's length and double it.

LeixlipRed · 20-05-2009 10:46PM

Sean_K wrote: »

Reminds me of the formula for the length of a piece of string.

Take half it's length and double it.

Yes, but is the string in a hole or half a hole?

ZorbaTehZ · 21-05-2009 09:36AM

Thoie wrote: »

http://www79.wolframalpha.com/input/?i=X^n+=+X/2X+(2(X)^n)

http://www79.wolframalpha.com/input/?i=X^n+%3D+X%2F(2X)+(2(X)^n)
Result: true!

jhayden · 21-05-2009 09:51AM

What everybody else was saying. Here is how I would do it.

X^n = X/2X [2(X)^n]
Multiply both sides by 2
2(X^n) = 2(X/2X [2(X)^n]
)
2(X^n) = 2x/2x[2(X)^n]
Simplify
2(X^n) = [2(X)^n]

LeixlipRed · 21-05-2009 10:40AM

Some people haven't copped on yet.

professorpete · 21-05-2009 10:49AM

silverside wrote: »

I don't think it works for x=0

I beg to differ!

the "equation" reduces to x^n = x^n; which is true for all x, n, and ^ for that matter! :pac:

LeixlipRed's got it right there!

..not a real professor..

alan4cult · 21-05-2009 11:05AM

professorpete wrote: »

I beg to differ!

the "equation" reduces to x^n = x^n; which is true for all x, n, and ^ for that matter! :pac:

LeixlipRed's got it right there!

..not a real professor..

x=0 and n=0?

professorpete · 21-05-2009 11:22AM

alan4cult wrote: »

x=0 and n=0?

damn.. hate when that happens!!

but isn't x^0 = 1 for all x?

that would take care of that, unless 0^0 is undefined? can't remember..

not trying to be smart honest!

jhayden · 21-05-2009 11:26AM

For x = 0 0/0 is undefined (so it doesn't work).

For n = 0

X^n
x^0 = 1

x/2x[2(X)^n] (I'm guessing square brackets means multiply by)
x/2x[2(1)]
2x/2x
1
n = 0 works.

professorpete · 21-05-2009 11:50AM

jhayden wrote: »

For x = 0 0/0 is undefined (so it doesn't work).

Sorry, beg to differ agian! X/2X appears in the equation, but since it reduces immediately, the division operation doesn't happen..so it does hold for x = 0.

it's an interesting format for the equation, but, however you dress it up, the equation (in it's simplest - ie true form) is

x^n = x^n

which, as they say is trivial..sorry, i know that's really anal but isn't that what mathematics is all about?!?!

what about the 0^0 thing though, anyone?

21-05-2009 12:07PM

professorpete wrote: »

what about the 0^0 thing though, anyone?

I believe 0^0 is defined to be 1.

x^0 is defined to be 1 for all x.

rjt · 21-05-2009 02:17PM

professorpete wrote: »

Sorry, beg to differ agian! X/2X appears in the equation, but since it reduces immediately, the division operation doesn't happen..so it does hold for x = 0.

This is just me being a pedant - but I don't believe this is true. More to the point 2X/X = 2 only when X is non-zero. For X=0 it's undefined. So we can't "reduce" it when X=0. (and the identity doesn't hold there, as the RHS is undefined!)

jhayden · 21-05-2009 02:22PM

professorpete wrote: »

Sorry, beg to differ agian! X/2X appears in the equation, but since it reduces immediately, the division operation doesn't happen..so it does hold for x = 0.

Sorry professorPete, I beg to differ also. As far as I remember in maths you always did brackets first, then multiplied then divided then added and finally subtracted. If that was the case the equation should have read 1/2 and not x/2x originally.

21-05-2009 05:02PM

Just a thought. Google Calculator defines 0^2 to be 0. Likewise, it defines 0^3 and 0^4 etc. to be 0. So, is 0^n (n>0; as 0^0 is defined as 1) defined to be 0? Wouldn't this violate Fermat's Last Theorem (i.e. 0^6 + 1^6 = 1^6; 0 + 1 = 1). Is this a fault, or just a matter of preference, of Google Calc? Or, is there a caveat in the Theorem not allowing 0? I can't see Andrew Wiles being too happy with that if it were true!

LeixlipRed · 21-05-2009 05:04PM

a,b and c have to be postive in FLT.

21-05-2009 05:27PM

LeixlipRed wrote: »

a,b and c have to be postive in FLT.

Ah right right, I see. Thanks for clearing that up.

professorpete · 21-05-2009 10:19PM

rjt wrote: »

This is just me being a pedant - but I don't believe this is true. More to the point 2X/X = 2 only when X is non-zero. For X=0 it's undefined. So we can't "reduce" it when X=0. (and the identity doesn't hold there, as the RHS is undefined!)

heh no worries i'm totally pedantic too!
2x/x is 2, always; there is no x, because x/x = 1

you don't have to worry about particular values of x, if you can show that the expression is equal to a constant

jhayden wrote: »

Sorry professorPete, I beg to differ also. As far as I remember in maths you always did brackets first, then multiplied then divided then added and finally subtracted. If that was the case the equation should have read 1/2 and not x/2x originally.

yep, that's true but it's the order you would evaluate the expression, given values for x and n. the brackets in this case don't really tell you what to do, ie you can, and should cancel off any inverses where you can, so with something like

x/2x[2(X)^n]

(excusing the inelegant bracket-wrangling!) the x's and the 2's cancel immediately, leaving x^n, innit?!

but anyway, it just occurred to me that since the expression does indeed reduce to

x^n = x^n
or
1 = 1

which, i'd say we can all agree is true... so that, i believe answers the original question, no? a way to check it - prove it!

rjt · 21-05-2009 10:34PM

professorpete wrote: »

heh no worries i'm totally pedantic too!
2x/x is 2, always; there is no x, because x/x = 1

you don't have to worry about particular values of x, if you can show that the expression is equal to a constant

x = 0
=> x = x + 0
=> x = x + x
=> x/x = x/x + x/x
=> 1 = 1 + 1
...

gra26 · 22-05-2009 10:33AM

YAY 1=2

ZorbaTehZ · 22-05-2009 10:05PM

professorpete wrote: »

heh no worries i'm totally pedantic too!
2x/x is 2, always; there is no x, because x/x = 1

Your logic fails at x = 0, you just simply cannot say that it is equal to 1, some examples why in the Wikipedia article, and quite a clever example by rjt above.

Does someone have a way to check this?

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