Does someone have a way to check this?

Kareir

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

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

LeixlipRed

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

I don't think it works for x=0

LeixlipRed

silverside wrote: »

I don't think it works for x=0

Haha, true.

Thoie

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

Thomas_S_Hunterson

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

Take half it's length and double it.

LeixlipRed

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

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

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

Some people haven't copped on yet.

professorpete

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

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

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

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

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?

professorpete

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

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

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.

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

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

LeixlipRed

LeixlipRed wrote: »

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

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

professorpete

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

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

YAY 1=2

ZorbaTehZ

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.