Consider the following

Consider the following.

>Nothing followed.

trivial, why'd you even waste the time it took to write down a question that simple?

2

[math] x^2-2x = x(x-2) [/math]
[math] 2+2=2\times 2 = 2^2 = 4 [/math]
[math]0= 0+0=0\times 0 [/math] wich may equal [math] 0^0 [/math] depending on how it's defined.

learn how to properly format ``quotes" please

[math]``\text{it's not hard}"[/math]

"For example," is not proof. - Jewish proverb

Also, the connection to the third bit (exponentiation) is missing.

You don't need it, except to check to see if the values you get from the first equality still hold.

Such a trivial exercise.

First, proving that 2 has this property is a trivial exercise left for the reader. Now, lets prove that this number is unique. Lets suppose there are two different numbers [math] x,y [/math] that have this property. Then:
[math] 2x = x^2 [/math]
[math] y^2 = 2y [/math]
[math] 2x y^2 = 2y x^2 [/math]
[math] xy^2 = yx^2 [/math]
[math] x = y [/math]

So there one and only one number with this property, which is 2.

He didn't give you an example, he solved the equation.
2x = x^2 has roots 2 and 0. These are the only numbers that even satisfy the first equality. Just check them.

Your exercise could be solved by 7th graders, stop trying to overcomplicate it

This is all well and good but why not play with the [math] x^x [/math] RHS a bit, by way of an alternative proof?

Are you scared? Do mathematicians have a difficulty with that one?

I gather that in the first place, the "necessary" part is the bits where you compare the addition-bit and the multiplication-bit, while the "sufficient" bit is where you just plug 2 ino the exponent bit and whoops oh yeah it works. Is that how it works?

>Now, lets prove that this number is unique. Lets suppose there are two different numbers x,yx,y that have this property. Then:
>2x=x22x=x2
>y2=2yy2=2y
>2xy2=2yx22xy2=2yx2
>xy2=yx2xy2=yx2
>x=y
You fucking brainlet, this doesn't work in the case of 0. In fact, 0 satisfies the first equality, it only fails because 0^0 is undefined.

I am the OP faggot and I wish to stress that this post is not me.

>This is all well and good but why not play with the RHS a bit, by way of an alternative proof?
Because it's stupid to solve a problem in a hard manner when there's an elementary school method available

>0^0 is undefined
nigger what

Brainlet detected

But isn't that a challenge, an alternate consideration?

What if one rephrases such-and-such, requring that the proof require such-and-such direct engagement with the exponentiation term? what then?

Are mathematicians scared, or unable? I'm given to understand that [math] x^x [/math] is a tricky critter.

0^0=1 you idiot

>he thinks you can create something about of nothing

creationists baka

brainlet detected

>What if one rephrases such-and-such, requring that the proof require such-and-such direct engagement with the exponentiation term? what then?
That's not how proofs work.

>Are mathematicians scared, or unable?
No
they're lazy

this is like asking why somebody would use a hammer when you challenge them to push a nail in with their thumb

Ok just because you're *really* triggering my autism ill humor you.

x^2=x^x
take logarithm of both sides
ln(x^2)=ln(x^x)
by properties of logarithm
2lnx=xlnx
2=x
1 is also a valid solution b/c ln(1)=0 but we don't care about it because it does not satisfy the first part 2x=x^2. that's all I'm willing to give you now go be a brainlet somewhere else

>Jewish proverb

Use our dedicated homework board for this stuff.