# FOUR DOES EQUAL THREE!

Remember that “a=0” is true in all cases for this equation.

Expand it into: a(4-3)-a(4-3)=0
Since a=0: a(4-4)-a(3-3)=0
Take out common factors: 4a(1-1)-3a(1-1)=0

This step is a bit tricky; but since a=0, and a(4-3)-a(4-3)=0, then a(3-3)=0. So, by replacing the 0 in the ending of the last equation so far shown with a(3-3), it won’t make a difference; and since 0 is not contributing anything to the equation on the left of each statement, then it will not matter if it’s gone. So it’s safe to say that I may move it over to the right.

So: 4a(1-1)=3a(1-1)=0
Divide each: 4a(1-1)/a(1-1)=3a(1-1)/a(1-1)=0
You end up with 4/1=3/1=0
Since it can also be written as 4=3=0, it is safe to say that: 4=3

I’m sure it’s flawed beyond recognition. Please correct it, k tnx.

uh what

assuming i know what i’m doing, it dies at step one.

you say “expand it”

so, i did, using FOIL, the way i was always taught to do polynomials

First: a4
Outer: -a
-3
Inner: -a4
Last:-a
-3

then i got

4a-3a-4a+3a=0

and thats just 0=0 since

4a-4a=0
3a-3a=0

maybe you meant to use two different variables, instead of both being a?

all of your expressions just factor into 0=0

stfu dev i didn’t say solve i said expand it into a(4-3)-a(4-3)=0.

i was just trying your proof, no need to get huffy :\

also, you divide by zero, right?

Yeah, the whole point was that it is impossible to divide by zero.

Why did you make another thread? -_-

step 1 is wrong. You don’t expand like that. You use foil to expand a factor. You also divided by 0. Simplify before moving things. a(1-1)=0. Therefore you can’t divide it out on both sides. You’d get an inditerminate or something like that.

Let me fix that.

“4a(1-1)/a(1-1)=3a(1-1)/a(1-1)”=0

4a(1-1) = 4a (0) = 0
a(1-1) = a (0) = 0

0/0 is impossible.

You have 0/0 on the other side, too.

Impossible.

Again- why did you make this thread? There was another thread, and it was already proven. Why did you think you would even come close to solving something that would actually be very revolutionary to mathematicians when you constantly come to us with problems that a middle schooler shouldn’t have problems with? Do you think before you post threads?

God damnit.

While we’re on the topic of math, mind explaining why x^0=1?

Anything to the power of 0 = 1.

I know, but mind explaining it? 0^0 != 1.

There’s no really explaining it, cause it’s defined by people to just <b>be</b> like that. x^4 * x^2 = x^(2+4) = x^6, and x^4 * x^0 = x^(4+0) = x^4, so x^0 is, <b>by definition</b>, equal to 1.

0^0 is undefined because it is in conflict with two definitions: that x^0 = 1, and that 0^x = 0.

<b>Anything to the power of 0 = 1.

.

I always thought of the pyramid triangle thingy.

x can be 10

10^5 = 1010101010
10^4 = 10101010
10^3 = 10
1010
10^2 = 10
10
10^1 = 10
10^0 = 10/10
10^-1 = 10/10/10
10^-2 = 10/10/10/10
10^-3 = 10/10/10/10/10

0^0 is usually left undefined. Much like 0/0

Although you could go ahead and say that 0^0 = 1, I don’t think it would do any harm. But it can also be argued that 0^0 = 0. So it’s usually left undefined because there’s no great reason to favor one or the other, and defining it either way doesn’t help solve any problems.

On the other hand, 0/0 is undefined because there is no reasonable value it could be. It can’t be 1 because then all numbers could be proven equal.

Well, the functions x^0 and 0^x converge to different values in the limit as x approaches zero.

I recall some formula I once found that tried to show why 0^0 !=1, but I don’t remember what it was. But I do remember my teacher saying something like 0 is not really 0, it really means something really small. So if you have something really small to something really small you can’t determine it’s answer. So it’s an inditerminate.

Actually, according to my graphing calculator:
lim <sub>x->0</sub><sup>+</sup> (0<sup>x</sup>) = 1
lim <sub>x->0</sub> (x<sup>0</sup>) = 1

However, a limit from the left does not exist for 0<sup>x</sup>, but they do both approach 1, even if 0<sup>x</sup> is not continuous.