1+1= two/to/too
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Fair enough, however I am OP: OVER POWERED!
The answer is no. The OP claims that what is 1+1. However, that is not true. What is what and 1+1 is 1+1, but what is not 1+1. Therefore his statement is false and the answer is no.
cornbeefhashvili wrote:
1+1=0
x = y
Then x2 = xy
Subtract the same thing from both sides:
x2 - y2 = xy - y2
Dividing by (x-y), obtain x + y = y
Since x = y, we see that 2 y = y
Thus 2 = 1, since we started with y nonzero
Subtracting 1 from both sides,
1 = 0
There are a lot of errors in this logic.
First of all, you divide by (x - y). Since x = y, I can apply the substitution property, so you are dividing by (x - x). You are dividing by zero! This is not allowed.
Second, you say near the end of your logic "since we started with y nonzero." When did you say that? That was not a part of the proof. You can't assume something partway through.
Finally, you didn't even finish the proof! You ended at 1 = 0. I know it's trivial to deduce from 1 = 0 that 1 + 1 = 0, but it's incomplete nonetheless.
Also, your use of a different font family makes me think you copied and pasted this proof from somewhere else. I googled "1+1=0 proof" and guess what the top result is? In Times New Roman font? 
"What is"
cmizgerd17 wrote:
cornbeefhashvili wrote:
1+1=0
x = y
Then x2 = xy
Subtract the same thing from both sides:
x2 - y2 = xy - y2
Dividing by (x-y), obtain x + y = y
Since x = y, we see that 2 y = y
Thus 2 = 1, since we started with y nonzero
Subtracting 1 from both sides,
1 = 0
There are a lot of errors in this logic.
First of all, you divide by (x - y). Since x = y, I can apply the substitution property, so you are dividing by (x - x). You are dividing by zero! This is not allowed.
Second, you say near the end of your logic "since we started with y nonzero." When did you say that? That was not a part of the proof. You can't assume something partway through.
Finally, you didn't even finish the proof! You ended at 1 = 0. I know it's trivial to deduce from 1 = 0 that 1 + 1 = 0, but it's incomplete nonetheless.
Also, your use of a different font family makes me think you copied and pasted this proof from somewhere else. I googled "1+1=0 proof" and guess what the top result is? In Times New Roman font?
Hey, I took a shot.
addition
And you missed
Good attempt though!
Apr 19, 2014
0
#90
But the statement is "What is 1+1?" not "What is 1+1."
It is a stretch to interpret this as a declaration about "What" by any standard use of the English language.
My answer is "troll".
We got it the first time
What is 1+1? Well since this is a word problem, is translates to an equal sign. So therefore, with what being our unknown, we can formulate the equation that x=1+1. Subtract 1 from both sides to get 1=x-1. Therefore, we substitute each 1 for x-1 to get x=(x-1)+(x-1), or x=x-1+x-1. From there, combine like terms to get x=2x-2. Plug in that value for x in the original equation of x=1+1 to get 2x-2=1+1. Add two to both sides and combine like terms to get 2x=4. Divide both sides by 2 to get x=2. Now from there, we can derive that 1+1=2.
However, Cornbeefofhashvili said that 1+1=0 and x=y so therefore 2x=xy. He's wrong there. It's actually x^2=xy since y=x and you're multiplying by y on both sides. From there, supposed that x=2. Then, 4=4y. 4x=x^2. Subtract 4x from both sides to get the quadratic x^2-4x=0. Then, factor out x to get x(x-4)=0. The possible values for x are 0 and 4. This works with any perfect square because you're squaring x on one side always. You will always get the same equation. Thus, x=0 or any perfect square. Therefore, 1+1=0 or any perfect square when we plug in x to the original equation.
Apr 19, 2014
0
#95
What is "1" and "+"? Are we wrong to assume this is addition within a ring? If this is just an abelian group, and "1" is the identity, then this is confusing notation. Please clarify.
The_Chess_Agent wrote:
... supposed that x=2. Then, 4=4y.
This is incorrect. If x = 2 and y = x, then 4y = 8.
The_Chess_Agent wrote:
... x=y so therefore 2x=xy.
Actually, what he meant was x^2 = xy. I'll admit it was a little unclear as he wrote it x2, but I'm pretty sure that's what he meant.
A duo
1+1=0
x = y
Then x2 = xy
Subtract the same thing from both sides:
x2 - y2 = xy - y2
Dividing by (x-y), obtain x + y = y
Since x = y, we see that 2 y = y
Thus 2 = 1, since we started with y nonzero
Subtracting 1 from both sides,
1 = 0
That means nothing to me.
My eyes glazed over on the second line
You are not the OP, so it doesn't matter to me what it means to you.