Prove 1 + 1 ≠ 2

underrated post

I thought by Z_2 you meant the ring of integers mod 2

...

That's a pretty mean meme, user. You hurt all my feels.

x = 1, y = 1
x=y
(x+y) = 0
1+1=0, 2=0
1+1/2=0/2
1=0 (add one both sides)
1+1=1

obv don't /0

...

You forgot to add a hypothetical 0001

2 = 1 + 1

2 ≠ 2 × 0 + 1 + 1

[eqn]
2^2 - 2^2 = 2^2 - 2^2 \\
2(2-2) = (2+2)(2-2) \\
2 = 2+2 \\
1 = 1+1\\
1 = 2
[/eqn]
Q.E.D.

easy. start with
7=7
then you divide by 0 and get
1+1 ≠ 2

I don't know where you guys went to school but the answer is pretty obvious, its 11

ghana baffle

Only in base 1, pleb

1 = 0.99e, .99e + .99e = 1.888e

prove me wrong

[math]\sum\limits_{n \in \mathbb{N}} n = 1 + \sum\limits_{n \ge 2} n = -\frac{1}{12} \Leftrightarrow 1 = -\frac{1}{12} -\sum\limits_{n \ge 2} n \Rightarrow 1+1=-\frac{1}{6} -\sum\limits_{n \ge 2} 2n[/math], and [math]|\sum\limits{n \le 2} 2n| \le \frac{1}{12} < \frac{1}{6}[/math], so [math]1+1

[math]\sum\limits_{n \ge 2} 2n[/math]*

>(x+y)=0
No

This.

1+1=window

you are throwing a set of solutions away
by factorising like that

2*0=4*0
2=4
uw0t

ANALYTIC CONTINUATIONS AREN'T EQUIVALENT TO SOLUTIONS REEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

2 = 6/3 = 3/3 + 3*1/3 = 1 + 3*0.333... = 1 + 0.999... != 1 + 1

cool but 1/3 != 0.333...

my bad,
2 = 6/3 = 3/3 + 3*1/3 != 1 + 3*0.333... !!= 1 + 0.999... !!!= 1 + 1

you could say 1/3 is roughly equal to 0.333... but not 1/3 = 0.333...

What if 5 = 7?

How would that work?

Prove 2+2=5

You do realize what kind of thread this is?

>1=1+1
>1=2

By transitivity of equality, you have proved that 1+1=2

>Thinking an analytic continuation correlates to a direct solution

1+1=x/0~2

Fine, I'll give a less elegant proof. This proof is done by contradiction, assuming [math]1+1=2[/math].

Let [math]a \neq 0 \neq b[/math]. Suppose [math]a=b[/math]. Now [math](a+b)(a-b)=a^2-b^2=a^2-ab=a(a-b)[/math], so we have [math]a=a+b=2a[/math], and so, by assumption, [math]1=2[/math]. Now, [math]1+1=2 \Leftrightarrow 1+1=1 \Leftrightarrow 1=0[/math], but [math]2[/math] is a SUCCessor of a natural number unlike [math]0[/math], which is a contradiction.

1+1=10 in base 2, you did not specify a base

Non-mathfag here, I have some qualms with your proof.

I fail to see how you demonstrated that a = a + b = 2a, as the preceding algebra doesn't relate to this at all. No natural number n exists in which n = 2n, and this isn't taken into account in your proof.

nerds doing math haha

Divide by (a-b).

1/3 ! = 0.333... ! = Γ(0.333...) = 0.89297...

Dividing by a-b is equivalent to dividing by 0 according to the supposition of the post, so I'm starting to think that maybe it was a troll proof at this point.

>a troll proof in a thread where you are supposed to prove 1+1=/=2
Hmmm... could it be?

But in base 2 1+1=10, and if you're talking about Z_2 then it's elements are congruence classes mod 2, and the fact we use 0 and 1 and not 2 and 3 is just a matter of convention, 0,2,4,6,8,... and 1,3,5,7,... respectively lie in the same congruence classes

An oldie but a goodie:

0 = 0 + 0 + ...
= 1 - 1 + (1-1) + ...
= 1 - ( 1 - 1 + (1-1) + (1-1) + ... )
= 1 - 0 = 1

Then add 1 to both sides