Prove me that 1+1 is not 2
d4nk m4tt3r
>>7900162
1+1= 11 cuz when u put em together
i think he means the actual people who say that 1 is not really 1
because + is just another way to write x
1+1 = 0 in [math]\mathbb{Z}_2 .[/math]
I wonder what alien math would be like. Maybe they refuse to use anything but finite fields or they consider the p-adics the natural completion of the rationals.
>>7900181
i am just here to lurkx but it is getting interesting
>>7900181
Everytime i think about it, i think they would have something so complex we haven't achieved it
1 + 1 = 6/3
>>7900181
>they consider the p-adics the natural completion of the rationals.
Whoa
>>7900162
It can be 10 in binary
>>7900208
>1 + 1 = 6/3
funny
>>7900162
If you have a number like 0,9999999 you would simply put 1. So if you add 1+1 it could actually be 0,9999+ 0,9999 (doesn't need to be same length obviously) you would have 1,9998 which you could (and usually would) put as a 2, but isn't really 2
>>7900274
Well, so 1 doesn't always represent 1? Sometimes I can get 1 + 1 (0,9999 + 0,9998) and that makes it a little more far than 2. And that continues happening forever while you are adding fractionals as integers.
>>7900181
I immediately thought of that example. Not sure if OP wants to hear that
From the Wildberger theorem:
[eqn]\infty \,=\, n_{\max}[/eqn]
Multiply on both sides by 1 + 1:
[eqn](1 \,+\, 1)\,n_{\max} \,=\, (1 \,+\, 1)\,\infty \,=\, \infty[/eqn]
Finally, divide on both sides by [math]n_{\max}[/math]:
[eqn]1 \,+\, 1 \,=\, \frac{\infty}{n_{\max}} \,\neq\, 2[/eqn]
Q.E.D.
>>7900308
and thats why we dont fuck with infinities
1 + 1
= 0.999... + 0.999 ...
= 1.999...8
=/= 2
QED
>>7900308
beautyful
>>7900325
I came here to post this
In this set of vectors in R2, addition is defined as (x1, x2) + (y1,y2) = (x1+y1, 1).
In this set, 1+1=1.
>>7900162
1 man + 1 woman = 1 couple
>>7900384
i think you can prove it without vectors
If 0.999... = 1
Then 0.99..8 = 0.999... = 1
etc
0 = 1
1+1 = 0+0 = 0
>>7901293
Yes, that is the most common and simple way to say it. Very good
>>7901293
>...8
>...1
>>7901389
He didn't get the infinity of 0.99...8
>>7901400
Exactly
>>7900162
[eqn]1+1=x[/eqn]
[eqn]x-x=0+1-1[/eqn]
[eqn]1-(-x+x)=1-0-1[/eqn]
[eqn]-x-x=2-0[/eqn]
[eqn]-2x=2[/eqn]
[eqn]x=-1[/eqn]
Thus, [math]1+1=-1[/math]
It's not about the content, it's about the presentation.
>>7900181
all you've proven is that 2 = 0 in that particular ring
1 + 1 = sqrt(4) = -+2
>>7901902
Nope, sqrt is defined to always return positive values.
>>7901927
Then why did I fail the test because of that shit
1+1=-1/12
>>7900166
>this is literally true for unary
1+1 = 1
Boolean algebra
/thread
>>7900170
It isn't though. It's just the image of the base element of a specific additive monoid defined by the Peano axioms.
Also, in the quotient group Z/2Z, you can say 1+1=0.
>>7901610
Clearly you can't prove it for every ring.
Take Z for example.
If you wanted it proven for a certain set of structures, you should have specified.
>>7901934
Hello kiddo. How's high school?
>>7900162
Assume 1+1=2. Then
(1+1)^2=2^2=4
1^2+2^2+1^2=4
And hence 2^2<4. But we can represent 2 as S(S(0))=S^2(0) where S is Piano's success function, which gives
(S^2(0))^2=S^4(0)=5<4
This statement clearly implies that infinity occurs in some neighberhood between 5 and 4. There can be at most countably many of these neighborhoods because the reals are countable and dense. But by assumption there's only one neighborhood containing 1+1, precisely {2}, rather than countably many. Contradiction establishes the result.
>>7902898
Not him, but I do better at maths here than in high school
According to Google, 1 + 1 = 0b10
>>7902905
>Piano's axioms
Kek
>>7900162
nah man, can't even do 1+1 because 1 is unique
>>7900162
1+1=69
Get it?
