/sci/ - Science & Math

>>11218624
11

It depends on the base you are using

In base 2 the answer is 10
In base 3 or above the answer is 2

>>11218624
Principa Mathematica contains a solution to this problem.

>>11218624
you are a computer scientist, aren't ya?

1.

>>11218624
Use the axiom of choice

>>11218624
I says right there, it equals ?.

>>11218624
i kek'd and even better are the replies.

1+1=
99999/99999+88888/88888=
0.999999999999+0.99999999999=
1.9999999999998=
4/2=
3.99999/1.99999=
(3+.999999)/(1+.999999)=
3/1+.999999/.999999=
3+1=
4

>>11218627
>t. JavaScript runtime

>>11218624
1+1 = 1 on a bun

As any undergraduate category theorist knows, this problem is merely a special case of the more general construction of the canonical coproduct
[eqn]\begin{matrix}
& & ? & & \\
& ^{l} \nearrow & ^{m} \Uparrow & \nwarrow ^{r} & \\
1 & \overset{i_L}{\rightarrow} & 1+1 & \overset{i_R}{\leftarrow} & 1
\end{matrix}[/eqn]
where [math]i_L[/math] and [math]i_R[/math] are the universal morphisms such that for any arbitrary object [math]?[/math] and morphisms [math]l,r[/math] from [math]1[/math] to [math]?[/math], [math]m[/math] is the unique arrow such that the entire diagram commutes.

The computation is left as an exercise to the reader, given that exercise is beneath the undergraduate category theorist who has ascended to a bird's-eye view of mathematics in its entirety.

>>11218624
1 + 1 = 0.999.. + 0.999.. = 0.999....9918

>>11220063
based

>>11220063
>As any undergraduate category theorist kno-
Lost hard

>>11220063
holy shit my sides are in orbit

>>11220063
lmao

>>11218624
assuming you are using + as logical OR and 1 as logical true, the answer is obviously 1

>>11220063
kek

>>11218627
>>11218639
the only correct answer is S(S(0))

Let [math]d \in \mathbb{N}, \phi \in D(\mathbb{R}^d)[/math] be a test function. Set [math] T(\phi) = T_1[/math]. Then [math]T /in D'(\mathbb{R}^d)[/math] because 1 is continuous, so it is in [math]L_{loc}^1[/math].
Now we compute:
[math] (T+T)(\phi) = T(\phi) + T(\phi) = \int_{\mathbb{R}^d} 1\phi(x)dm^n(x) + \int_{\mathbb{R}^d} 1\phi(x) dm^n(x)[/math].
|1| = 1 is continuous and [math] \phi \in C_c^\infty(\mathbb{R}^d[/math] so the integral exists and we can use linearity:
[math] = \int_{\mathbb{R}^d} (1 + 1)\phi(x)dm^n(x) = \int_{\mathbb{R}^d} 2 \phi(x)dm^n(x) = T_2(\phi)[/math].
Therefore the answer, in a distributionsl sense, is [math] T_2[/math] which we shall identify with [math]2[/math]

>>11218624
here is a video for you. it explains how this works really well

https://www.youtube.com/watch?v=Zh3Yz3PiXZw

Why are you all retarded, it's pretty clear that 1 + 1 = 1 + 1.

>>11220833
tautology for the win
now i have something else for you: the i in iphone stands for iphone