Help me with my homework.

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1 + 1 = 1 + 1

It's trivial.

Google gives me 2.

what is "1", how do we define "1"? some people might consider the value represented by "1" to be something other than 1

remember to show your work

10

>>9080959
Off by one.

>>9080861
Well first, we need to define 1, along with the other natural numbers. To do so, we first define the [math]\textit{successor}[/math] function, [math] S [/math]: [eqn] S(n)=n\cup\{n\} [/eqn] where [math] n [/math] is any set. Now, call a set [math]A~\textit{inductive}[/math] if the following conditions hold: [eqn] \varnothing\in A \\ \forall n\in A,~S(n)\in A [/eqn] i.e. [math] A [/math] is closed under the successor function. Now we can define [math]\mathbb{N}[/math]:[eqn] \mathbb{N}=\bigcap\limits_{A~\text{inductive}} A [/eqn] We can see that [math] \mathbb{N} [/math] consists exactly of the elements [eqn] \varnothing,~S(\varnothing),~S(S(\varnothing)),~\dots[/eqn] and we rename each by the number of times the successor function is taken, so that [eqn]
\varnothing=0,~S(\varnothing)=1,~S(S(\varnothing))=2,~\dots[/eqn] Now we can begin defining addition on [math]\mathbb{N}[/math]. For each [math]n\in\mathbb{N}[/math], we define [eqn] A_n(0)=n \\ A_n(S(m))=S(A_n(m))~\forall m\in\mathbb{N} [/eqn] Finally, let [math]\textit{addition}[/math] be the binary operation given by [eqn] +:=\{((m,\,n),\,k):m,n\in\mathbb{N},\,A_m(n)=k\} [/eqn]Therefore, [math]1+1=A_1(1)=A_1(S(0))=S(A_1(0))=S(1)=2[/math]. Q.E.D.

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>>9080861
window...

>>9080861
1+1
=1 1
=11
t. world class mathematics professor

We know that

[math]
e^x = 1 + x + \dots
[/math]

Thus
[math]
e \approx 1 + 1
[/math]
[math]
1+1 \approx 2.7 \approx 3
[/math]

There you go
[math]
1+1 = 3
[/math]

>>9080861
1 + 1
= .999999999999[...] + .9999999999999[...]
= 1.888888888888888888888[...]

It's just science