/sci/ - Science & Math

[Anonymous No.8408820]

PROVE that 1+1=2. Actually PROVE it.

[Anonymous No.8408822]

>>8408820
But it doesn't.
You ask the impossible.

[Anonymous No.8408832]

>>8408822
you get it

[Anonymous No.8408833]

>>8408820
1=1
11=11
1+1=11
x+1=11 iff x=1

[Anonymous No.8408845]

Set of natural numbers is the semigroup. Binary operation is defined at this set. 1 + 1 = 2 as defined. QED

[Anonymous No.8408847]

>>8408820
I have discovered a truly marvelous proof this, which this chinese cartoon board is too small to contain.

[Anonymous No.8408854]

>>8408820
>He hasn't taken group theory

[Anonymous No.8408856]

>>8408820
Lets refer to the sucessor function as S(x). In the We usually define addition in the following way:

x + 0 = x
x + S(y) = S(x + y)

We can use this to compute at the most elementary level, the value of 1+1 as follows:

1 + 1 = 1 + S(0)
Given that 1 is defined to be the sucessor of 0 (at least in a system of naturals were you include 0)

Note that 2 is defined as the sucessor of 1, 3 of 2, 4 of 3, etc. These are symbolic definitions. We agree that 2 = S(S(0)) and so on. So we have

1 + S(0) = S(1 + 0) by the second axiom of addition. Then

S(1+0) = S(1) by the first axiom of addition.

Just to be clear, I will compute another sum.

1 + 2 = 1 + S(1)
= S(1+1)
= S(1+S(0))
= S(S(1+0))
= S(S(1))

etc.

[Anonymous No.8409492]

>>8408820
1+1 := 2
therefore 1+1 = 2
QED

[Anonymous No.8409503]

>>8408820
Okay, so if OP is sucking on 1 dick, and 1 more dick comes along, if you count the dicks OP is sucking you'll discover that OP is sucking on 2 dicks. I learned math in kindergarten the hard way.

[Anonymous No.8409602]

>>8409503
Can you really suck on two dicks at once? OP could put two in his mouth, but how does he close his mouth over both of them?
This suggests 1+1=0
OP please show us that video you referenced of you sucking two cocks at once so we can resolve this

[Anonymous No.8409604]

>>8408856
winrar
peano/10

[Anonymous No.8409629]

>>8408820
half of my calculus lectures is bullshit like this already. or proof that x*0=0*x=0
just kill me

[Anonymous No.8409665]

We have a+a=b

let a=1 and b=2 QED

[Anonymous No.8409669]

>>8408820
Defining 1 and 2 as Neumann ordinals

1 := {{}}
2 := {{},{{}}}

For all natural numbers x, x + 1 = x u (x}

1 + 1 = {{}} u {{{}}}
1 + 1 = {{},{{}}}

Therefore 1 + 1 = 2

qed

[Anonymous No.8409673]

>>8409629

proving stuff in lower level math, like really close to the definitions is just using said definitions.

like x*0 = 0 because identity, similarly x*0 = 0*x because commutativity(I think is what it's called) so 0*x = 0.

You really don't have to think, since you are playing really close to the definitions just use it.

[Anonymous No.8409676]

>>8409673

cont'd

uhh I don't think it's called identity, but x*0 = 0 is an axiom for sure. I forgot the name.

[Anonymous No.8409692]

>>8408820
Why? The "fact" that * | * should be * * when we remove the |-bar
is almost axiomatic in nature. Self-evidently true because of human perception...

Basic semantics (i.e. meaning of words/symbols) also come into play.

We simply call * for '1', * * for '2' and so on. If you let the act
of removing the |-bar to represent mathematical addition it should be obvious why 1+1 = 2. No need to prove it.

[Anonymous No.8409696]

>>8409676
[math]x \cdot 0 = 0 [/math] is typically not taken as an axiom as far as I'm aware.

[eqn]x \cdot 0 = x \cdot (0 + 0)[/eqn]
(0 is the neutral element of addition, so [math]0 = 0 + 0[/math].
[eqn]x \cdot 0 = x \cdot 0 + x \cdot 0[/eqn]
Multiplication is distributive over addition.

Now just add [math]-(x \cdot 0)[/math] to both sides.
[eqn]0 = x \cdot 0[/eqn]
QED

[Anonymous No.8409708]

>>8408820
>1+1=2
'=' is assignment, you fickle flailing boxshit
'==' is comparative equivalence
GET /grammars
EAT /shit

[Anonymous No.8409861]

>>8408847
underrated post

[Anonymous No.8409877]

>>8408820
Take one apple and then another one how many apple you have now xD? Not sure if op is troll but this is basic physics

[Anonymous No.8409898]

>>8409708
Get fucked programmer.

Assignment in mathematics is given by
:=
or
[math]\triangleq[/math]

While = is an equivalence relation.

[Anonymous No.8409914]

>>8408820
Well...
You have one Apple (1).
You buy an Apple (1).
You now have one Apple and the other Apple you bought. You now have two (2) Apples.

[Anonymous No.8409934]

>>8408847
kek

[Anonymous No.8409941]

>>8408847

[Anonymous No.8409944]

>>8408833
>chemistry guy is talking about proteins
>picture is obviously an ellagitannin
>not a single amino acid in sight
Sasuga /sci/

[Anonymous No.8410000]

>>8408833
>the complex zeta function is the same as the real zeta function

[Anonymous No.8410004]

>>8408820
Add -1 to both sides by the additive property of equality. 1=1 Q.E.D

[Anonymous No.8410009]

>>8410000
I got quads, nice.

[Anonymous No.8410189]

>>8408856
Good job but you forgot to prove recursion with two variables. ;^)

[Anonymous No.8410197]

>>8409696
>>8409696
>x⋅0=0x⋅0=0 is typically not taken as an axiom as far as I'm aware.

Actually it is if you want to define multiplication using recursion. First-order theories of N may assume multiplication is distributive and cancellative but usually one proves these properties using recursion.

[Anonymous No.8410206]

This is grade 1 shit. You have 1 Apple, your friend Billy gives you 1 Apple, holy shit now you have 2 Apples!

You mathfags can be really dumb sometimes. Stop being so far up your own asses and use some common sense.

[Anonymous No.8410230]

>>8409492
Wrong

[Anonymous No.8410256]

>>8409914
>>8410206
Not a proof. That's a speculative theory based on limited empirical evidence.

Does it hold true for oranges? Or non-fruit?
How can you be sure?

[Anonymous No.8410264]

Let us freely generate the set Q from a base set B = {0}, closed under the successor operation S(x) = x + 1. S is the set of natural numbers.

Define the following sentence in a first-order language which uses only (a) the universal quantifier, (b) the binary equivalence predicate, and (c) the functionally complete set of sentential connectives {NOT, , AND, IFF}, and (e) a single constant c, and (f) a single one-place function S.

FOR ALL a ( = c a IFF = S(c) S(a) )

We then define a structure M whose domain is the set of all natural numbers, Q, and the equivalence predicate is the identity equivalence over the natural numbers, the constant c is 1, and the function S is the successor function. We trivially have that M satisfies our given sentence. QED.