I am a busser and a host right now. One job plus one job equals two jobs. But I am One. Therefore 1 and 1 make 1

Furthermore, Thucydides is also One, ergo we are Three but One. And semck is also One. All Are One. One Ande One Ise The Infinite Sete

actually you are part time busser and part time host, together you are 1.

but yeah, my 2+2=5 thing is legitimate logic

Mathematical symbols have no inherent meaning. The only meaning they have is what is given to them. It is most useful if everyone agrees on the meaning of a symbol, such as +. You can redefine a symbol if you wish, but doing so doesn't make you clever or the resulting equation insightful.

@semck

I say "crypto-fascist" is the next webDip meme.

is that anything like a cryptologist?

Not sure, ask Strauss.

a=b
a^2=ab
a^2-b^2=ab-b^2
(a+b)(a-b)=b(a-b)
a+b=b
2b=b
2=1
1+1 = 2
so 1+1=1

inb4 divide by zer0

shush you

@smeck, yes; but there is a possible solution.
And given that the op has a different native language (which i've only found out later) it would b reasonable to assume this is a translation issue - you don't like my translation?

Also it feels rather wrong,
but you can add vectors in a similar fashion, and when i can't put a little arrow over the 1⃗ or underline it or bold it... i might use bad notation to represent a magnitude sum.

1⃗ + 1⃗ = 1⃗? Oh wow, unicode is awesome!

i think i like damian's explaination the best. the concept of one-ness is rather simple, and it is clear that one ball of silly putty +one ball of silly putty make one ball of silly putty...

a⃗ + b⃗ = c⃗

'a + b = b'
'2b = b'

or

'a + b = b'
(subtract b from both sides)
a = 0

orathaic,

That would only be true of a vector if the vector was zero.

@semck, what if instead i'm talking about |a⃗| + |b⃗| = 1

where |b⃗| =1 and |a⃗| = 1 ? (this is the same magnitude trick i did with sets, excpet the Union operator combines with magnitudes differently.)

1 (unit) + 1 (unit) = 1 (binary exponent)

orathaic,

If you're talking about |a⃗| + |b⃗| = 1, and |b⃗| =1 and |a⃗| = 1, then your equation is incorrect. If |b⃗| =1 and |a⃗| = 1, then |a⃗| + |b⃗| = 2.

That's true, but |(a⃗ + b⃗ )| can = 1
And this has the same issue as my set cardinality.

Orathaic,

Yes, that's true. Unfortunately enough, in the equations

|b⃗| =1 and |a⃗| = 1 and |(a⃗ + b⃗ )| = 1, the placement of the norm symbols is vital, and in no accepted notation could the three equations be written in the notation of the OP. So, it's not an example of OP's claim.

Well seen as the op seems to have lost interest... I doesn't really matter;

Still you have to redefine the addition of vector to get the right equation; but in principle you could very well define a system where the opening equation is true.

True. But then, you could also define a system where "1 + 1" means "cheese," where "=" means "tastes", and where "1" (by itself) means "good." Then the original statement would also be true.

1 man + 1 woman = 1 argument about
1 man + 1 broken lawnmower = 1 accident waiting to happen
1 good guy + 1 great story teller = 1 religion

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