## 1 + 1 = ?

For the discussion of math. Duh.

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bureau
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### 1 + 1 = ?

Sorry if this has been posted before. Is it true that it's possible for 1 + 1 to not equal 2?

auteur52
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### Re: 1 + 1 = ?

That depends on your definitions of "1", "2", "+", and "=".

Qoppa
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### Re: 1 + 1 = ?

In a field of characteristic 2 (such as Z/2), 1+1=0.

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_=0,w=-1,(*t)(int,int);a()??<char*p="[gd\~/d~/\\b\x7F\177l*~/~djal{x}h!\005h";(++w<033)?(putchar((*t)(w??(p:>,w?_:0XD)),a()):0;%>O(x,l)??<_='['/7;{return!(x%(_-11))?x??'l:x^(1+ ++l);}??>main(){t=&O;w=a();}

jazznaz
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### Re: 1 + 1 = ?

Under modulo arithmetic (mod 2) , 1 + 1 = 0...

I'm not great on math theory, but that may be equivilent to the above statement...

heyitsguay
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### Re: 1 + 1 = ?

1+1=100010101 base i, but that's not unique.

Chet
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### Re: 1 + 1 = ?

I think the correct answer is:
"Not in the usual sense!"

Cleverbeans
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### Re: 1 + 1 = ?

1+1 can equal absolutely anything if you're willing to change the meaning of "equals" based on context. It might seem artificial at first however different flavors of equality exists for the natural, rational, real and complex numbers for "normal" math. Here is a link to the wikipedia article on the subject.

http://en.wikipedia.org/wiki/Equivalence_relation
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Qoppa
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### Re: 1 + 1 = ?

jazznaz wrote:Under modulo arithmetic (mod 2) , 1 + 1 = 0...

I'm not great on math theory, but that may be equivilent to the above statement...
It is. As I pointed out, Z/2 (the integers mod 2) is a field of characteristic 2.

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_=0,w=-1,(*t)(int,int);a()??<char*p="[gd\~/d~/\\b\x7F\177l*~/~djal{x}h!\005h";(++w<033)?(putchar((*t)(w??(p:>,w?_:0XD)),a()):0;%>O(x,l)??<_='['/7;{return!(x%(_-11))?x??'l:x^(1+ ++l);}??>main(){t=&O;w=a();}

antonfire
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### Re: 1 + 1 = ?

It's worth noting that in all of these cases, you still have 1+1=2. It's just that it also happens to be true that, modulo 2, 2=0.

A case where you take 1+1 to be something other than 2 is when you redefine the addition operation as a +' b=max(a,b) and multiplication as a *' b=a+b.

Also, I'm pretty sure there's been a thread about exactly this question, but I can't find it.
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### Re: 1 + 1 = ?

Basically, if you see someone asserting that "1+1=2", there's a few possibilities

1) They've given you a several-line proof that seems to "prove it", especially if the first line is "a=b". Chances are they divided by zero somewhere in there but hid it by dividing by "a-b", or something similar.

2) They know some high-level mathematics and are redefining the terms "one" "plus" "equals" or "two".
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chapel
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### Re: 1 + 1 = ?

2 + 2 = 5 for large values of 2, as the old joke goes.

keeperofdakeys
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### Re: 1 + 1 = ?

how about 1 + 1 = 10

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### Re: 1 + 1 = ?

keeperofdakeys wrote:how about 1 + 1 = 10

That's just a different representation of 2.

It's all a matter of definitions, which is to say, semantics. For any reasonable definition of 2, 2=1+1, if it even makes sense to define 2 in your field ring group semigroup magma.
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### Re: 1 + 1 = ?

jazznaz wrote:Under modulo arithmetic (mod 2) , 1 + 1 = 0...

I'm not great on math theory, but that may be equivilent to the above statement...

It doesn't change the equivalence though:
1 + 1 = 0 mod 2
2 = 0 mod 2
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Fimbulfamb
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### Re: 1 + 1 = ?

I believe 1+1 was proven to be 2 around 1900 by Bertrand Russel.

Cpt. Red
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### Re: 1 + 1 = ?

Charles Seife wrote:Dividing by zero...allows you to prove, mathematically, anything in the universe. You can prove that 1+1=42, and from there you can prove that J. Edgar Hoover is a space alien, that William Shakespeare came from Uzbekistan, or even that the sky is polka-dotted. (See appendix A for a proof that Winston Churchill was a carrot.)

Cass
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### Re: 1 + 1 = ?

keeperofdakeys wrote:how about 1 + 1 = 10

Thats correct, thought only in base 2.

Correct me if im wronge but if you defined a vector space of all real numbers and defined your axioms could you not have 1+1 = anything you want. as long as 1+(-1) = 0 ?
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Qoppa
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### Re: 1 + 1 = ?

Well no. Vector addition has to be well defined, which means that adding two vectors should produce the same result each time.

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_=0,w=-1,(*t)(int,int);a()??<char*p="[gd\~/d~/\\b\x7F\177l*~/~djal{x}h!\005h";(++w<033)?(putchar((*t)(w??(p:>,w?_:0XD)),a()):0;%>O(x,l)??<_='['/7;{return!(x%(_-11))?x??'l:x^(1+ ++l);}??>main(){t=&O;w=a();}

shinjak
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### Re: 1 + 1 = ?

If the additive structure of your vector space is going to have very little to do with the usual additive structure on the reals, then you might as well ditch the underlying set and call the elements of your vector space something altogether different.

A_of_s_t
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### Re: 1 + 1 = ?

Fimbulfamb wrote:I believe 1+1 was proven to be 2 around 1900 by Bertrand Russel.

It was. I have a picture of his original work. Its a halarious paper long proof.
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### Re: 1 + 1 = ?

A_of_s_t wrote:
Fimbulfamb wrote:I believe 1+1 was proven to be 2 around 1900 by Bertrand Russel.

It was. I have a picture of his original work. Its a halarious paper long proof.

He also spent a few hundred pages just laying the groundwork for it.
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chapel
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### Re: 1 + 1 = ?

It can get proved assuming the Peano Postulates. The real quick and dirty of the proof is that you define addition so that a+b = (a+c)' when b = c' where x' is the successor to x (and so is an element of the naturals so long as x is), and then let 1' = 2. So,a = b = 1 implies 1 + 1 = 1' = 2. There are some key details missing, but that sketch should be good enough that someone with a list of the Peano Postulates should be able to make a real proof.

By the way, google Principia Mathematica and 1 + 1 = 2 (I went looking for the original work). Someone asked for a proof that 1 + 1 = 2 on Yahoo questions and at least a dozen people said, "If you have 1 something and someone gives you another something, then you have two somethings," before calling the guy an idiot for not being able to come up with their obviously correct proof.

PM 2Ring
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### Re: 1 + 1 = ?

Sir_Elderberry wrote:
A_of_s_t wrote:
Fimbulfamb wrote:I believe 1+1 was proven to be 2 around 1900 by Bertrand Russel.

It was. I have a picture of his original work. Its a halarious paper long proof.

He also spent a few hundred pages just laying the groundwork for it.

Sort of. Russell & Whitehead were attempting to formalize mathematics, from the ground up, using symbolic logic (and basic set theory). Their plan was flawed, but they didn't know about Godel's theorems*. Their proof for 1+1=2 isn't really that long, but they did have to formalize a fair amount of more fundamental material before they got to 1+1=2. So the 1+1=2 result was just an incidental result they derived for fun. The fact that it occurs after a few hundred pages illustrates that this "basic mathematical fact" isn't as basic as the lay person may think.

Disclaimer: I haven't read Russell & Whitehead's Principia Mathematica. And I don't intend to in a hurry. Reading page after page of symbolic logic makes my brain turn to jelly.

[*] But if it wasn't for Principia Mathematica, Godel may have not discovered his theorems. Although Godel wasn't the only one looking into these issues: there were also people like Church, Tarski & Turing looking at the limitations of formal mathematical procedures.

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### Re: 1 + 1 = ?

Well, yes, I know that he wasn't out to just prove 1+1=2. But nonetheless, the proof of that proposition was only possible after a volume or two of logic and formalization. Pretty hardcore stuff.

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### Re: 1 + 1 = ?

PM 2Ring wrote:So the 1+1=2 result was just an incidental result they derived for fun. The fact that it occurs after a few hundred pages illustrates that this "basic mathematical fact" isn't as basic as the lay person may think.

Well, what they really proved was the fact that logic and a few basic axioms of set theory result in a structure that contains a set corresponding to how everyone had already been treating integers for thousands of years.

The axiomatization and formalization of mathematics wasn't so much an effort to prove facts everyone already knew, as a way to provide firm logical groundwork consistent with those facts, from which to prove other things that people didn't already know, but which fit with what they did.
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codyhotel
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### Re: 1 + 1 = ?

I'm amazed that no one has put this, maybe out of fear, but i'll risk it.

Obviously,

1+1=1.99999...8
from a previously (over)discussed theorem.
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auteur52
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### Re: 1 + 1 = ?

codyhotel wrote:I'm amazed that no one has put this, maybe out of fear, but i'll risk it.

Obviously,

1+1=1.99999...8
from a previously (over)discussed theorem.

No. 2=1.999....

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### Re: 1 + 1 = ?

auteur52 wrote:
codyhotel wrote:I'm amazed that no one has put this, maybe out of fear, but i'll risk it.

Obviously,

1+1=1.99999...8
from a previously (over)discussed theorem.

No. 2=1.999....

Considering that 1.9999....8 doesn't make any sense in any notation I'm aware of, yes.
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Deep Fried Pickles
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### Re: 1 + 1 = ?

I have a friend who can use circular logic to prove that 1 + 1 is 0

Freiberg
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### Re: 1 + 1 = ?

Deep Fried Pickles wrote:I have a friend who can use circular logic to prove that 1 + 1 is 0

Off topic, but he actually proved that 1 + 1 = 0/0

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### Re: 1 + 1 = ?

(x-x)(x+x) = x²-x²
(x-x)(x+x) = x*(x-x)
x+x = x
2x = x
2 = 1
1+1 = 1, 2, 3, or 4.

1+1 is clearly ambiguous.

spot the error haha

another one:

(-1) * (-1)= 1

log (-1 * -1) = log 1
log -1 + log -1 = log 1 = 0
log -1 = - (log -1)
(log -1)/(log -1) = -1
1 = -1

1+1 = 0, +-2

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### Re: 1 + 1 = ?

Divisions by zero.
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### Re: 1 + 1 = ?

log (-1 * -1) = log 1
log -1 + log -1 = log 1 = 0
log -1 = - (log -1)
(log -1)/(log -1) = -1
1 = -1

1+1 = 0, +-2

I like this one. I haven't seen it before.

The trick is it is NOT true that log xy = log x + log y for all real x and y. Namely, log -1 = pi * i, so log (-1)(-1) = 0 /= pi * i + pi * i. The identity does seem to hold when working in mod 2-pi arithmetic.

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### Re: 1 + 1 = ?

Tac-Tics wrote:The trick is it is NOT true that log xy = log x + log y for all real x and y.

Sure it is. It's well known that log 1 = 2pi*i, among other things. There's nothing wrong with the rule that log xy = log x + log y, so long as you don't assume that the result is the only value log xy equals.

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### Re: 1 + 1 = ?

Buttons wrote:
Tac-Tics wrote:The trick is it is NOT true that log xy = log x + log y for all real x and y.

Sure it is. It's well known that log 1 = 2pi*i, among other things. There's nothing wrong with the rule that log xy = log x + log y, so long as you don't assume that the result is the only value log xy equals.
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### Re: 1 + 1 = ?

Macbi wrote:
Buttons wrote:Sure it is. It's well known that log 1 = 2pi*i, among other things. There's nothing wrong with the rule that log xy = log x + log y, so long as you don't assume that the result is the only value log xy equals.
If you don't treat your functions like functions the Mathematical Fairy stabs you to death with set squares.

Thank you Macbi!

This is the math forum, not the physics forum. Functions are single-valued. Always. If you need to choose bewteen two alternative "branches", then pick on and be done with it!

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### Re: 1 + 1 = ?

You'd rather throw out the rule log xy = log x + log y than accept that ex is not injective? Inverses of noninjective functions are multivalued. There's nothing physics-y about that.

I'm all for narrowing the range in a given situation, of course. But I would hardly say that log x = 2pi*i "has no solutions" in general.

Tac-Tics
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### Re: 1 + 1 = ?

Buttons wrote:You'd rather throw out the rule log xy = log x + log y than accept that ex is not injective? Inverses of noninjective functions are multivalued. There's nothing physics-y about that.

I'm just saying that a "multivalued" function isn't a function. It's a relation or something else, but I'd rather give up the log xy = log x + log y rule than screw around with multivalued functions. Both result in the same answer, but one has a strong formal foundation and the other is just lazy mathematicians playing loose-and-fast with notation.

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### Re: 1 + 1 = ?

[imath]e^z[/imath] is a continuous homomorphism from the additive group [imath]\mathbb{C}[/imath] to the multiplicative group [imath]\mathbb{C}^{*}[/imath] with kernel the integer multiples of [imath]2 \pi i[/imath], hence restricts to an isomorphism
$\mathbb{C}/2 \pi i \mathbb{Z} \to \mathbb{C}^{*}.$
There's nothing mysterious about this. The preimage of any element of [imath]\mathbb{C}^{*}[/imath] is a coset of the kernel like it always is, and this is perfectly rigorous; mathematicians play around with preimages all the time.

FatPhil
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### Re: 1 + 1 = ?

Tac-Tics wrote:
Macbi wrote:
Buttons wrote:Sure it is. It's well known that log 1 = 2pi*i, among other things. There's nothing wrong with the rule that log xy = log x + log y, so long as you don't assume that the result is the only value log xy equals.
If you don't treat your functions like functions the Mathematical Fairy stabs you to death with set squares.

Thank you Macbi!

This is the math forum, not the physics forum. Functions are single-valued. Always. If you need to choose bewteen two alternative "branches", then pick on and be done with it!

Functions may be single-valued, but it's perfectly valid for the codomain to be a set of equivalence classes, rather than a single value. Of course, you've then got to use the members of those sets appropriately, but if you're prepared to do the pedantry to start the process, you've probably got the skills to pull it off.