Does 1 + 1 always equals to 2?

All,

I dont have direct geometry related question but indirect one. My co-worker and I had a debate about 1 + 1 = 2. He says in geometry it is not always the case but I have always believed math is absolute subject and it cant be disputed, and only absolute subject so it was hard for me to digest. My co-worker mentioned that in geometry if we take pythagorean theorem it is actually 1 + 1 = 1.4.

So my question to you is, can 1 + 1 = anything but 2 ever? Is math an absolute subject or not? If it is absolute then how can one branch of math dispute another branch of math?

I very much apprecaite your help. I actually dont believe that saying 1 + 1 = 1.4 is a correct way of looking at pythagorean theorem but we couldn't come any conclusions. So I am really looking forward for correct answer I can change my belief but I need a satisfactory answer. Could you please help?

Re: Does 1 + 1 always equals to 2?

Quote:

Originally Posted by

**naik007** [...] if we take pythagorean theorem it is actually 1 + 1 = 1.4.

You're co-workers statement is erroneus. It's on the contrary

$\displaystyle 1 +1 \approx 1.4^2$

Re: Does 1 + 1 always equals to 2?

Your question simply doesn't make sense until you are more precise. If you mean "arithmetic in base 10", then, yes, 1+ 1= 2. But if you are talking, say, about arithmetic in base 2, 1+ 1= 10. As for saying "1+ 1 is not equal to 2 in geometry", again, it doesn't make sense without your friend saying what he is talking about. "Geometry" typically deals with figures not numbers, except in terms of formulas for area, volume, etc. in which we typically use "arithmetic in base 10". But it is impossible to confirm or deny your your friend's statement without knowing what your friend is talking about.

Re: Does 1 + 1 always equals to 2?

Re: Does 1 + 1 always equals to 2?

in general, it depends on what you mean by "1", "+" and "2". these symbols can have different meanings in different contexts.

in the "usual context", where 1 and 2 are taken to be "natural numbers" and + represents the usual notion of addition, then in fact, 1+1 and 2 are "the same number", by DEFINITION.

a somewhat "less trivial fact", is that: 2+2 = 4:

2+2 = 2+s(1) where s(1) just means: "the next number after 1" so s(n) = n+1.

addition is DEFINED by the rule:

a+s(b) = s(a+b), and a+0 = a (because 0 isn't "s of" anything. hey, we have to start SOMEWHERE).

in this case, a = 2, and b = 1.

thus 2 + s(1) = s(2+1). now to "evaluate" this, we need to know what "2+1" is. so we have to go again:

2+1 = 2+s(0) = s(2+0) = s(2) = 3 (because 3 = s(2) = s(s(1)) = s(s(s(0)), see? this just means that 3 = 1+1+1, not that surprising).

so now we know what 2+1 is, so we have that:

2+2 = 2+s(1) = s(2+1) = s(3) = 4 (again, 4 is just "short-hand" for 1+1+1+1).

i'm glossing over some technicalities, like the fact that (1+1)+1 = 1+(1+1) (which is really just saying that s(1+1) = 1+s(1), which is just our "rule" with a = b = 1).

the point being: IF we adopt certain "RULES" as true, then certain other things become CONSEQUENCES of those rules. but "other rules" are possible: we could, for example, define:

1+1 = 0 (in which case we don't get "very many numbers" by adding 1's, we only get 0 and 1). does this "makes sense"?

suppose "1" means "flip a switch", "+" means "and then..." and "0" means "don't flip a switch (don't do anything)".

then 1+1 = 0 simply means:

"flipping a switch twice, does the same thing as doing nothing", something you can readily verify with any common on/off switch.

Re: Does 1 + 1 always equals to 2?

Emakorav - Thank you! Those links were helpful.

Devano Understood - pretty much as long as the values of 1, + and = are what they are in arithmatic the answer will remain 2. I diddnt understand the 1 + 1 = 0 part, If I understood correctly, I assume you are using binary here with boolean operators as per your description of the swtich but I think here there is no plus, your example uses binary with boolean operators, and, nand.... so we cant really use plus to describe this. Also, flipping the switch twice is not same as 1 + 1. When you flip a switch twice you are doing boolean and and that is 1 and 0(notice I diddnt use plus) = 0. In boolean and operator, only 1 + 1 = 1 every other combinataion gives the result of 0. Makes sense? I am OK with Binary but my overall question is whether 1 + 0 can equal to anything else or not.

Re: Does 1 + 1 always equals to 2?

MathCrusader, Thank you sir. That is how I understod it and that is what I was saying but my co-worker is pretty intelligent man and he said 1 + 1 = 1.4, could you please confirm this with example or reference. Because square of 1 is 1 and hence taking the A^2 + b^2 = C^2 the C will be equal to 1.4 correct? I mean I understood it exactly as you mentioned but this is the argument that can come up hence I wanted to clarify.

I had looked this up and I agree with what you said.

Square root of 2 - Wikipedia, the free encyclopedia

Re: Does 1 + 1 always equals to 2?

Halls of Ivy,

Understood your confusion. As I mentioned, my friend is saying as per geometry 1 + 1 = 1.4 and I disagree. That is what my friend is saying. As per taking binary, I am very good at binary, We cant really say that in binary 1 + 1 = 10, I mean we can but we cant talk different number system and use it in context of base 10. What I mean is in binary it is 0001 + 0001 = 0010 which taking it in binary it is two. We cant read that in base 10. If you convert that to base 10 then our answer is same 1 + 1 = 2. Basically my question is whether in any math, any math(geometry, calculus, algebra, arithmatic), can you or can you not say if 1 + 1 = 2. Can this statement be disputed in any of the branches of math? Example my co-worder used was for phythagorean theorem where he said 1 + 1 = 1.4. Do you agree?

Re: Does 1 + 1 always equals to 2?

Quote:

Originally Posted by

**naik007** I dont have direct geometry related question but indirect one. My co-worker and I had a debate about 1 + 1 = 2. He says in geometry it is not always the case but I have always believed math is absolute subject and it cant be disputed, and only absolute subject so it was hard for me to digest. My co-worker mentioned that in geometry if we take pythagorean theorem it is actually 1 + 1 = 1.4.

Quote:

Originally Posted by

**naik007** MathCrusader, Thank you sir. That is how I understod it and that is what I was saying but my co-worker is pretty intelligent man and he said 1 + 1 = 1.4, could you please confirm this with example or reference. Because square of 1 is 1 and hence taking the A^2 + b^2 = C^2 the C will be equal to 1.4 correct? I mean I understood it exactly as you mentioned but this is the argument that can come up hence I wanted to clarify.

Your co-worker is clearly confused. If we have a isosceles right triangle with unit legs, then the hypotenuse is $\displaystyle \sqrt2$.

So by the pythagorean theorem $\displaystyle 1+1=1^2+1^2=(\sqrt2)^2=2$.

Re: Does 1 + 1 always equals to 2?

On the other hand, it could be that he was arguing, geometrically, that a vector of length 1, added to a vector of length 1, which is perpendicular to the first vector, has length $\displaystyle \sqrt{2}$. That would be a correct calculation but it would still not be a case of "$\displaystyle 1+ 1= \sqrt{2}$" because when you add vectors you are NOT adding their lengths.

Re: Does 1 + 1 always equals to 2?

Quote:

Originally Posted by

**naik007** Emakorav - Thank you! Those links were helpful.

Devano Understood - pretty much as long as the values of 1, + and = are what they are in arithmatic the answer will remain 2. I diddnt understand the 1 + 1 = 0 part, If I understood correctly, I assume you are using binary here with boolean operators as per your description of the swtich but I think here there is no plus, your example uses binary with boolean operators, and, nand.... so we cant really use plus to describe this. Also, flipping the switch twice is not same as 1 + 1. When you flip a switch twice you are doing boolean and and that is 1 and 0(notice I diddnt use plus) = 0. In boolean and operator, only 1 + 1 = 1 every other combinataion gives the result of 0. Makes sense? I am OK with Binary but my overall question is whether 1 + 0 can equal to anything else or not.

i was speaking of "xor", not "nand". in some forms of boolean algebra, "+" means "or", in which case 1+1 = 1 (and 1+0 = 1, and 0+1 = 1, and 0+0 = 0), and "and" is usually represented by $\displaystyle \cdot$.

but one doesn't need to refer to "binary" logical operations for a similar example:

consider ordinary integers, where all we are interested in is the PARITY (even-ness or odd-ness).

a little thought shows that:

even + even = even

even + odd = odd

odd + even = odd

odd + odd = even.

now let "0" stand for any even number, and let "1" stand for any odd number. the above rules then become:

0 + 0 = 0

1 + 0 = 1

0 + 1 = 1

1 + 1 = 0

so that 1 + 1 = 0, just says the the sum of any two odd integers is even.

this system is also called: the integers mod 2, and has a more precise definition which i won't bore you with.

*******

as others have pointed out, your co-worker has a mis-understanding of what the pythagorean theorem says. it is NOT:

A + B = C, but rather:

A^{2} + B^{2} = C^{2}, and indeed:

1 + 1 = 1^{2} + 1^{2} = (√2)^{2} = 2

(where A,B and C are the lengths of the sides of a right triangle, with C the longest side).

*******

the Peano axioms of arithmetic (which are generally recognized as encoding our "usual notions" of what "natural numbers" mean), force upon us 1+1 = 2, every time, no exceptions. it is a remarkable fact that people had been using arithmetic for centuries, but that a logical basis for them was not fully realized until fairly recently (Peano devised his axioms in 1889, based on a similar set of rules published by Richard Dedekind in 1888. Peano seems to have been unaware of a similar attempt to formalize arithmetic by Gottlob Frege in 1879).

the basic idea of natural numbers in Peano's system is fairly intuitive:

1) we start with 0.

2) we have an equivalence defined on natural numbers by equality (a is equivalent to b if and only if a and b are the same natural number).

3) every natural number possesses "a next natural number", called it's successor. this successor is unique, and no two different numbers have the same successor. thus this encodes the intuitive idea of "counting".

4) 0 is not the successor of any other number.

5) if something is true of 0, and true of every successor of any natural number, it is true for all natural numbers (although this appears to say nothing more than: "the natural numbers are precisely the numbers we can count to from 0", it actually says far more...it allows us to use the principle of induction to prove statements about natural numbers).

under these assumptions one can PROVE that: 1 + 1 = the successor of the successor of 0, more commonly known as 2 (and where 1 is also "the successor of 0"), given the following definition:

a + 0 = a

a + s(b) = s(a + b)

1 = s(0)

2 = s(s(0)).

of course, the cavemen probably knew this:

| and | is ||.

Re: Does 1 + 1 always equals to 2?

Quote:

Originally Posted by

**HallsofIvy** On the other hand, it could be that he was arguing, geometrically, that a vector of length 1, added to a vector of length 1, which is perpendicular to the first vector, has length $\displaystyle \sqrt{2}$. That would be a correct calculation but it would still not be a case of "$\displaystyle 1+ 1= \sqrt{2}$" because when you add vectors you are NOT adding their lengths.

HallsofIvy, Thank you! Yes, I meant to mention vector just after this. Thanks for confirming what I thought about vector too.

Re: Does 1 + 1 always equals to 2?

Devano,

Thank you! Understood about XOR operator. You are correct there. Thanks for confirming what I initialy thought.

Re: Does 1 + 1 always equals to 2?

Plato, Thank you! Much appreciated.

Re: Does 1 + 1 always equals to 2?

Everybody on this site - Thanks for help. You guys are very helpful. I am much appreciative.