# Complex numbers and 1+1 = 2.

1. Nov 20, 2007

### Oerg

Hello

Me and my friend were discussing about an irrelevent subject when he said that 1+1 is not always equals to 2 mathematically due to complex number variations. Can anyone highlight the influence of complex numbers on this equation?

2. Nov 20, 2007

### HallsofIvy

Staff Emeritus
"complex number variations"? I don't see that this has anything to do with "complex numbers" so maybe he just meant "complicated".

In, for example, the binary number system 1+ 1= 10- but that's a "cheat" because "10 base 2" is just another way of representing 2 (base 10). You can also argue that, in the "group of integers modulo 2", 1+ 1= 0. Again, the point is that "1" does not mean the same thing that it does in ordinary arithmetic.

3. Nov 20, 2007

### Oerg

well, we were talking online, so i couldnt have seen wrongly, he meant complex numbers, and he said that a teacher in school once taught him that 1+1 does not always equal 2. He did not quite remember the reason, but he said it was something to do with complex numbers.

4. Nov 20, 2007

### PhY

I'm pretty sure 1+1 base 10 is Equal to 2.
Base 10 is the Natural base, your friend brings 1 bottle to a party with 1 bottle, party has 2 bottles.
If it was complex Numbers wouldn't you need to add an "i" to the equation?
Also 10 is the default base an other space would need to be specified...because they are not natural.
1+1 in binary is equal to 10<2>.....so does that mean that 1+1 is not equal to 2 ? no.
It means that 1+1<base2> is not equal to 2, but then again , 10 specifically means On and Off which translates into 2 states.

I haven't done complex numbers yet (i'll do it in about a month), but I did go over it and it did include an 'i' in every complex number equation.

What i mean to say is 1+1 only counts for base10, putting it into any other context would need a Declaration of some sorts. . maybe i(1+1)....maybe sin(1+1), maybe 1+1B or maybe 1+1H. Else it would be very ambiguous.

5. Nov 20, 2007

### Count Iblis

You can define another number system containing only the numbers 0 and 1. In that system you can define addition such that 1+1=0

6. Nov 20, 2007

### rock.freak667

maybe by vector addition 1+1=1.9 ...as in a 1N force acting at some angle to another 1N force and the resultant(+) is given by 1+1 which may be less than 2

7. Nov 20, 2007

### k3N70n

In $$Z_2$$

8. Nov 21, 2007

### wysard

I remember this one a bit from high school.

The teacher put a series of equations on the board that proved that 1 + 1 = 4 instead of 2. The problem involved "complex" numbers because, as I recall from many years ago it involved the square root of a negative number in the process.

Tell your friend to write down the equations and look at them closely. As memory serves there is a hidden element in one of the equations where one of the negative numbers is divided by zero and given the value 1 instead of zero. When the whole system is multiplied back out the error is magnified.

1 + 1 = 2.

If it doesn't, then you can bet your bottom dollar there is an error in the math somewhere.

9. Nov 21, 2007

### Chronos

It's an old algebra trick: a=b, a+b=2a, axb=a^2: (a+b)(a-b) = 2a(a-b), (a^2-b^2) = 2(a^2-b^2), 1=2.

10. Nov 21, 2007

### Integral

Staff Emeritus
2 vectors (or imaginary numbers) with magnitude = 1 can add up to anything up to 2.

11. Dec 14, 2007

### yasiru89

Without any dubious algebra, or resorting to a different modulo, there arises no need for imaginary quantities to be introduced to the decimal equation 1 + 1 = 2
Why do I refrain from saying complex? Because on the standard scale the complex number system encompasses all the others, thereby making integers complex by definition. Indeed, the complex field is a stable one as Gauss proved all those years ago.

12. Dec 14, 2007

### daniel_i_l

It all depends on what you mean by 1, + and 2. If you're talking integers in number theory where 1 is the unit, 2 is the integer after 1, and + works according to all the axioms of number theory then 1+1 = 2. It all depends on the system you're working with and how you define things. It's as simple as that.

13. Dec 14, 2007

### Invictious

Yes 1+1 is always equal 2, and never will change, in our real number system.

Whether it is equal to 3, or 8, or 10, or 2i, or whatever, it does not really apply to our daily life, unless you DO work with complex numbers or computers.

As far as we are concerned, just so long that is true, we are fine.

14. Dec 15, 2007

### ice109

and how did you make that happen?

15. Dec 15, 2007

### CompuChip

Simply multiplied (a + b) = 2a
on both sides with (a - b)

16. Dec 17, 2007

### ice109

and the last part? a=a+b some how?

17. Dec 17, 2007

### CompuChip

(a+b)(a-b) = 2a(a-b)
Working out the brackets on the left hand side gives
(a^2-b^2)
Working out the brackets on the right hand side gives
2 (a^2 - a b)
but since a = b, a b = b^2 so it is also
2 (a^2 - b^2)

Now dividing both sides by (a^2 - b^2) = 0 we get 2 = 1.

18. Dec 17, 2007

### malty

But isn't (a-b)=(a-a) = 0 so how can you multiply by zero???????

(a+b)(a-b) = 2a(a-b)
(2a)(0)=0

19. Dec 17, 2007

### CompuChip

Of course you can multiply by zero:
(2a)(0) = 0
0 = 0

that last equation is true.
But you are close, the problem is indeed that in the last step it is written:
2 x 0 = 1 x 0
and we try to divide out zero (written very complicated as (a^2 - b^2)) to conclude 2 = 1.
The entire point of this "calculation" is that the reader is easily confused by this, and overlooks the fact that there is a division by zero, because it is obscurely written.

Of course this example is constructed, but in general one should watch out that something similar doesn't accidentally happen somewhere in a large calculation

(I put the above text in white, so people who don't know this one yet (though I doubt there are many of them) can more easily skip over it.)