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1+1 Dosen't Always Equal 2?! What?!


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While looking back on my old posts, I found an old thread that claimed that in some types of calculation, 1 plus 1 doesn't necessarily equal 2. I'm sorry, but every time I've counted on my fingers, 1 plus 1 equaled 2. Can someone please explain this to me in the closest thing to layman's terms as possible? Thanks.

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Well, for a not-quite-correct response:

Assume x = 1

x2 = x (multiply by x)

x2 -1 = x -1 (subtract 1)

(x - 1)(x + 1) = x-1 (factor x2 -1)

x + 1 = 1 (divide by [x - 1])

1 + 1 = 1 (remember, x = 1)

Not that this is correct, of course.

This is why we don't allow people to divide by zero.

Anyway, it would be nice to have some context for that last thread of yours. But you see it often during improper rounding. For example:

0.74+0.67 = 1.41. If we round all these to the nearest integer we get 1+1=1.

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Indeed, because dividing by x- 1 is dividing by zero :P

This chain of equation is only defined when x != 1

Are you sure about that?

Assume x = y

x2 = xy (multiply by x)

x2 -y2 = xy -y2 (subtract y2)

(x - y)(x + y) = y(x - y) (factor)

x + y = y (divide by [x - y])

2y = y (remember, x = y)

2 = 1 (divide by y)

Of course, there's another division by zero there.

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Are you sure about that?

It's all about restrictions.

When you do algebra like this, when you are making divisions, you must note /when/ you are allowed to do divisions.

In this case, you can only do that division when x != y

I've lost so many points in my grade 9-10 exams over forgetting my restrictions it stopped being funny :P

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In truth, there's nothing stopping 1+1 from equaling 2, nor 0/0 from being defined. It's merely because we make some basic assumptions that are generally useful, but we may also consider an algebra in which those assumptions are not true. You end up with something like this: http://en.wikipedia.org/wiki/Wheel_theory

Further reading, if you're interested (I'm not :) ): http://www2.math.su.se/reports/2001/11/2001-11.pdf

That said, unless you're into Wheel Algebra (i.e. insane, or worse, a mathematician), it's probably safe to assume that 0/0 is undefined and that 1+1 does, in fact, equal 2. I'd rather stick to those assumptions, and to simple stuff like Minkovsky spacetime and Schrodinger equations (oh, and don't ask me to touch Strings with a 10-meter, 4-dimensional pole. :) I'm not crazy enough for that yet...).

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In mathematics not everything is easily understandable.

Have a look at this formula: x = 1 - 1 + 1 - 1 + ... (infinite times)

So what's the value of x? 1? 0? We can't decide. But we can still make calculations with this formula.

Now look at that:

2x = x + x

2x = (1 - 1 + 1 - 1 + ...) + (1 - 1 + 1 - 1 + ...)

2x = 1 - 1 + 1 - 1 + ... + (1 - 1) + (1 - 1) + ...

2x = 1 - 1 + 1 - 1 + ... + 0 + 0 + ...

2x = 1 - 1 + 1 - 1 + ...

2x = x?

or what about this?

x - x = (1 - 1 + 1 - 1 + ...) - (1 - 1 + 1 - 1 + ...)

x - x = 1 - 1 + 1 - 1 + ... - 1 - (-1) + (-1) - (-1) + ...

x - x = 1 - 1 + 1 - 1 + ... - 1 - (-1) + (-1) - (-1) + ... (nothing changed, it's just to show the colors)

x - x = 1 - (1 - (-1)) + (1 + (-1)) - (1 - (-1)) + ...

x - x = 1 - 0 + 0 - 0 + ... - 1

x - x = 0?

But that means x must have the value of 1. But we already know x can be 0, too!

Crazy, I know!

Edit:

They use this formula in quantum mechanics. There the value of x is 1/2. (That's even more crazy!)

Edited by *Aqua*
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Wouldn't x=0 in that case? 0-0=0, 2*0=0. But then, I'm studying physics, not mathematics. Sure, there are mathematical lectures, but I'm not that good at understanding the theory being this all (though I do manage to apply those mathematics when I need them). :)

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Binary for Dummies:

1+1=10

Why?

Binary was originally designed for computers (I think.) An electrical circuit can be one of two things: on or off. Either the circuit is complete, or it isn't. To get around the issue of only having two potential outcomes to an input, eight bits (aka a single on/off) were grouped together to form a byte (pronounced as "bite.") Solving this issue relies on exponents of base two (since there can only be two options for an individual bit.)

1 0 0 1 1 0 1 1

2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0 (I wish I could make true exponents on this forum)

so, you get ((2^0)*1)+((2^1)*1)+((2^2)*0)+((2^3)*1)+((2^4)*1)+((2^5)*0)+((2^6)*0)+((2^7)*1)

Simplifying a bit (pun intended,) you get (1*1)+(2*1)+(4*0)+(8*1)+(16*1)+(32*0)+(64*0)+(128*1)

Taking it farther, you get 1+2+0+8+16+0+0+128

So your final solution is 155.

Lets use 1+1=10 as another example. In binary, it looks like 00000010. I'm going to skit the work for the first six zeros (as it would be a headache to type all of them. In decimal, 1+1=2. Now it is a matter of conversion. The figure "2" is not a valid figure in binary. However, we can use binary to arrive at this solution.

1 0

2^1 2^0

So we get: ((2^1)*1)+((2^0)*0)

Which equals: (2*1)+(1*0)

Which in turn equals: 2+0

Which finally equals 2, however, two is represented as "10" in binary.

You can go beyond 2^7; I stopped there because binary is mostly used in computers, and, IIRC, early computers only used eight bits to a byte. Today, however (and someone correct me if I'm wrong) we go up to 2^63 (64 bit processes.)

Edit: And for some reason, its not maintaining the spacing for the examples.

Edited by rpayne88
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Y'know, dyscalculiacs like me want to kill such threads with fire :sticktongue: I have enough problems memorising my own phone number. When someone drops a bombshell like this i want to bite him. Math is already hard, please don't confuse me any further with knowledge that 2+2 not always equals 4 :confused:

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Y'know, dyscalculiacs like me want to kill such threads with fire :sticktongue: I have enough problems memorising my own phone number. When someone drops a bombshell like this i want to bite him. Math is already hard, please don't confuse me any further with knowledge that 2+2 not always equals 4 :confused:

Not an issue in binary. To my computer, "2" and "4" don't exist.

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Well, changing between number radices can be fun :)

We can even say 31 = 25 ! But we should have the number bases at least abbreviated near it when working with different radices :P in this case, i have 31oct = 25dec :P (octal and decimal) - now you can understand why some geeks get mixed up between halloween and christmas ;)

Edited by sgt_flyer
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More precisely, 0.(9) is equal to 1. But this is obvious at high school level if you've got a good teacher. Basically, you can express any number like 0.(x) as x/9. Simple enough. Apply this to x=9 and you get... 9/9, which equals 1.

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To be clear, I'm fine with binary. There is apparently a form of calculation called Z2, where 1+1=0. That is the sort of thing I'm confused about.

Ah, you mean ℤ2? That's a so called group which only consists of the numbers 0 and 1. Whatever you do with these numbers, on the result you must calculate the remainder and use that as the result. And that will always be 0 or 1. You can't break out of it. That's a group axiom (closure).

All numbers you know are groups. I can prove it to you but it will require a bit lengthy and boring explanation.

In other mathematical craziness:

0.99999... is exactly equal to 1.

The idea behind 0.99... is that it is the next smaller neighboring number to 1. The difference between these two numbers is infinite small so we can conclude that there is no difference between them. This applies to all number pairs with an infinite small difference, e. g. 0.2499... = 0.25 and 0 = 0.000...0001.

Proof:

x = 0.99...

10x = 9.99...

10x - x = 9.99... - 0.99...

9x = 9

x = 1

=> 0.99... = 1

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The idea behind 0.99... is that it is the next smaller neighboring number to 1. The difference between these two numbers is infinite small so we can conclude that there is no difference between them. This applies to all number pairs with an infinite small difference, e. g. 0.2499... = 0.25 and 0 = 0.000...0001.

I don't think this is quite accurate. 0.999... is not the next smaller neighboring number to 1, it is 1 expressed in a different way. As your proof ably demonstrates.

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