# I can prove that 1 = 2

Suppose that a = b… then:

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

Can anyone spot the trick?

Sure,
(a + b)(a - b) = b(a - b)
a and b are both 0.

so 1a = 2a but that doesn’t mean 1=2.

I agree the trick is around this line, but can you explain why this entails that a and b are both 0?

I see the trick as being the move between these two lines:

(a + b)(a - b) = b(a - b)
a + b = b

(a-b) equals zero, so this step requires dividing by zero, which is undefined.

Carleas for the win!!!

Anybody got any other math tricks?

(Please no more 0.999… = 1).

Gib, I don’t think that’s a trick.

What do you mean, Meno_?

I mean the idea of .99999999=1 has not yet been solved as yet, not from what has been worked on in that forum here, nor anywhere else. Because it deals with the existence of absolutes, it will never be solved.

And since 1=2 has generally been solved as a paradox, and shown as a paradox, once the shroud of mystery surrounding it disappears, it clears it up.

Oh, you mean ‘trick’ as in a form of deception, something that once unveiled removes the magic. I see. In that case, I agree that 1 = 0.999… is not a trick, it’s just a fact. And I agree that its standing as a trick will forever be debated by the infinitesimalists. Still, I’m sick of debating it. I want something new, something to ponder my brains over.

You got anything?

Also not a trick, but 1 + 2 + 3 + 4 + 5 + 6 + 7 + … = -1/12

There are a lot of these sums of infinite series that have weird and counterintuitive answers, and the answers are apparently born out in certain areas of physics.

No, Gib, I did not use the the word ‘trick’ I just said that 1=.999999999~
Is not a trick, as such- because, the inference of Your example implying that therefore- it is a trick invalid… It is based on the binominal extension of the basic logic of the rule of exclusion, therefore not prevy to analysis of higher order mathematical adaptation to physical sciences.genetally. specific functions may do. And I am only thinking in the terms suggested, and they seem credible.

So Your example may predict a solution as it was described by Carleas , but the later appears to have closed that probability.

Well yes the step only makes sense if both are zero. I didn’t win?

They could both be 1, and a - b would be 0. Dividing by a - b would count as dividing by 0.

I’m interested in this:

1 + 2 + 3 + 4 + 5 + 6 + 7 + … = -1/12

Are we saying that if you add all the positive integers, you get negative a twelfth?

Well but if theyre both 1 the last line would be different. Mathematics always must work both ways! So I just looked at what you concluded and used that to see what your premises were.

Maybe you overlooked that if a equals b a - b is always 0?

Look:

"Suppose that a = b… then:

a^2 = ab
a^2 - b^2 = ab - b^2
(a + b)(a - b) = b(a - b)
a + b = b
a + a = a
2a = a"

So far so good

“2 = 1”

It doesn’t follow unless you said that a is not zero.

It doesn’t follow period. That’s why it’s a trick.

Yeah, my first response was thats an error not a trick. But I got creative.

Thats why I lost.

It does not follow.
a+b!=(a + b)(a - b)

b(a - b)!= b

You’re right, but where in the steps did it say:

a + b = (a + b)(a - b)

b(a - b) = b

Gib, I know you can do better than this.
You have a strong mind, you just need to take your time and focus.

The whole thing was supposed to be a system of qualities.
So I should be able to match any equality to another entity in the system.
You wrote a + b = b, and that (a + b)(a - b)=b(a - b), that a+b=(a + b)(a - b), that a +b = b.
So it should logically follow that b(a-b)=b.
But it fails in all cases except zero.
For instance, 4(4-4)=4.
That is false. It always will equal zero.