lordoflight wrote:Okay, looking at it further, you put

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

Which does not follow.

It would produce nonsensical results such as this:

(3+4)(3-4)=4(3-4)

-7=-4

gib wrote:Suppose that a = b...

I don't know why this wasn't an open/shut thread. A nice little trick to figure out, for sure, but it doesn't warrant in depth discussion unless you don't understand it. It certainly doesn't throw numbers into doubt.

gib wrote:I'm interested in this:

1 + 2 + 3 + 4 + 5 + 6 + 7 + .... = -1/12

This one might warrant discussion though.

It happens when you mess around with infinite series, notably "s = 1-2+3-4+5-6+7-..." and what happens when you add it to itself in a certain way (i.e. add the 1st term to the 2nd, the 2nd to the 3rd and so on) and noting that it results in "t = 1-1+1-1+1-1+1-...", which you have to accept equals a 1/2, because it's an average of whether you "stop" at the last even number in the series or the last odd number in the series.

Accepting these, you can say that 2s = t = 1/2, therefore s = 1/4.

Then you can subtract "s" from the infinite series in question of "u = 1+2+3+4+5+6+7+...", the first term from the first term, the second from the second and so on - nothing special here. You get "u-s = 0+4+0+8+12+0+16+...", so you can factor out the 4 and get "u-s = 4(1+2+3+4+5+6+7+...)", which is "u-s=4u".

This simplifies to "-s = 3u", we know s = 1/4 so "-1/4 = 3u", making our infinite series "u = -1/12".

Obviously it rides on the problem of "t". I don't see why you can't add "s" to itself in the way I described above... infinite series extend infinitely so it doesn't matter what order you add each number together. The rest is just standard rearrangement and substitution.

I don't think the infinite series "t" has a valid answer, which makes possible such things as "1+2+3+4+5+6+7+..." = -1/12. I'm pretty sure you can make it equal other different values too.