I can prove that 1 = 2

[gib]

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?

[barbarianhorde]

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.

[Carleas]

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

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.

[gib]

Carleas for the win!!!

Anybody got any other math tricks?

(Please no more 0.999... = 1).

[Meno_]

Gib, I don't think that's a trick.

[gib]

Gib, I don't think that's a trick.

What do you mean, Meno_?

[Meno_]

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.

[gib]

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?

[Carleas]

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.

[Meno_]

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?

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.

[barbarianhorde]

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

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.

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

[gib]

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?

[barbarianhorde]

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?

[barbarianhorde]

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.

[gib]

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

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

[barbarianhorde]

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

Thats why I lost.

[lordoflight]

Suppose that a = b... then:

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

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

b(a - b)!= b

[gib]

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

[lordoflight]

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.

[gib]

You wrote a + b = b Yep, and that (a + b)(a - b)=b(a - b) Yep, that a+b=(a + b)(a - b) Nope, that a +b = b. Yep

[lordoflight]

You wrote a + b = b Yep, and that (a + b)(a - b)=b(a - b) Yep, that a+b=(a + b)(a - b) Nope, that a +b = b. Yep

That was a typo. I wrote everything from memory.

You said (a + b)(a - b) = b(a - b). You said b(a-b)=b. But it doesn't. Can't. Unless b is zero.

There is no "trick" being used here. The original post just doesn't make sense to begin with.

Here is where you said b=b(a-b)

"(a + b)(a - b) = b(a - b)
a + b = b"
You dropped the b down underneath b(a-b). Implying it is the same.

[gib]

You said (a + b)(a - b) = b(a - b). Yep. You said b(a-b)=b. Nope... geez, you aren't very good at this. But it doesn't. Can't. Unless b is zero. D'uh!

There is no "trick" being used here. The original post just doesn't make sense to begin with.

Everybody knows that it doesn't make sense. By 'trick', did you think I meant I actually made 1 equal to 2?

Saying there is no trick is what people do when they know there is a problem in the logic but can't identify where it is... so they just say: er, uh, ee... it's just wrong!

You need to be like Carleas... point out exactly where in the proof (or "proof") I made an invalid move.

Here is where you said b=b(a-b)

"(a + b)(a - b) = b(a - b)
a + b = b"
You dropped the b down underneath b(a-b). Implying it is the same.

Huh? I didn't drop any b. I "dropped" (a - b) under each side of the equation to get a + b = b. Nowhere did I get b = b(a - b).

[lordoflight]

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

The crux of what you are doing, is exposing a flaw in mathematics. You set the value of the equations to zero, so it erodes any qualities the equations originally had.
See my thread called "the number delusion".

The same thing happens when you scale a 3 dimensional shape to zero. You can scale it to 10%, 50%, or even a negative, and still retain the shape. But once you set it to zero information is lost and the shape no longer exists. That is the crux of what you are doing here.

[Silhouette]

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

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.

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.

[gib]

Okay, looking at it further, you put

(a + b)(a - b) = b(a - b)
Which does not follow.

Sure it does... so long as a = b.

It would produce nonsensical results such as this:
(3+4)(3-4)=4(3-4)
-7=-4

You can't do that! a must equal b!

The crux of what you are doing, is exposing a flaw in mathematics. You set the value of the equations to zero, so it erodes any qualities the equations originally had.
See my thread called "the number delusion".

I don't know about any flaw in mathematics... just a trick in a pseudo-proof.

The same thing happens when you scale a 3 dimensional shape to zero. You can scale it to 10%, 50%, or even a negative, and still retain the shape. But once you set it to zero information is lost and the shape no longer exists. That is the crux of what you are doing here.

Not sure what you mean by "set the value of the equations to zero." There was a divide by zero, and if a = b, then some of the equations resolved to zero, but not all. Equations should be permitted to equal zero sometimes, but it's the divide by zero which really screws things up.