1=2

Let there be two equal numbers: a=b.
Multiply everything by a: a^2=ab
add (a^2-2ab) two each side: 2a^2-2ab=a^2-ab
therefor: 2(a^2-ab)=a^2-ab
you get after dividing: 1=2

Can you spot the mistake?

Originally posted by yevgenip
Let there be two equal numbers: a=b.
Multiply everything by a: a^2=ab
add (a^2-2ab) two each side: 2a^2-2ab=a^2-ab
therefor: 2(a^2-ab)=a^2-ab
you get after dividing: 1=2

Can you spot the mistake?

a^2=ab so a^2-ab=0...

That's Right! You can't divide by zero.
This was realy an easy one, does anyone have a more challenging riddle of the same kind?

this was brought up by my best friend EE

Ok here goes;

Make a sequence of numbers and call it ( * );
1, -1/2, 1/2, -1/3, 1/3, -1/4, 1/4, ...

Now look at the sum of this sequence. Clearly all terms except the 1 cancel out (you substract something, then add the same thing again), and the sum is 1.

Now we make the sum a bit more interesting;
(1-1/2) + (1/2-1/3) + (1/3-1/4) + (1/4-1/5) + ... = 1

We note that 1/7-1/8 = (8-7)/(8*7) = 1/(8*7), similarly for the other terms we create this formula;
S = 1/2 + 1/6 + 1/12 + 1/20 + ... = 1.

Now to make it even more interesting;
You can see that 1/(3*4) = -2/3 + 3/4, and 1/(8*9) = -7/8 + 8/9 and so on.

Now we substitute that in S;
S = 1/2 + 1/6 + 1/12 + ... = 1/2 -1/2 +2/3 -2/3 +3/4 -3/4 +4/5 -...=0
That is strange, for jsut a minute ago we saw that S=1, tohugh now we get S=0 !

It might seem that the last term doesn't have a sucessor to cancel with, but that's an illusion, for there is no last term. Every number there is followed immedeately by a negative term that cancels it out.

So the sum of 1/2 + 1/6 + 1/12 +... =0. Kinda strange since all terms are greater then 0.

Now we change the order of ( * ), we take one positive term, and then two negative ones, then a positive one, and so on. Call this ( ** );

1, -1/2, -1/3, 1/2, -1/4, -1/5, 1/3, ...

The sum of this sequence is also 1, since addition is commutative (a+b=b+a for all a,b) and the sum of ( * ) was 1.
Now we have;

1= 1+ (-1/2) + (-1/3) + 1/2 + ...=
((1-1/2)-1/3) + ((1/2-1/4)-1/5) + ((1/3-1/6)-1/7) +...=
(1/2-1/3) + (1/4-1/5) + (1/6-1/7)+...=
1/6 + 1/20 + 1/42 + ... <
1/2 + 1/6 + 1/12 + 1/20 +...

So 1/2 + 1/6 + 1/12 + ... is not 1, not 0, but more then 1!

Now we change ( * ) again. Now we take one positive term, followed by 3 negative terms. Then we have (you can verify it);

1= (1-1/2-1/3-1/4) + (1/2-1/5-1/6-1/7) + (1/3-1/8-1/9-1/10)+ ... =
-2/(2*3*4) + -2/5*6*7 + -2/8*9*10 + ... < 0

So now the sum is less then 0!

Originally posted by TheMaster37
S = 1/2 + 1/6 + 1/12 + 1/20 + ... = 1.
Now to make it even more interesting;
You can see that 1/(3*4) = -2/3 + 3/4, and 1/(8*9) = -7/8 + 8/9 and so on.
Now we substitute that in S;
S = 1/2 + 1/6 + 1/12 + ... = 1/2 -1/2 +2/3 -2/3 ...[text shortened]... ge, for jsut a minute ago we saw that S=1, tohugh now we get S=0 !

There is a fallacy in the above section. In the second equation (1/2 -1/2 +2/3 -2/3 +3/4 -3/4 etc) the sum diverges, as each term approaches 1 or -1.
Similarly, let M= (1/2 - 1/2) + (1/2 -1/2) + (1/2 - 1/2) ... =0
Since addition is associative, regroup without changing the order, M= 1/2 +(-1/2+1/2) +(-1/2+1/2) +(-1/2+1/2)...= 1/2

The rest gets a little to convoluted for me.

IIRC, re-ordering the terms of a series is inadmissible unless the series is absolutely convergent. That is, SIGMA |a(n)| is convergent.

Ah good, both right 🙂

I love doing that trick for highschool students, they simply don't get it :p

Originally posted by TheMaster37
Ah good, both right 🙂

I love doing that trick for highschool students, they simply don't get it :p

Challenge for The Master: do the same trick with a finite sequence 😀🙄