I always thought I was pretty good at algebra, especially the simple stuff, but I saw this yesterday and I still can't figure out where the trick is: (Where a2 is 'a' squared, b2 is 'b' squared and 2b is 2 times 'b'. I can't figure out superscript on here and pasting from word loses the formatting). The steps are numbered to make any potential answers easier to understand. 1. if a = b 2. a2 = ab 3. a2 -b2 = ab- b2 4. (a+b)(a-b) = b(a-b) 5. a+b = b 6. 2b = b 7. 2 = 1 What can't I see?

(a+b)(a-b) = 0 b(a-b) = 0 a + b = 2b (or a + b =2a) The break is between statement 4 and 5. That's my go. Haven't done it since '73, though.

(a+b)(a-b)=a2 +ab -ab -b2 =a2 - b2 But if a and b =0 then any amount of a = a. Seems fair, but doesn't really explain it. Don't see where your break is though between 4 & 5.

To get from step 4 to step 5, you are dividing both sides by (a-b). As it is given that a=b, then a-b=0, this means you are dividing both sides by zero. There are lots of these look-crazy-at-first-glance equations around, but they all invariably involve dividing by zero, as that produces the required, odd, result!

a and b cannot equal zero. Well, not unless you wish to argue zero equals zero, which seems a waste of time.

Agreed with what has been said and will add, zero will throw up daft solutions and also infinity will. I dont really count them as solutions though. 6. 2b=b this doesnt look right does it, two of something equalling one of something.

Equations 1-4 are generic, ie: apply for all values of a and b. It starts to go wrong (the trick begins) at step 5, which is true only when a and b = 0 and NOT for all integer values of a and b. Carrying on you get a divide by 0 from steps 6 to 7.

Looking back at this now, it is irrelevant if a = 0 or any other value, because at step 3. both sides are equal to zero anyway. From there we are playing with either infinity or zero so all answers are true.

The value of A or B doesn't matter. If A = B then (A - B) = 0 and when you divide by 0 and accept the answer is a real number (which is what your equation is saying), then you can make any integer = any other integer.