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# 1=2? Best anwer to whoever finds the mistake first. (1=2 doesn't count as the mistake!)?

Let x=1 and x=y

x=y

xy=y^2 multiply both sides by y

xy-x^2=y^2-x^2 subtract x^2 from both sides

x(y-x)=(x+y)(y-x) factor

x=x+y divide both sides by (y-x)

1=1+1 since x=1 and x=y

1=2

Happy Hunting

### 8 Answers

- CHESSLARUSLv 71 decade agoFavorite Answer
In the fifth line of your "demostration" you are dividing by ZERO because x=y hence x-y=0

That is the mistake!

Good luck!

- RaymondLv 71 decade ago
You are dividing by 0.

PS: (just trying to annoy everyone with a long post)

Let's see, though, what happens in the Z1 ring.

The Z1 ring contains only one element, which is usually written as the number 0.

Any integer is congruent to 0 when applied to this ring (how? the basis of the ring is 1, so divide the integer by 1: what is the remainder?)

A "1" is any number which, when multiplied to any element in the ring, will leave the element unchanged. What is our candidate for the number 1 in the ring?

In Z1, we have little choice : let's pick... hmmm... Zero!

0 multiplied by any element in the ring (let pick at random... hmmm... ah yes! zero) gives 0 -- thereby showing that the randomly selected element was left unchanged. Therefore 0 really is a 1.

If we can show we can divide by (x-y) when x=y in this ring, then we truly have x = x+y. A division in a ring is re-defined as a multiplication by a multiplicative inverse : a number which, when multiplied by the element, will give 1.

Let's see, I wonder what is the multiplicative inverse of 0 in this ring... hmm... zero?

Let us check: If I multiply 0 by its inverse 0, I get 0 which was shown, above, to be the same as 1 in this ring. Therefore 0 really is the inverse of 0. Therefore, in this ring we can divide by zero.

And we can show that 1 = 2 (although we would say 1 is congruent to 2).

- Joni DaNerdLv 61 decade ago
You're dividing by zero in the third last step.

the first step says x = y so when you divide by y-x you're dividng by zero.

And if x = y you cannot have x = x + y unless x = y = 0, and you say in the beginning that x = 1.

This is an old chestnut that should be consigned to the garbage dump.

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- Jim BurnellLv 61 decade ago
x=x+y divide both sides by (y-x)

x = y, so you're dividing by zero (not allowed).