This particular math "formula" has been around for a long time and often gets the untrained layman to believe that 1 plus 1 can equal 1. The formula is wrong and if you read on you will find out why.
x = y |
x^{2} = xy |
x^{2}-y^{2} = xy-y^{2} |
(x+y)(x-y) = y(x-y) |
(x+y)(x-y) = y(x-y) |
Therefore: (x+y) = y
If x = 1 and y = 1, (1+1)=1!
Now, let us break this down and see where the writer gets it wrong and confuses the layman.
x = y | At this point in the equation everything is OK. We know that no matter what, both x and y have the same value. |
x^{2} = xy | Here we are also fine, for if one value is squared and the other is multiplied by itself AND both variables have the same value then squaring and multiplying by oneself will result in the same answer. |
x^{2}-y^{2} = xy-y^{2} | Since we have already proven that squaring a number and multiplying it by oneself are the same thing we know this line is also correct. |
(x+y)(x-y) = y(x-y) | This is the line where the trick is first put in place. Because any time you subtract a number from itself you will end up with a value of zero. This line of the equation is only accurate as long as the (x-y) items remain in the equation. IF you were to factor both sides down and remove the (x-y) you would have (x+y) = y, which we know to be false. No matter what values you assign to x and y (which must be the same as evidenced by line 1 above) this line will be correct because you are multiplying both sides by zero which, as we all know, will result in 0 = 0. |
(x+y)(x-y) = y(x-y) | After reading line 4 above you should now understand the trickery at work here. Not only are both denominators equal to zero, but, each numerator also works to zero! If you remove the zeroes the equation would no longer be accurate. |
I hope this little bit of math mayhem brightens your day. Feel free to tease your students and see how long it takes them to figure it all out. |