What happened?
A famous mathmetician sent a letter containing a riddle to one of his best friends.
It read:
"My friend, help me.
If x=1, y=1,
then it's must be true that x-y=0.
Then 5(x^2-y^2)=0 must be correct.
And also 2(x-y)=0 must be correct.
Because they both equal 0, then 5(x^2-y^2)=2(x-y) must be correct.
But then if you divide both sides by x-y,
you end up with 5(x+y)=2.
But 5(2)=2, 10=2 is not correct.
What happened during this process?
Dear friend, I hope you can be even sharper than I."
Can you see what went wrong??
It read:
"My friend, help me.
If x=1, y=1,
then it's must be true that x-y=0.
Then 5(x^2-y^2)=0 must be correct.
And also 2(x-y)=0 must be correct.
Because they both equal 0, then 5(x^2-y^2)=2(x-y) must be correct.
But then if you divide both sides by x-y,
you end up with 5(x+y)=2.
But 5(2)=2, 10=2 is not correct.
What happened during this process?
Dear friend, I hope you can be even sharper than I."
Can you see what went wrong??
Labels: SharedPuzzle
posted by James Yu at
3:38 PM
9 Comments:
lolzz he divided by zero!!!! LOLLLL
and then x was added with y (1+1=2)
He screwed the F**K up.... he divided by Zero...
This is where he went wrong:
(5(x^2-y^2))/(x-y) does not equal 5(x+y).
You should at that point divide by 5 getting x^2-y^2 = 2/5(x-y)
Then get x's and y's on the same side:
x^2-2/5x=y^2-2/5y
therefore:
x=y
1=1
Okay, the one above me is incorrect.
5(x^2-y^2)/x-y is 5(x+y).
Because x^2-y^2 is a difference of two squares so it factors into
(x-y)(x+y).
So 5(x-y)(x+y)/x-y the x-y's cancelles out to 5(x+y).
But the problem is that you can't divide by 0.
Divide by zero
1st, u dont devide by variabls, 2nd, u dont devide untill u distribute, cuz u dont mess with a equation untill its a polynomial, and the distrubuating is what makes it equal. if u distributed first, then devided. you get 0=0
5x^2-5y^2=2x-2y
devide.
5/2x^2-5/2y^2=x-y
0=0
and the logis, then when u devide the x-y from both sides, is disapears, if false. it just would go to x+y
x=1;y=1
1/1 =1
x/x=x
-y/-y=y
so it would be
5(x^2+y^1)=2(x+y)
5x^2+5y^2=2x+2y
5(1)^2+5(1)^2=2(1)+2(1)
5+5=2+2
10=4
witch is a false statement. but u get it from doing stuff out of order
but still its funny u double messed up lol, u thought x/x=0;1/1=0
and u messed with an equation the wrong way XD
btw i would know, i'm curently takeing alg 2, so i'm a lil more fresh on the topic then most of you
well to the person above i'm taking algebra 1 and knew the answer almost imediately...
he just didn't do his order of operations correctly,
5(x^2-y^2)=2(x-y) (parenthesis since there's nothing to do w/ exponents)
and since x-y=0,x^2-y^2 would also equal 0 so,
5(0)=2(0)
0=0
I am glad I wasn't in those algebra classes. It had nothing to do with the order of operations and there is absolutely nothing wrong with dividing by variables. James Yu had it right with the explanation of a difference of squares. This is an old puzzle that I have seen in a couple different forms. The ones I saw however, were proofs that 1=2. The answer, though, was the same: you can't divide by 0.
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