الملتاع
14-03-2005, 12:29:50 AM
.The following is a "proof" that one equals zero
Consider two non-zero numbers x and y such that
x = y
Then x^2 = xy*
:Subtract the same thing from both sides
x^2 - y^2 = xy - y^2
Dividing by (x-y), obtain
x + y = y
Since x = y, we see that
2y = y
Thus 2 = 1, since we started with y nonzero
,Subtracting 1 from both sides
0=1
?What's wrong with this "proof"
:Presentation Suggestions
This Fun Fact is a reminder for students to always check when they are
.dividing by unknown variables for cases where the denominator might be zero
:The Math Behind the Fact
The problem with this "proof" is that if x=y, then x-y=0
.(Notice that halfway through our "proof" we divided by (x-y
=====
* "x^2 means "x to the power 2
copieeeeeed
Consider two non-zero numbers x and y such that
x = y
Then x^2 = xy*
:Subtract the same thing from both sides
x^2 - y^2 = xy - y^2
Dividing by (x-y), obtain
x + y = y
Since x = y, we see that
2y = y
Thus 2 = 1, since we started with y nonzero
,Subtracting 1 from both sides
0=1
?What's wrong with this "proof"
:Presentation Suggestions
This Fun Fact is a reminder for students to always check when they are
.dividing by unknown variables for cases where the denominator might be zero
:The Math Behind the Fact
The problem with this "proof" is that if x=y, then x-y=0
.(Notice that halfway through our "proof" we divided by (x-y
=====
* "x^2 means "x to the power 2
copieeeeeed