05-18-2013, 08:54 PM
(This post was last modified: 07-05-2013, 10:14 PM by MUHAMMAD FAISAL RAFIQUE.)

The following is a "proof" that one equals zero.

Consider two non-zero numbers x and y such that

x = y.

Then x2 = xy.

Subtract the same thing from both sides:

x2 - y2 = xy - y2.

Dividing by (x-y), obtain

x + y = y.

Since x = y, we see that

2 y = y.

Thus 2 = 1, since we started with y nonzero.

Subtracting 1 from both sides,

1 = 0.

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).

NOTE:- MAYRAY DOSRAY TOPICS PE THREADS PARHNAY K LEYAY NICHAY LINK PE CLICK KRAIN

http://forum.paksc.org/search.php?action...178ffc079a

Consider two non-zero numbers x and y such that

x = y.

Then x2 = xy.

Subtract the same thing from both sides:

x2 - y2 = xy - y2.

Dividing by (x-y), obtain

x + y = y.

Since x = y, we see that

2 y = y.

Thus 2 = 1, since we started with y nonzero.

Subtracting 1 from both sides,

1 = 0.

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).

NOTE:- MAYRAY DOSRAY TOPICS PE THREADS PARHNAY K LEYAY NICHAY LINK PE CLICK KRAIN

http://forum.paksc.org/search.php?action...178ffc079a