An easy question

Desert Sleet · 03-05-2005, 02:05 AM

<blockquote id="quote">quote<hr height="1" noshade id="quote">Originally posted by bilal azhar
 if somebody proves that 2=1 then he should be given a noble prize.
The general misconception among the people are that ac****ants are very good in maths but the fact is something else(i am not saying this simply because of my above post).

Take the example of above prove,nobody is being able to detect a simple basic mathematical mistake in the above prove.

bilal
<hr height="1" noshade id="quote"></blockquote id="quote">
Well, not sure what the official rebuttle for this is, but from what I remember from high school algebra there is only one number that can make this statement true
<blockquote id="quote">quote<hr height="1" noshade id="quote">2x = x <hr height="1" noshade id="quote"></blockquote id="quote">
and based on that number you cannot perform this operation
<blockquote id="quote">quote<hr height="1" noshade id="quote">2x/x = x/x <hr height="1" noshade id="quote"></blockquote id="quote">

There is only one number which can proove this statement true and its 0.
<blockquote id="quote">quote<hr height="1" noshade id="quote">(x+y)(x-y) = y(x-y)<hr height="1" noshade id="quote"></blockquote id="quote">
At this point, if you apply the rule, x = y, both sides of the = are 0, and of course every thing after that is correct, if you assume x = 0.
<blockquote id="quote">quote<hr height="1" noshade id="quote">x+y = y
x+x = x
2x = x<hr height="1" noshade id="quote"></blockquote id="quote">
<blockquote id="quote">quote<hr height="1" noshade id="quote">2x/x = x/x

2 = 1<hr height="1" noshade id="quote"></blockquote id="quote">
At this point you have just divided 0 by 0, on both sides of the equal sign. Now, I have no idea if that mean they're both 0, undefined, or 1.

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If I could... Then I would... Turn back time!!