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This question already has an answer here:

Well I was wondering how is this possible.

let say:

4=4

-> R.H.S

$=4-(9/2)+(9/2)$

$=\sqrt{ (4 - ( 9/2 )) ^ 2}+ (9/2)$

$=\sqrt{ 16 - 36 +( 9/2 )^2}+ (9/2)$

$=\sqrt{ - 20 +( 9/2 )^2}+ (9/2)$

$=\sqrt{ -45 + 25 +( 9/2 )^2}+ (9/2)$

$=\sqrt{ 5^2 - 2x(9/2)x5 +( 9/2 )^2 }+ (9/2)$

$=\sqrt{ (5 - (9/2))^2 }+ (9/2)$

$=5- (9/2) + (9/2)$

=5

going through this I found that every number can be proven equal to any number?????? still scratching my head.....

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marked as duplicate by Douglas S. Stones, dfeuer, Dominic Michaelis, Davide Giraudo, user61527 Aug 28 '13 at 20:51

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I had faced this problem about 15 years ago :). I assured myself saying that in the penultimate step you should take 9/2-5 instead of 5-9/2. $\endgroup$ – user67773 Aug 7 '13 at 12:07
  • $\begingroup$ See also: $2+2 = 5$? error in proof $\endgroup$ – Martin Sleziak Nov 19 '16 at 13:21
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This is old.

$$x \ne \sqrt{x^2} = |x|$$

for $x < 0$.

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  • $\begingroup$ @@vedran can you elaborate please.... $\endgroup$ – Timon Aug 7 '13 at 12:09
  • $\begingroup$ $4 - 9/2 = -1/2 \ne 1/2 = \sqrt{(4 - 9/2)^2}$, so your second line is wrong. $\endgroup$ – Vedran Šego Aug 7 '13 at 12:20
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    $\begingroup$ The mistake is going from the first line to the second. $4 - \frac 92 \ne \sqrt{\left(4 - \frac 92\right)^2}$. $\endgroup$ – Tunococ Aug 7 '13 at 12:22

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