mean of two consecutive number helps proving both number equals.. [duplicate]

Well I was wondering how is this possible.

let say:

-> 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.....

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

• 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. – user67773 Aug 7 '13 at 12:07
• See also: $2+2 = 5$? error in proof – Martin Sleziak Nov 19 '16 at 13:21

$$x \ne \sqrt{x^2} = |x|$$
for $x < 0$.
• $4 - 9/2 = -1/2 \ne 1/2 = \sqrt{(4 - 9/2)^2}$, so your second line is wrong. – Vedran Šego Aug 7 '13 at 12:20
• The mistake is going from the first line to the second. $4 - \frac 92 \ne \sqrt{\left(4 - \frac 92\right)^2}$. – Tunococ Aug 7 '13 at 12:22