/sci/ - Science & Math

The root of a square is the absolute value

>>3561297
The following statement over Real Numbers is false:
(for all x) sqrt(x^2) = x

The following statement over Real Numbers is true:
(for all x) sqrt(x^2) = |x|

There's nothing wrong with this.
2 + 2 = 5

>.5^2 = (-.5)^2

ok

>0.5 = -0.5

nope

from this point:
(4-9/2)^2 = (5-9/2)^2
you should do
(4-9/2)^2 - (5-9/2)^2 = 0
remember arithmetical identity
a^2 - b^2 = (a-b)*(a+b)
so we have
(4-9/2-5+9/2)*(4-9/2+5-9/2)=0
9*0=0
0=0
nothing wrong, OP ^^

4-9/2 is -.5
5-9/2 is .5
squaring them gives the same result
what you should have said is
±(4-9/2) = ±(5-9/2)
9/2-4 = 5-9/2

What's wrong with it? It purposely obfuscates the problem using fractions so the average 'HURR HOW DO I MATH' person cannot follow it. Simplify it and you can see the glaring error in it, it boils down to:
-.5^2 = 0.5^2
Now, this is true, but taking the square root of each side leads to:
+/- .5 = +/- .5
When you hand-pick which one is a + or - and purposely choose ones that do not work (in case I need to spell it out, you CANNOT do this), you get:
-.5 = .5
or
0 = 1
therefore
2+2 = 4
and by additive property of 0
2+2 = 4+0
but since 0 = 1
2+2 = 4+1
2+2 = 5

I guess it wasn't stated enough times, so
(-0.5)^2!=0.5^2

>>3561409
>but since 0 = 1

nope.mp3

>>3561441
but 2 factorial is just 2

>how would i avoid making this mistake

if you get 2+2=5, you probably made a mistake

For this shit , this is what i shall give : NO.exe