sqrt(x) = -sqrt(x)
prove me wrong
[math]\mathsf{It\text{'}s~now~safe~to~turn~off~your~computer.}[/math]
[math]\mathsf{It\text{'}s~now~safe~to~turn~off~your~computer.}[/math]
[math]\mathsf{It\text{'}s~now~safe~to~turn~off~your~computer.}[/math]
[math]\mathsf{It\text{'}s~now~safe~to~turn~off~your~computer.}[/math]
[math]\mathsf{It\text{'}s~now~safe~to~turn~off~your~computer.}[/math]
[math]\mathsf{It\text{'}s~now~safe~to~turn~off~your~computer.}[/math]
>>8431349
True when x =0.
>>8431349
y=sqrt(x) is really just y^2=x which is reflective about the x axis, so yes that is true.
>>8431387
>here let me add infinitely many extraneous solutions to the equation and pretend that it's the same thing
>>8431397
What's wrong with doing that?
>>8431397
I see you've only taken calc II
when you're at a non-brainlet level math course you'll learn a few things
>>8431349
sqrt(x)= -sqrt(x)
sqrt(x)/sqrt(x)= -1
1 =/= -1 on the basis of e1, e2 of R^2
>>8431401
wow I'm so glad I have somebody who passed a lower division undergrad math course around to set me straight
you're really smart man, i hear that's a hard class. Pretty badass. Post more!
>>8431399
because the supposed solutions to the equation are extraneously created by operating on the functions. they are not solutions to the original equation, which has no solutions
>>8431387
>>8431441
because you're a brainlet that hasn't taken a higher level course to explain the fundamentals of math
The corresponding statement would actually be
y^2=(-y)^2
which under the operator of a sqrt leads to
+y, -y= -y, +y
Which do not give proper matching corresponding values, but this is just inuitive and not rigorous
Another explaination for a brainlet?
y^2=(-y)^2
ln(y^2)=ln((-y)^2)
(i) 2ln(y)=2ln(-y)
However (i) is a contradiction as 2ln(y) =/= 2ln(-y) for any real number except 0
Thus we can conclude there isn't enough information to conclude sqrt(x)= -sqrt(x) on the standing basis of e1, e2 of R^2 except for the zero vector
>>8431438
:^)
Roots are tuples. Understand what you are misrepresenting yet?