Math Memes

fuck off you mong

What the fuck is the π button for?

3.1415926535

>When you take the square root of a number, the result can be both positive and negative

[math] \displaystyle
\sqrt {x^2} \ne \pm x, \quad \sqrt {x^2} = \left | x \right |
\\
|x| =
\begin{cases}
\;\;\; x & ,x \geq 0 \\
-x & ,x < 0
\end{cases}
[/math]

this phenomenon is insufferable

people don't deserve their ego most of the time

Not even the guy you replied, but he said that the result CAN, not WILL be both positive and negative

You only need to remember the first 5 decimal places mate.

no, it can't be negative

The answer is simple. The square root function is not well defined and the fact that no one makes a big deal out of this in lower education is a sign of educational malpractice.

You could "prove" the same thing even without complex numbers. Simply set

(-1)^2 = 1^2
-1 = 1

Adding complex numbers to your argument is really only tricking brainlets into thinking that something deep is going on.

The confusion you guys are having comes from the fact that the sqrt function is not surjective, while the square function happens to be bijective, so we would kind of want -sqrt(3) and sqrt(3) to correspond to 3, but we don't do that for real analysis reasons.

is wrong in saying that -sqrt(3)=sqrt(3) of course, but is right about everything else.

Perhaps if you actually read what he said it would've made a bit of sense but you are too thirsty to post your sexy latex all over to try and proove people wrong

Why is 6 afraid of 7?
Because 7 8 9!
:)

shit did i say bijective i meant injective don't mind me

Shitty Norman math meme

where is the original proof of OPs pic?

>trip

Pls kill yourself.

[eqn](1+9^{-4^{6\times 7}})^{3^{2^{85}}}[/eqn]

I was just explaining what the button did.

PhD Trippel Integrals

[math]
He is just multiplying by -1. See.
-1 * \sqrt{3}
\sqrt{(-1)^2} * \sqrt{3}
Then:
\sqrt{(-1)*(-1)*(3)}
[/math]

He is just multiplying by -1.
See.
[math]
-1* \sqrt{3} = \sqrt{(-1)^2} * \sqrt{3} = \sqrt{(-1)*(-1)*(3)}
[/math]

...it's not injective either. "[math]f: X \rightarrow Y[/math] is injective" means [math]\forall a,b \in X \quad f(a)=f(b) \Rightarrow a=b[/math].
I think.

You can't do:
[math]
\sqrt{(-1)*(-1)*(3)} \neq \sqrt{(-1)} \sqrt{(-1)} \sqrt{3}[/math]

The number inside the root is 3, who can be represented by -1*-1*3, but it doesn't mean you can split all around.

Can you prove that this works for all positive reals?

It completely depends on what your domain is. If your domain is natural numbers, squaring is injective.
If your domain is the whole numbers, it is neither surjective nor injective.
When using real numbers, it is still neither.
With complex numbers, squaring is indeed bijective, because you can get the square root of 1, -1, i, and -i. So you can get a result for every square root and also a result for every square.

Well, there are two different proofs for this. One uses approximations of infinite sums and one uses the Zeta-function of Riemann. I'll give you the easier with the sums.

First some definitions:
1-1+1-1+1-1...=a
1-2+3-4+5-6...=b
1+2+3+4+5...=c

Now we add b to b and add an offset of one, so the pairings are the following:
2b=1+(-2+1)+(3-2)+(-4+3)...

That is equal to 1-1+1-1+1-1...=a
So 2b=a

a had an approximate result of 0.5, thus b is approximately 0.25.

Now calculate c-b.
c-b = (1-1)+(2+2)+(3-3)+(4+4)...=0+4+0+8+0+12+0+16...=4+8+12+16...=4*(1+2+3+4...)=4c

So,
c-b=4c
-b=3c
-1/4=3c
-1/12=c=1+2+3+4...

QED.

>a had an approximate result of 0.5, thus b is approximately 0.25.
You cant be this retarded

this is not a proof, this is retarded and chances are you watched it on some shit "math" channel

Third step is wrong, you can not take i, out of a positive number.
Sqrt (i)*sqrt (-x)=sqrt (-1*-x) is not valid.

underrated meme

1.0
Engineers always laugh last.

Explain

You can't just couple values together of a non-converging series like a or b. That's equivalent to sampling a sinewave every half period and starting on a zero. And then conclude sin(x) converges.

youtube.com/watch?v=xgBGibfLD-U

Man that guys fucking face. It's like the cameraman is holding a gun to his mother's head or something offscreen.

I cry every time.

HAHAHAHAHAHAHHAHAHAHAHAHAHAHAHHAHAHAHAHAHAHAHAHHAHAHAHAHAHAHAHAHHAHAHAHAHAHAHAHAHHAHAHAHAHAHAHAHAHHAHAHAHAHAHAHAHAHHAHAHAHAHAHAHAHAHHAHAHAHAHAHAHAHAHHAHAHAHAHAHAHAHAHHAHA

youtube.com/watch?v=w-I6XTVZXww

s=1-1+1-1+1-1+1-1+...
1-s=1-(1-1+1-1+1-1+...)=1-1+1-1+1-1+...=s
1-s=s => 1=2s => 1/2=s
QED.

You didn't have to write all that to prove it.
(-1)^2(sqrt3)^2=3

I'm sorry, I forgot you in my proof that a can be approximated with .5

The confusion stems from the result sqrt(3)=-sqrt(3) and this can only be resolved by saying that (-x)^2=x^2