Math Memes

File: mathsqrt.jpg (31KB, 654x960px) Image search: [Google]

31KB, 654x960px

Can anyone explain this?????????!!111

>>8591463
Second to third line is wrong. We do not generally have root(ab) = root(a)*root(b) if root a or b is complex.

>>8591463
[math]\displaystyle \sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b}[/math] only works for positive real numbers.

>>8591463
Quite simple, OP. When you take the square root of a number, the result can be both positive and negative. For example:
[code]
sqrt(16) = +-4; 4^2 = 16, -4^2 = 16
[/code]
The bottom line is [code]- sqrt(3)[/code], which is the same as -1 * sqrt(3). Since the root of 3 is positive and negative (the roots, rather), a -1 thrown in the mix would only serve to change the plus-or-minus to minus-or-plus, which is the same thing. Therefore, sqrt(3) does equal -sqrt(3)

>>8592961
this is b8, do not reply

>>8591461
i don't get the series = 1/12 meme

>>8593878
Then you're underage and/or American because it comes from [math]\displaystyle \zeta(-1)[/math].

>>8593883
Maybe you're the retard

>>8593962
>I get proven wrong with mathematical facts
>I must protect my honour by calling them a retard, I'm never wrong

File: HowInternetArgumentSeemToWork-66065.jpg (43KB, 499x306px) Image search: [Google]

43KB, 499x306px

>>8593962
fuck off you mong

>>8593991
What the fuck is the π button for?

>>8593991
3.1415926535

>>8592961
>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]

>>8593986
this phenomenon is insufferable

people don't deserve their ego most of the time

>>8594063
Not even the guy you replied, but he said that the result CAN, not WILL be both positive and negative

>>8594046
You only need to remember the first 5 decimal places mate.

>>8594223
no, it can't be negative

>>8591463
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.

>>8594223
>>8594063
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.

>>8592961 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!
:)

>>8594302
shit did i say bijective i meant injective don't mind me

>>8594303
Shitty Norman math meme

>>8591461

where is the original proof of OPs pic?

>>8594307
>trip

Pls kill yourself.

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

>>8594231
I was just explaining what the button did.

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PhD Trippel Integrals

>>8591463
[math]
He is just multiplying by -1. See.
-1 * \sqrt{3}
\sqrt{(-1)^2} * \sqrt{3}
Then:
\sqrt{(-1)*(-1)*(3)}
[/math]

>>8594475
He is just multiplying by -1.
See.
[math]
-1* \sqrt{3} = \sqrt{(-1)^2} * \sqrt{3} = \sqrt{(-1)*(-1)*(3)}
[/math]

>>8594304
...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.

>>8591463
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.

>>8591473
Can you prove that this works for all positive reals?

>>8594505
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.

>>8593878
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.

>>8595162
>a had an approximate result of 0.5, thus b is approximately 0.25.
You cant be this retarded

>>8595162
this is not a proof, this is retarded and chances are you watched it on some shit "math" channel

>>8591463
Third step is wrong, you can not take i, out of a positive number.
Sqrt (i)*sqrt (-x)=sqrt (-1*-x) is not valid.

>>8594430
underrated meme

>>8594430
1.0
Engineers always laugh last.

>>8594430
Explain

>>8595162
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.

>>8595672
https://www.youtube.com/watch?v=xgBGibfLD-U

>>8595692
Man that guys fucking face. It's like the cameraman is holding a gun to his mother's head or something offscreen.

>>8594430
I cry every time.

>>8594303
HAHAHAHAHAHAHHAHAHAHAHAHAHAHAHHAHAHAHAHAHAHAHAHHAHAHAHAHAHAHAHAHHAHAHAHAHAHAHAHAHHAHAHAHAHAHAHAHAHHAHAHAHAHAHAHAHAHHAHAHAHAHAHAHAHAHHAHAHAHAHAHAHAHAHHAHAHAHAHAHAHAHAHHAHA

>>8594313
>>8593878
https://www.youtube.com/watch?v=w-I6XTVZXww

>>8595162
>>8595191
>>8595679

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.

>>8591463
You didn't have to write all that to prove it.
(-1)^2(sqrt3)^2=3

>>8595190
>>8596289

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

>>8596297
The confusion stems from the result sqrt(3)=-sqrt(3) and this can only be resolved by saying that (-x)^2=x^2