A question about Square roots

>>8504060
the square root of 1 is 1 sir

File: 10397839_901626653278832_667037626012979884_n.jpg (5KB, 200x193px) Image search: [Google]

5KB, 200x193px

>>8504060
>What am i missing ?
A brain...
√1 does not equal -1
What you're thinking about is the solutions to x^2=1

>>8504060

I think you're confusing normal roots with the quadratic formula.

To be more specific, the square root of 1 is 1, but it is also -1.

√4 = 2 because 2 * 2 = 4.
But -2 * -2 is also 4, the √4 has two answers.

√1 = 1 and -1

Therefor 1 = -1

I'm missing some more complicated maths

>>8504073
the square root is conventionally defined as the positive solution to x^2=a

2 * 2 = 4 = -2 * -2
∴ 2 * 2 = -2 * -2
divide both sides by 2

2 = - 2
Is this wrong ?

>>8504080

Yes.

>>8504073
if 1=-1
then √1=√-1

Thus no need for the complex numbers. You sir has just revolutionized maths!

>>8504080

-2*-2/-2=2

>>8504083
oops i meant

-2*-2/2=2

>>8504080

yes, if you divide both side by 2 you get

2=(-1)*(-2) => 2=2

Alright, i see, Cancel the two but still left with -1.

I'm still not understanding why
if √4 = 2
and √4 = -2

-2 = √4 = 2
Solve this equation. Am i writing this wrong?

Should it be -2 = -√4 = √4 = 2
by solving the equation we add whatever to every 'side'.

>>8504095
the square root of 4 is 2 and only 2

>>8504060
square roots are defined as positive

>>8504097
you're saying 2 times itself is the only number that will be 4.

but -2 times itself is also 4. therefor 2 is not the only number that is the square root of 4

>>8504101
evaluate this with math and show me where its wrong
-2 = -√4 = √4 = 2

>>8504109
right about here
>-√4 = √4

a number inside the square route cannot be negative, therefore the "1=-1" is entirely incorrect

>>8504132

well, you can always go to Complex analys.

√-1=√(i^2)=±i

not 100% sure if you can do this, but atleast it works.

>>8504060
>If the √1 = ± 1
its not. sqrt(1) = 1.
you confuse it with x^2 = 1 -> x = +1 or -1.

>>8504060
[math] \displaystyle
\sqrt {x^2} \ne \pm x, \quad \sqrt {x^2} = \left | x \right |
\\
\left | x \right | =
\left \{
\begin{align}
x, & \hspace{1em} x \geq 0 \\
-x, & \hspace{1em} x < 0
\end{align}
\right .
[/math]

sqrt() uses the principal branch, so for positive real numbers sqrt(x) is going to be positive

>>8504102
Thats not what he's saying. He's saying that "the square root" of a number y is always defined to be the positive real number x such that x^2 = y. There are other values for x that also work, but if you're talking about the canonical function, then there is only one answer.

Saying sqrt(1) = 1 is only correct if you use the definition that I mentioned. If you're talking about all possible roots, then you need to say this:
"Sqrt(1) = 1 or -1"

You can't just pick one value, otherwise it leads to the contradiction in OP that 1 = -1.

>>8504060
It just means that the square root of one is 1 or -1
They don't equal each other

>>8504060
That whole matter confused me as well. The principle square root is 1.

what the fuck is this troll thread?

>>8504060
Ebin bait. The square root application is clearly defined as the POSITIVE number that solves x^2 = a.

x -> sqrt(x)
R+ -> R+

File: x^2=4 solutions.png (73KB, 1312x1258px) Image search: [Google]

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If you plot x^2 (the black curve), and see where its y-coordinate (ordinate) equals 4, there are two possible values for the corresponding x-coordinate: +2 and -2.

>>8504082
Actually, your statement is true because you began with a false premise.

File: infiniteseries.jpg.CROP.original-original.jpg (92KB, 590x337px) Image search: [Google]

92KB, 590x337px

>>8505361
Yes, and the positive one is √4. The negative one is -√4.

It's decided that way in normal mathematics. Of course, you can decide to have it your way where negative numbers are simply non-existent, but I don't think there's much use to it.

It's more-or-less the same reason pic related is true if you allow certain things not permitted in normal mathematics.

>>8504204
> "the square root" of a number y is always defined to be the positive real number x such that x^2 = y

This, OP.
Think about it this way: √1 = 1 ; but it also works with -1. The first solution has a name, and its the most commonly used result. I can't remember the name of it (I think it was the arithmetic solution, but im not really sure) and, for example, it's the one showed in the calculator.

>>8504060
What you're talking about isn't a function so the equality does not hold.

>>8504060
F(a)=F(b)
a=b
5*0=753769348957*0
5=753769348957

>>8507502
Didn't define f/10