A question about Square roots

I Believe I'm missing something.

If the √1 = ± 1
does this mean 1 = -1 ?

if = means equality then..

1 = √1 = -1

What am i missing ?

the square root of 1 is 1 sir

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

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

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 ?

Yes.

if 1=-1
then √1=√-1

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

-2*-2/-2=2

oops i meant

-2*-2/2=2

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'.

the square root of 4 is 2 and only 2

square roots are defined as positive

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

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

right about here
>-√4 = √4

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

well, you can always go to Complex analys.

√-1=√(i^2)=±i

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

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

[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

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.

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

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

what the fuck is this troll thread?

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

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

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.

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

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.

> "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.

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

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

Didn't define f/10