What is the Square Root Property in Mathematics?

[cbarker1]

Dear everyone,

I have a question about a property of square root.
$${\frac{1}{x}\sqrt{x^2}}$$$\implies$ $\sqrt{\frac{x^2}{x^2}}$=$\left| 1 \right|$

Is that property of a square root? Since
$$\sqrt{x^2}$$= $\left| x \right|$.

 

[chisigma]

Dear everyone,

I have a question about a property of square root.
$${\frac{1}{x}\sqrt{x^2}}$$$\implies$ $\sqrt{\frac{x^2}{x^2}}$=$\left| 1 \right|$

Is that property of a square root? Since
$$\sqrt{x^2}$$= $\left| x \right|$.

The question can be a little controversial...

a) You can consider the square root of a positive number as a one value function and in this case is $\displaystyle \sqrt{1} = 1$...

b) You can consider the square root of a positive number as a two values function and in this case is $\displaystyle \sqrt{1} = \pm 1$...

Kind regards

$\chi$ $\sigma$

 

[MarkFL]

I have often seen this treated as:

\(\displaystyle \frac{\sqrt{x^2}}{x}=\frac{|x|}{x}=\text{sgn}(x)\equiv\begin{cases}-1, & x<0 \\[3pt] 1, & 0<x \\ \end{cases}\)

 

[topsquark]

Please correct me if I'm wrong, but I learned the convention that if you take the square root of a number then it's considered positive. But if you take the square root of a variable, then it's plus or minus.
\(\displaystyle \sqrt{4} = 2 \text{ and } \sqrt{x^2} = \pm x\).

Math conventions in Physics always get me into trouble. (Doh) I had an interview with a Professor from NCU and overheard a "discussion" between her and one of her grad students. It seems that she insisted that \(\displaystyle 0^0 = 1\) and was rather aggressive about it. So, of course, when I walked in for the interview I just had to ask if \(\displaystyle 0^0 = 1\) was a new convention as it should be undefined. (Wasntme) She made sure that I understood that she felt I needed "remedial Math" before I could work on a dissertation.

-Dan

 

[I like Serena]

Dear everyone,

I have a question about a property of square root.
$${\frac{1}{x}\sqrt{x^2}}$$$\implies$ $\sqrt{\frac{x^2}{x^2}}$=$\left| 1 \right|$

Hi Cbarker1,

I'm afraid that it is not generally true.
$$\sqrt{\frac 1{x^2}} = \left|\frac 1 x\right|$$
This is not the same as $\frac 1 x$.

What you write is only true if it is given that $x \ge 0$.

 

[MarkFL]

Please correct me if I'm wrong, but I learned the convention that if you take the square root of a number then it's considered positive. But if you take the square root of a variable, then it's plus or minus.
\(\displaystyle \sqrt{4} = 2 \text{ and } \sqrt{x^2} = \pm x\).

Math conventions in Physics always get me into trouble. (Doh) I had an interview with a Professor from NCU and overheard a "discussion" between her and one of her grad students. It seems that she insisted that \(\displaystyle 0^0 = 1\) and was rather aggressive about it. So, of course, when I walked in for the interview I just had to ask if \(\displaystyle 0^0 = 1\) was a new convention as it should be undefined. (Wasntme) She made sure that I understood that she felt I needed "remedial Math" before I could work on a dissertation.

-Dan

It is my understanding that:

\(\displaystyle |x|\equiv\sqrt{x^2}\)

It is also my understanding that $0^0$ is indeterminate.

 

[I like Serena]

Please correct me if I'm wrong, but I learned the convention that if you take the square root of a number then it's considered positive. But if you take the square root of a variable, then it's plus or minus.
\(\displaystyle \sqrt{4} = 2 \text{ and } \sqrt{x^2} = \pm x\).

From wikipedia:

Every non-negative real number a has a unique non-negative square root, called the principal square root, which is denoted by √a, where √ is called the radical sign or radix.​

Is see no ambiguity.
The symbol √ represents the principal square root, which is by definition non-negative for real numbers.

Math conventions in Physics always get me into trouble. (Doh) I had an interview with a Professor from NCU and overheard a "discussion" between her and one of her grad students. It seems that she insisted that \(\displaystyle 0^0 = 1\) and was rather aggressive about it. So, of course, when I walked in for the interview I just had to ask if \(\displaystyle 0^0 = 1\) was a new convention as it should be undefined. (Wasntme) She made sure that I understood that she felt I needed "remedial Math" before I could work on a dissertation.

$0^0$ is more ambiguous.

It depends on the type of the exponent.
If the exponent is an integer, $0^0$ is usually defined as 1.

But if we're talking about continuous exponents, and if the form $0^0$ arises as a limit of $f(x)^{g(x)}$, it must be handled as an indeterminate form.

If not explicitly defined in the text, it could depend on the field you're working in.
Which field did the particular Professor work in?

 

[Dethrone]

I thought this was quite interesting:

Q: What does 0^0 (zero raised to the zeroth power) equal? Why do mathematicians and high school teachers disagree? | Ask a Mathematician / Ask a Physicist

 

[cbarker1]

If I confused any one, sorry. I picked out the certain step.
I am Deriving the cubic formula. I have trouble figuring out what how the Quadratic formula forms into the "R" and how the missing one half.

Here is the work shown:

$u^3-\frac{p^3}{27u^3}+q=0$
I multiply the 27u^3 both sides that yields
$27u^6-27u^3q+p^3=0$
divided by 27
$u^6-u^3q+\frac{p^3}{27}=0$
I use the Quadratic Formula:
$u^3=\frac{-q\pm\sqrt{(-q)^2+4(1)(\frac{p^3}{27}})}{2(1)}$
Therefore, $u^3=\frac{-q}{2}\pm\frac{\sqrt{\frac{27q^2+4p^3}{27}}}{2}$
If we let $R=\left(\frac{-q}{2}\right)^{\!{2}}+\left(\frac{p}{3}\right)^{\!{3}}$, then $u^3=\frac{-q}{2}\pm\sqrt{R}$.

 

[I like Serena]

Therefore, $u^3=\frac{-q}{2}\pm\frac{\sqrt{\frac{27q^2+4p^3}{27}}}{2}$
If we let $R=\left(\frac{-q}{2}\right)^{\!{2}}+\left(\frac{p}{3}\right)^{\!{3}}$, then $u^3=\frac{-q}{2}\pm\sqrt{R}$.

Since the denominator $2$ is positive, we can bring it into the square root without sign problems.
$$\frac{\sqrt{\frac{27q^2+4p^3}{27}}}{2}
=\sqrt{\frac{27q^2+4p^3}{2^2\cdot27}}
=\sqrt{\frac{q^2}{2^2}+\frac{p^3}{27}}
=\sqrt{\left(\!\frac{q}{2}\!\right)^{\!2}+\left(\!\frac{p}{3}\!\right)^{\!3}}
=\sqrt{\left(\!\frac{-q}{2}\!\right)^{\!2}+\left(\!\frac{p}{3}\!\right)^{\!3}}
$$

 

[topsquark]

Which field did the particular Professor work in?

String Theory. From what I've studied about it I've never even seen 0^0 come up. Physicists are funny creatures...they define one thing in one paper and change it to suit their needs in the next. (I try hard not to be that bad. That's one of the reasons I try to take Math classes from a Math professor, not a Physics one.)

-Dan