MHB Radicals and using the definition

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The discussion highlights the necessity of using the definition $\sqrt{x^2} = |x|$ when evaluating $\sqrt{(x+y)^2}$, which results in $|x+y|$. This is crucial because $|x+y|$ can differ from $x+y$ when $x+y$ is negative, particularly when both $x$ and $y$ are negative. In contrast, $\left(\sqrt{(x+y)}\right)^2$ is only valid if $x+y \geq 0$, as the square root of a negative number is undefined in the real number system. The distinction lies in the fact that squaring a number always yields a non-negative result, which complicates the reverse operation of taking the square root. Therefore, the two expressions are not equivalent in all cases, especially when dealing with negative sums.
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can you tell me if there's a necessity to use the definition:

$\displaystyle \sqrt{x^2}=|x|$

to this,

$\displaystyle \sqrt{(x+y)^2}$

if yes, why? if not why?

and how it is different to

$\displaystyle \left(\sqrt{(x+y)}\right)^2$

thanks!
 
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bergausstein said:
can you tell me if there's a necessity to use the definition:

$\displaystyle \sqrt{x^2}=|x|$

to this,

$\displaystyle \sqrt{(x+y)^2}$

if yes, why? if not why?

and how it is different to

$\displaystyle \left(\sqrt{(x+y)}\right)^2$

thanks!
You already know that any positive real number has two distinct square roots. One is positive and one is negative. Now suppose you want a machine (a function to be more precise) which takes a positive real number as the input and returns you the positive square root of the inputted number. Call this machine $M$.

It is then easy to see that $M(x^2)=|x|$.

It so happens that the standard notation for $M$ is actually $\sqrt{(\,)}$. It is as simple at that.

You can similarly have a machine $N$ which returns the negative square root of a given number.

It is again easy to see that $N(x^2)=-|x|$.

You can show further that $N\equiv-\sqrt{(\,)}$.

Tell me if you have any more questions.

To answer your question about how $\sqrt{(x+y)^2}$ is different from $x+y$, note that $\sqrt{(x+y)^2}$ is $|x+y|$.
Can you think of numbers $x$ and $y$ where $|x+y|\neq x+y$?
 
To answer your question about how $\sqrt{(x+y)^2}$ is different from $x+y$, note that $\sqrt{(x+y)^2}$ is $|x+y|$.
Can you think of numbers $x$ and $y$ where $|x+y|\neq x+y$?

when x and y are negative numbers. am I right? can you show me some examples.

i thought that

$\sqrt{(x+y)^2}$ is the same as $\left(\sqrt{(x+y)}\right)^2$
 
bergausstein said:
when x and y are negative numbers. am I right? can you show me some examples.

i thought that

$\sqrt{(x+y)^2}$ is the same as $\left(\sqrt{(x+y)}\right)^2$
First, the fact that this is a sum is irrelevant. It is simply a matter of \sqrt{a}= |a| where a= x+ y. Yes, if x and y are both negative, then x+ y is negative so \sqrt{(x+ y)^2}= |x+ y|= -(x+ y). But they don't have to both be negative, just that x+ y be negative.

For example, it x= -30 and y= 5, then x+ y= -25 so that (x+ y)^2= (-25)^2= 625 and then \sqrt{(x+ y)^2}= \sqrt{625}= 25= -(x+ y).

Yes, it is still true that \sqrt{(x+ y)^2}= \left(\sqrt{x+ y}\right)^2 as long as x+ y\ge 0. If x+ y< 0 then \sqrt{x+ y} does not even exist (as a real number). If we extend to the complex numbers, the square root function is no longer singly valued so \sqrt{a^2}= |a| is no longer true.
 
Imagine you have an elite calculator that can understand verbal instructions, and give verbal answers to certain mathematical questions.

So, if you tell this calculator: "tell me the square root of 9", it replies, "Three".

Now let's give this super-duper android some stuff to do.

Our idea is simple: first we'll give it a number, then ask it's square, then ask for the square root of the square. In diagram form:

$a \to a^2 \to \sqrt{a^2}$

Then, we'll do the steps in the reverse order (because our android is just THAT good):

$a \to \sqrt{a} \to (\sqrt{a})^2$.

We'll ask our cyborg friend to tell us what the "current state" is, after each step. Ok, ready? Let's go!

"Android, the input $a$ is $9$."

Our android does the first routine:

"9...calculating...81...calculating...9"

Next he (she? who knows?) does the second routine:

"9...calculating...3...calculating...9".

Well, both methods seem to give the same answer. Huh.

Let's try a different number:

"Android, the input $a$ is $-4$.

Androidess whirrs:

"-4...calculating...16...calculating...4".

Now for the second routine:

"-4...calculating...calculating...calc...ERROR! ERROR! routine undefined...a34eeee00x1...coredmp."hello.world"/daisy...dai...(bleeeeeeep)"

What went wrong? Android got confused when computing $\sqrt{-4}$.

Now we could get around this with $hotfixpatch/complex.numbers$, in which case Android might respond (with perhaps a bit less bravado):

"-4...calculating...(switch to patch mode)...calculating...$2i$...calculating...-4".

Now our two routines give different answers. So there must be something different about:

$\sqrt{a^2}$, and:

$(\sqrt{a})^2$

having to do with whether or not $a < 0$.

You see, squaring is "sneaky", it always spits out a positive number, even if we start with a negative one. So when we "unsquare" (take the square root), we might not get out what we started with:

$2 \to 4 \to 2$ (OK!)
$-2 \to 4 \to 2$ (what the...?)

Trying to "unsquare" a negative number leads to a peculiar problem: we feel that for $k > 0$ that $\sqrt{-k}$ ought to be "the same size" as $\sqrt{k}$, but neither $\sqrt{k}$ nor $-\sqrt{k}$ seems to do the trick. So whatever $\sqrt{-k}$ is, it's NOT on the normal "number line", it's off in some other direction.
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...

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