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.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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