MHB Radicals and using the definition

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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.
 
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Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.

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