Welcome to our community

Be a part of something great, join today!

radicals

bergausstein

Active member
Jul 30, 2013
191
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!
 
Last edited:

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
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 $|x+y|\neq x+y$?
 

bergausstein

Active member
Jul 30, 2013
191
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 $\left(\sqrt{(x+y)}\right)^2$
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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 [tex]\sqrt{a}= |a|[/tex] where a= x+ y. Yes, if x and y are both negative, then x+ y is negative so [tex]\sqrt{(x+ y)^2}= |x+ y|= -(x+ y)[/tex]. 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 [tex](x+ y)^2= (-25)^2= 625[/tex] and then [tex]\sqrt{(x+ y)^2}= \sqrt{625}= 25= -(x+ y)[/tex].

Yes, it is still true that [tex]\sqrt{(x+ y)^2}= \left(\sqrt{x+ y}\right)^2[/tex] as long as [tex]x+ y\ge 0[/tex]. If x+ y< 0 then [tex]\sqrt{x+ y}[/tex] 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 [tex]\sqrt{a^2}= |a|[/tex] is no longer true.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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.