Why Does the Equation ##\sqrt{a^2} = a## Lead to Confusion in Mathematics?

[Mr Davis 97]

If we define $|a| = \sqrt{a^2}$, then why can't we do something like $\sqrt{a^2} = (\sqrt{a})^2 = a$? Or equivalently $\sqrt{a^2} = (a^2)^{1/2} = a^{2/2} = a$? Isn't this a contradiction?

Also, how would this relate to showing that $\sqrt{|a|} = |\sqrt{a}|$ is true or false?

 

[Buzz Bloom]

Hi Mr Davis:
When you take a square-root the result may be + or -. Therefore the definition of √a² should be (+/-) a.

Regarding

Also, how would this relate to showing that √|a|=|√a|\sqrt{|a|} = |\sqrt{a}| is true or false?

I am not sure how this topic is taught this century, but I would say the answer is neither "true" or "false", but rather the answer is "incomplete", since it is also correct that

√|a| = -|√a|.​

Hope this helps.

Regards,
Buzz

 

[FactChecker]

The $\sqrt{ a }$ symbol with no sign in front of it is understood by convention to be the positive value. If you want the negative value, you must put a minus sign in front. It could have been done other ways, but it would have no advantage or would be ambiguous.
Many of your other questions are ignoring that the square root of a negative number is a problem. So you can not say that √(a²) = (√ a )² = a if a is negative.

 

[Buzz Bloom]

So you can not say that √(a²) = (√ a )² = a if a is negative.

Hi @FactChecker:

Would you agree that for a < 0 it is OK to say:

(√ a )² = - √(a²) ?​

Regards,
Buzz

 

[FactChecker]

Hi @FactChecker:

Would you agree that for a < 0 it is OK to say:

(√ a )² = - √(a²) ?​

Regards,
Buzz

Depends on whether complex numbers are allowed. Staying within the reals, I would say that the left side is undefined.

 

[Buzz Bloom]

Hi @FactChecker:

Another question occurred to me.

What does the notation (+/-)√a means by current conventions?

Is is ambiguous? Or does it mean both + and - values are possible as correct interpretations? Or does it mean that either one or the other value, or both values may be a correct interpretation?

Regards,
Buzz

 

[FactChecker]

Hi @FactChecker:

Another question occurred to me.

What does the notation (+/-)√a means by current conventions?

Is is ambiguous? Or does it mean both + and - values are possible as correct interpretations? Or does it mean that either one or the other value, or both values may be a correct interpretation?

Regards,
Buzz

Good question. In the places I have seen it, I always interpreted it as "try both options and use any and all that work". But I don't know if that is always how it should always be interpreted.

 

[Mark44]

Therefore the definition of √a² should be (+/-) a.

I agree with FactChecker. By tradition, the square root symbol represents the positive square root of a number. (I am considering only the square root of a nonnegative real number.)
$\sqrt{(-4)^2} = |-4| = 4$.

It is not $\pm 4$.

 

[Stephen Tashi]

Hi Mr Davis:
When you take a square-root the result may be + or -. Therefore the definition of √a² should be (+/-) a.

That is not correct. Even people who define the phrase "square root" so it allows some numbers to have two different square roots (such as the current Wikipedia article https://en.wikipedia.org/wiki/Square_root ) also define the notation "$\sqrt{}$" to mean the principal square root. Many math texts disagree with the Wikipedia definition and define the (verbal) phrase "square root" so that a number has at most one square root.

Computing $\sqrt{a^2}$ is a different task than solving the equation $x^2 = a^2$.

The solutions to $x^2 = a^2$ are members of the set $\{-a, a\}$ and this fact may be abbreviated by the notation $x = \pm a$.

 

[Mr Davis 97]

I am still confused. The manipulation $\sqrt{x^2} = x$ seems valid by the rules of exponents, just as the manipulation $(x^3)^{1/3} = x$ is valid. Under what conditions is this valid or invalid, and does it also depend on the underlying field (real or complex)?

 

[FactChecker]

Don't worry about complex numbers. The problem is when x is negative. $\sqrt{x^2}$ is always positive.

 

[Mr Davis 97]

Don't worry about complex numbers. The problem is when x is negative. $\sqrt{x^2}$ is always positive.

So when exactly does the exponent rule $(x^a)^{1/a} = x^{a/a} = x$ hold true? We established that it doesn't when a = 2, but it does when a = 3.

 

[FactChecker]

It is the √ symbol that causes the problem. I believe that it implies the positive value. When you use general exponents, you are free to interpret as you need, but when there are multiple roots, you should be clear about which roots you mean.

 

[Stephen Tashi]

I am still confused. The manipulation $\sqrt{x^2} = x$ seems valid by the rules of exponents,

What rules of exponents are you referring to? If $x = -2$ then $\sqrt{x^2} = -x$.

Under what conditions is this valid or invalid,

What do you mean by "this"? Are you asking about the validity of $\sqrt{x^2} = x$ or about the validity of $(x^3)^{1/3} = x$ or about the general thought that $(x^a)^b = x^{ab}$?

$\sqrt{x^2} = x$ is valid for those $x \ge 0$.

and does it also depend on the underlying field (real or complex)?

In the complex domain, a non-zero number has n distinct n-th roots. Since a non-zero $x$ has 3 distinct cube roots, the notation $(x^3)^{1/3} = x$ is ambiguous unless we define which cube root is to be taken. If we use the convention that $b^{1/3}$ shall mean the real number (i.e. the one with zero imaginary part) that is a cube root of $b$ then $(x^3)^{1/3}$ is not valid as a generality for all complex numbers $x$. It is valid for those $x$ in the subset of the complex numbers that consists of the real numbers (i.e. those with no imaginary part).

In the domain of real numbers, $(x^3)^{1/3} = x$ is valid. However, in the domain of real numbers, $(x^a)^b = x^{ab}$ is not valid as a generality for all $x,a,b$. For example, with $x = -1, a = 2/3, b = 1/a = 3/2$ we have $x^a = ( (-1)^2) ^{1/3} = 1^{1/3} = 1$ and $(x^a)^b = (1)^{3/2} = 1$ but $x^{ab} = x^1 = (-1)^1 = -1$.

So when exactly does the exponent rule $(x^a)^{1/a} = x^{a/a} = x$ hold true? We established that it doesn't when a = 2, but it does when a = 3.

In the domain of real numbers, we can give sufficient conditions for it to hold true, but it would be harder to describe all possible instances when it does hold true. It is sufficient that $a$ be an odd integer