Finding square root of number i.e. ##\sqrt{\dfrac{16}{64}}##

[chwala]

Homework Statement

I am finding a common mistake made by my students on finding square roots of numbers like;

Find $\sqrt{\dfrac{16}{64}}$

Relevant Equations

square roots.

The correct answer is; $\sqrt{\dfrac{16}{64}}=\dfrac{4}{8}$ .

I do not seem to understand why some go ahead to simplify $\dfrac{4}{8}$ and getting $\dfrac{1}{2}$ which is clearly wrong. I do not know if any of you are experiencing this... I guess more emphasis on my part. Cheers!

Your insight is welcome guys.

 

[WWGD]

This is going to kill you: If you creatively cancel the 6 on the numerator with the 6 in the denominator, you will get the right answer somehow.
Edit: Still, why do you believe 1/2 is wrong, given it's equal to 4/8?
Simplify inside the root, and you will see how to obtain 1/2. 16 in the numerator vs 16(4) on the denominator.

 

[PeroK]

Homework Statement:: I am finding a common mistake made by my students on finding square roots of numbers like;

Find $\sqrt{\dfrac{16}{64}}$
Relevant Equations:: square roots.

The correct answer is; $\sqrt{\dfrac{16}{64}}=\dfrac{4}{8}$ .

I do not seem to understand why some go ahead to simplify $\dfrac{4}{8}$ and getting $\dfrac{1}{2}$ which is clearly wrong. I do not know if any of you are experiencing this... I guess more emphasis on my part. Cheers!

Your insight is welcome guys.

Surely $\dfrac 4 8 = \dfrac 1 2$?

 

[chwala]

Surely $\dfrac 4 8 = \dfrac 1 2$?

That is where i have a problem with; my understanding is that we should be able to move from squares to square root in both ways i.e

$\sqrt {a^2}=a$ such that $a× a=a^2$
For eg.

$\sqrt{\dfrac{16}{64}}=\sqrt{\dfrac{4^2}{8^2}}=\dfrac{4}{8}$

$\dfrac{1}{2} ×\dfrac{1}{2} ≠\dfrac{16}{64}$

 

[chwala]

Surely $\dfrac 4 8 = \dfrac 1 2$?

This is where i need clarity...yes the two are equivalent but not in the context of the problem. I stand guided sir.

 

[PeroK]

That is where i have a problem with; my understanding is that we should be able to move from squares to square root in both ways i.e

$\sqrt {a^2}=a$ such that $a× a=a^2$
For eg.

$\sqrt{\dfrac{16}{64}}=\sqrt{\dfrac{4^2}{8^2}}=\dfrac{4}{8}$

Why do you think $\dfrac 4 8 \ne \dfrac 1 2$?

 

[chwala]

Why do you think $\dfrac 4 8 \ne \dfrac 1 2$?

$\dfrac{1}{2} ×\dfrac{1}{2} ≠\dfrac{16}{64}$

 

[PeroK]

$\dfrac{1}{2} ×\dfrac{1}{2} ≠\dfrac{16}{64}$

Nonsense!

 

[chwala]

Nonsense!

I see we can simplify $\dfrac{16}{64}=\dfrac{1}{4}=\dfrac{1}{2} × \dfrac{1}{2}$

...yes, i have been wrong on this... if the numbers are rational and simplifiable then we can simplify first before applying square roots.

My understanding has always been;

$\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$ ... by considering $a$ and $b$ as independent numbers ...totally missed out on the simplification bit.

 

[nuuskur]

$\sqrt{a^2} = |a|$ and $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$ is valid for nonnegative numbers.

 

[chwala]

I see we can simplify $\dfrac{16}{64}=\dfrac{1}{4}=\dfrac{1}{2} × \dfrac{1}{2}$

...yes, i have been wrong on this... if the numbers are rational and simplifiable then we can simplify first before applying square roots.

My understanding has always been;

$\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$ ... by considering $a$ and $b$ as independent numbers ...totally missed out on the simplification bit.

Lastly on this, is there a limit (rule) to the point at which one can simplify when finding square roots? you realize like in the current example;
$\dfrac{1}{2}$ can also be written as $\dfrac{0.5}{1}=\dfrac{0.25}{0.5}$ and so on...

What about simplifying upwards I.e

$\dfrac{1}{2}=\dfrac{1000}{2000}=\dfrac{10000}{20000}$ ...

would all this be mathematically correct?

 

[WWGD]

Lastly on this, is there a limit (rule) to the point at which one can simplify when finding square roots? you realize like in the current example;
$\dfrac{1}{2}$ can also be written as $\dfrac{0.5}{1}=\dfrac{0.25}{0.5}$ and so on...

What about simplifying upwards I.e

$\dfrac{1}{2}=\dfrac{1000}{2000}=\dfrac{10000}{20000}$ ...

would all this be mathematically correct?

Yes. Ultimately,
$\frac {a}{b}=\frac{c}{d}$ iff $ad=bc$

 

[Mark44]

The correct answer is;
$\sqrt{\dfrac{16}{64}}=\dfrac{4}{8}$ .

I do not seem to understand why some go ahead to simplify $\dfrac{4}{8}$ and getting $\dfrac{1}{2}$ which is clearly wrong.

This has already been responded to by several other members, but it bears repeating, for emphasis.
The above is correct, despite your assertion that it is "clearly wrong."

$\frac 4 8 = \frac {4 \cdot 1}{4 \cdot 2} = \frac 4 4 \cdot \frac 1 2 = \frac 1 2$.

The only difference between $\frac 4 8$ and $\frac 1 2$ is that the latter is simplified as much as possible.

Another take is that $\sqrt{\frac{16}{64}} = \sqrt{\frac 1 4} = \frac 1 2$.

BTW, @chwala, I'm very surprised that you're having problems with something so elementary, given that you have posted questions about calculus and differential equations (both ODE and PDE) and statistics.

 

[nuuskur]

$\dfrac{1}{2}=\dfrac{1000}{2000}=\dfrac{10000}{20000}$ ...

would all this be mathematically correct?

Yes, $x\cdot 1 = x$ for every $x$. In particular, $\frac{a}{b} = \frac{a}{b} \cdot \frac{c}{c}$.

 

[chwala]

This has already been responded to by several other members, but it bears repeating, for emphasis.
The above is correct, despite your assertion that it is "clearly wrong."

$\frac 4 8 = \frac {4 \cdot 1}{4 \cdot 2} = \frac 4 4 \cdot \frac 1 2 = \frac 1 2$.

The only difference between $\frac 4 8$ and $\frac 1 2$ is that the latter is simplified as much as possible.

Another take is that $\sqrt{\frac{16}{64}} = \sqrt{\frac 1 4} = \frac 1 2$.

BTW, @chwala, I'm very surprised that you're having problems with something so elementary, given that you have posted questions about calculus and differential equations (both ODE and PDE) and statistics.

...at times I make mistakes, just being human ...then I learn from my mistakes...and become better...

of course it's pretty obvious that $\dfrac {1}{2}=\dfrac{4}{8}$...my mind was way off! but hey, it happens to all of us! Cheers!

 

[chwala]

Yes. Ultimately,
$\frac {a}{b}=\frac{c}{d}$ iff $ad=bc$

This implies that there is no restriction to the form of solution that a student may use...as long as the fractions are equivalent...but more generally, simplifying downwards would be the ideal norm.

 

[WWGD]

This implies that there is no restriction to the form of solution that a student may use...as long as the fractions are equivalent...but more generally, simplifying downwards would be the ideal norm.

Yes, correct. A mathematical expression may take many equivalent forms. Not always clear when/if two such forms are equivalent.

 

[Math100]

Homework Statement:: I am finding a common mistake made by my students on finding square roots of numbers like;

Find $\sqrt{\dfrac{16}{64}}$
Relevant Equations:: square roots.

The correct answer is; $\sqrt{\dfrac{16}{64}}=\dfrac{4}{8}$ .

I do not seem to understand why some go ahead to simplify $\dfrac{4}{8}$ and getting $\dfrac{1}{2}$ which is clearly wrong. I do not know if any of you are experiencing this... I guess more emphasis on my part. Cheers!

Your insight is welcome guys.

The answer is $0.5$ if you enter $\sqrt{\dfrac{16}{64}}$ into any online calculator. So I wouldn't say $\frac{1}{2}$ is wrong. And $\frac{4}{8}=\frac{1}{2}$. It's the same thing.