Question

Why is the cube root of a decimal number larger than the square root of the same decimal number?
Let's say the number is 0.4225. It's square root is 0.65 while its cube root is 0.75037. Why is that?

Answer 1

Are you there @emcrazy14 ?

Answer 2

maybe first think why the "cube of a number" is less than the "square of a number" when the number is between 0 and 1

Answer 3

or start here :

if the number is between 0 and 1,
why the "square of a number" is less than the "actual number" ?

Answer 4

consider a concrete example :

why \(\large 0.8^2\) is less than \(\large 0.8\) ?

Answer 5

@ganeshie8 : Okay, so the logic here is that since squaring or taking cube of a number between 0 & 1 yields a smaller number than the original number, in the same way the square root & the cube root tend to increase?

Answer 6

before that, do we see why squaring a number between 0 & 1 gives a smaller result ?

Answer 7

Yes. \[0.8^{2}\] is like solving for 80% of 0.8 which will definitely be less than 0.8.

Answer 8

Excellent!
so from that how are we going to conclude that \(\large 0.8^3 \lt 0.8^2\) ?

Answer 9

Multiplying a number with a number smaller than 1 but great than 0 decreases it
\[5 \times \frac{1}{2}=2.5<5\]
The same is true for numbers smaller than 1 but greater than itself
\[\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}<\frac{1}{2}\]
Now suppose
\[x=(\frac{1}{2})^6=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \]
\[\sqrt{x}=\sqrt{(\frac{1}{2})^{6}}=(\frac{1}{2})^{3}=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\]
and
\[x^\frac{1}{3}=((\frac{1}{2})^6)^\frac{1}{3}=(\frac{1}{2})^2=\frac{1}{2} \times \frac{1}{2}\]
Notice in square root, we are multiplying 3 times, so it will become ever smaller
in cube root it is multiplied 2 times, so it will be as small

Answer 10

but greater than 0*

Answer 11

hint:
I think that we have to specify if that number is greater than1 or less than 1.
If:
\[\Large \begin{gathered}
x = \sqrt N \hfill \\
y = \sqrt[3]{N} \hfill \\
\end{gathered} \]
then:
\[\Large \begin{gathered}
\log x = \frac{1}{2}\log N \hfill \\
\log y = \frac{1}{3}\log N \hfill \\
\end{gathered} \]
so, if N>1, then
\[\Large \log x > \log y \Rightarrow x > y\]

Answer 12

Wow. Thank you so much guys. It makes sense now. :) @Michele_Laino @Nishant_Garg @ganeshie8

Answer 13

:)