## anonymous one year ago 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?

1. arindameducationusc

Are you there @emcrazy14 ?

2. ganeshie8

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

3. ganeshie8

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

4. ganeshie8

consider a concrete example : why $$\large 0.8^2$$ is less than $$\large 0.8$$ ?

5. anonymous

@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?

6. ganeshie8

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

7. anonymous

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

8. ganeshie8

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

9. anonymous

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

10. anonymous

but greater than 0*

11. Michele_Laino

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 $\Large \log x > \log y \Rightarrow x > y$

12. anonymous

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

13. Michele_Laino

:)