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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?
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?

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arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0Are you there @emcrazy14 ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2maybe first think why the "cube of a number" is less than the "square of a number" when the number is between 0 and 1

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2or start here : if the number is between 0 and 1, why the "square of a number" is less than the "actual number" ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2consider a concrete example : why \(\large 0.8^2\) is less than \(\large 0.8\) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@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?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2before that, do we see why squaring a number between 0 & 1 gives a smaller result ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes. \[0.8^{2}\] is like solving for 80% of 0.8 which will definitely be less than 0.8.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Excellent! so from that how are we going to conclude that \(\large 0.8^3 \lt 0.8^2\) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Multiplying 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1hint: 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 \[\Large \log x > \log y \Rightarrow x > y\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wow. Thank you so much guys. It makes sense now. :) @Michele_Laino @Nishant_Garg @ganeshie8
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