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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?

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  1. arindameducationusc
    • one year ago
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    Are you there @emcrazy14 ?

  2. ganeshie8
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    consider a concrete example : why \(\large 0.8^2\) is less than \(\large 0.8\) ?

  5. anonymous
    • one year ago
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    @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
    • one year ago
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    before that, do we see why squaring a number between 0 & 1 gives a smaller result ?

  7. anonymous
    • one year ago
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    Yes. \[0.8^{2}\] is like solving for 80% of 0.8 which will definitely be less than 0.8.

  8. ganeshie8
    • one year ago
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    Excellent! so from that how are we going to conclude that \(\large 0.8^3 \lt 0.8^2\) ?

  9. anonymous
    • one year ago
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    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
    • one year ago
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    but greater than 0*

  11. Michele_Laino
    • one year ago
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    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
    • one year ago
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    Wow. Thank you so much guys. It makes sense now. :) @Michele_Laino @Nishant_Garg @ganeshie8

  13. Michele_Laino
    • one year ago
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    :)

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