Cube or square

Algebra Level 1

$\large A = \left(\frac 35\right)^3 \quad \quad \quad B = \left(\frac 52\right)^2$

Compare these terms.

B > A A > B

2 solutions

Md Zuhair
Oct 5, 2016

We can see that $(3/5)^3$ = $(0.6)^3$ and and $(5/2)^2$ = $(2.5)^2$ , So now we can see that $(0.6)^3$ will tend more towards $0$ or for numbers lesser that $1$ we can see that their square tends more towards $0$ and for $(5/2)^2$ = $(2.5)^2$ will tend towards obviously towards number greater than $3$ . So $(2.5)^2$ > $(0.6)^3$

How can you directly judge that ${\left(2.5\right)}^2$ is greater that ${\left(0.6\right)}^3$ ? Can you provide a legit explanation for your conclusion?

Tapas Mazumdar - 4 years, 8 months ago

@Tapas Mazumdar Is it ok now?

Md Zuhair - 4 years, 8 months ago

Just fine. You can say that for whatever positive exponent a real number between $0$ and $1$ is raised it will always tend to zero as powers get large. Not just 'squares' (which you have incorrectly mentioned) or cubes.

Tapas Mazumdar - 4 years, 8 months ago

I think you are Bangali , Same here :)

Md Zuhair - 4 years, 8 months ago

Yes man. :)

Tapas Mazumdar - 4 years, 8 months ago

Viki Zeta
Oct 5, 2016

$\displaystyle 27 < 125 \\ \displaystyle 3^3 < 5^3 \\ \displaystyle \left(\dfrac{3}{5}\right)^3 < 1 \\ \displaystyle 4 < 25 \\ \displaystyle 2^2 < 5^2 \\ \displaystyle 1 < \left(\dfrac{5}{2}\right)^2 \\ \displaystyle \implies \left(\dfrac{3}{5}\right)^3 < 1 <\left(\dfrac{5}{2}\right)^2 \\ \displaystyle \implies \left(\dfrac{3}{5}\right)^3 < \left(\dfrac{5}{2}\right)^2 \\ \displaystyle \implies A < B \\ \displaystyle \implies B > A$

Cube or square

2 solutions

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