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Algebra Level 2

What is bigger, $\displaystyle{\frac{2017^2-2016^2}{2017\cdot2016}\;\mathrm{or}\;10^{-3}}?$

They are equal

10^{-3}

\displaystyle{\frac{2017^2-2016^2}{2017\cdot 2016}}

1 solution

Brian Charlesworth
May 29, 2017

$\dfrac{2017^{2} - 2016^{2}}{2017 \times 2016} = \dfrac{(2017 - 2016)(2017 + 2016)}{2017 \times 2016} \lt \dfrac{2 \times 2017}{2017 \times 2016} = \dfrac{1}{1008} \lt \dfrac{1}{1000} = 10^{-3}$ .

Thus $\boxed{10^{-3}}$ is the greater of the two given values.

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