If $3$ and $- 4$ are integers then $3 + (- 4) = - 1$ is also an $()$ .

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Solution

If we add or subtract the integers than the result is always an integer.

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[\frac{1}{1}] + [\frac{1}{2}] + [\frac{2}{2}] + [\frac{1}{3}] + [\frac{2}{3}] + [\frac{3}{3}] + [\frac{1}{4}] + [\frac{2}{4}] + [\frac{3}{4}] + [\frac{4}{4}] + [\frac{1}{5}] + \dots

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[\frac{1}{1}] + [\frac{1}{2}] + [\frac{2}{2}] + [\frac{1}{3}] + [\frac{2}{3}] + [\frac{3}{3}] + [\frac{1}{4}] + [\frac{2}{4}] + [\frac{3}{4}] + [\frac{4}{4}] + [\frac{1}{5}] + \dots

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(\sqrt[3]{4} + \sqrt[3]{2} - 2) (\sqrt[3]{4 a} + \sqrt[3]{4 a} + \sqrt[3]{2 b} + c) = 20

μ = {x / x is an integer, - 4 \leq x \leq 6}

P^{'} = {- 4, - 3, 0, 1, 2, 4, 6}

Q^{'} = {- 3, - 2, 1, 3, 6}

P

P = {- 2, - 1, 3, 5}

[x]

x

[[\frac{3}{4}] + [\frac{3}{4} + \frac{1}{100}] + [\frac{3}{4} + \frac{2}{100}] + \dots + [\frac{3}{4} + \frac{99}{100}]]

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