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17 Αυγ 2017 · You can use this property to show that 1/2 is therefore not an integer (in fact, you can use this to show immediately that there is no integer $n\in \mathbb{Z}$ such that $0<z<1$ by essentially the same argument!)
30 Δεκ 2014 · The conjugate expression of $\sqrt[3]{a} \pm\sqrt[3]{b}$ is $\sqrt[3]{a^2} \mp\sqrt[3]{ab}+\sqrt[3]{b^2} $. You can use that to rationalise the denominators. The expression inside the parentheses is $\sqrt[3]{77}$ so that finally you get $77$.
Number theory introduction. An integer n is even if, and only if, n equals twice some integer: is even ⇔ ∃ an integer k such that n = 2k. An integer n is odd if, and only if, n equals twice some integer plus 1: is odd ⇔ ∃an integer k such that n = 2k + 1. Reasoning examples: Is 0 even? Yes, 0 = 2·0.
Example 1.2. Find all positive integers d such that d divides both n2+1 and (n+1)2 +1 for some integer n. Solution. Let d | (n 2+ 1) and d | [(n + 1) + 1], or d | (n2 + 2n + 2). Then d | [(n2 + 2n + 2) − (n2 + 1)], or d | (2n + 1) ⇒ d | (4n2 + 4n + 1), so d | [4(n2+2n+2)−(4n2+4n+1)], or d | (4n+7). Then d | [(4n+7)−2(2n+1)],
Show that $a_n=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$ would not contain a natural number for all n 2 elementary proof for $1 + \frac{1}{3} + \frac{1}{5} + … + \frac{1}{2n+1}$ is not an integer.
Prove the statement: For all integers \(a\text{,}\) \(b\text{,}\) and \(c\text{,}\) if \(a^2 + b^2 = c^2\text{,}\) then \(a\) or \(b\) is even. Hint . A proof by contradiction would be reasonable here, because then you get to assume that both \(a\) and \(b\) are odd.
Theorem: If n is an integer and n2 is even, then n is even. Proof: By contradiction; assume n is an integer and n2 is even, but that n is odd. Since n is odd, n = 2k + 1 for some integer k. Then n2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1. Now, let m = 2k2 + 2k. Then n2 = 2m + 1, so by definition n2 is odd. But this is impossible, since n2 ...
