So far I have:
Assume $n$ is odd, $n = 2k+1$ for some integer $k$.
Then\begin{align}3n&= 3(2k+1)\\ &= 6k + 3\end{align}
I'm unsure where to go from here.
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Can you find an integer $z$ such that your $3k$ is $3z+1$? – Ethan Bolker Jul 30 '19 at 18:32
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$6k$ is even. Add $3$ to it gives.... – David G. Stork Jul 30 '19 at 18:34
6 Answers
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Use the fact that $6k+3=2(3k+1)+1$.
José Carlos Santos
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1Because the odd integers are precisely those that can be written as $2N+1$. – José Carlos Santos Jul 30 '19 at 18:35
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and so that means I can then write 2N + 1 where the N is actually the 3k + 1 ,,,which means its odd and the proof is true? – Amanda_C Jul 30 '19 at 18:37
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1A proof is neither true nor false. But if you add that $6k+3=2(2k+1)+1$, then it will be a correct proof. – José Carlos Santos Jul 30 '19 at 18:39
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Since $2n$ is even, we have $3n=2n+n$ as a sum of an odd number and an even number, which is obviously odd.
szw1710
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Take $$n=2k+1$$ this is an odd number, then $$3(2k+1)=6k+3=(6k+2)+1$$ this is also an odd number.
Dr. Sonnhard Graubner
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Let $n \in \mathbb{Z}$ be odd, so $n = 2m + 1$ for some $m \in \mathbb{Z}$. We have: \begin{align*} 3n = 3(2m+1) = 6m + 3 = 2(3m + 1) + 1, \end{align*} where $3m + 1 \in \mathbb{Z}$ by closure. Let $z = 3m + 1$, so we have \begin{align*} 3n = 2z + 1. \end{align*} Thus, $3n$ is odd.
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$6k + 3$ can be written as $6k + 2 + 1$, now factoring we get $2(3k+1) + 1$, let $(3k+1) = A$, then $2A + 1$ is odd. in general, multiplying an odd integer with an odd integer always yields an odd integer. It is a good idea to prove it, you can also prove it with induction.
Donlans Donlans
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Let $n$ be an odd integer. Then $3n$ is the product of two odd integers, and thus odd.
DDS
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