Prove that $\exists n,m \in \mathbb{N}$ such that $$3n+5m=12 $$ This is clearly false, but I am not sure how to conduct a proof stating it is false. Should I just give examples with $n = 1,2,3$ and then pick $m$s afterwards?
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You can work with some inequalities. Since you are working with natural numbers, it is easy to exhaust the other values. – Sigur Nov 02 '13 at 16:30
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1$3\mid 5m = 12-3n \Longrightarrow 3 \mid m \Longrightarrow m \ge 3 \Longrightarrow 3n = 12-5m <0$. – njguliyev Nov 02 '13 at 16:35
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The equality sign should be replaced by an inequality sign ($\neq$) in your title. – John Bentin Nov 02 '13 at 19:40
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Since it asks for solutions in $\mathbb{N}$, you could cycle through all the (finitely) many possible solutions and show none satisfies the equation - i.e. you exhaustively show all possible solutions fail.
However, another simple argument is to suppose that some $m$, $n$ exist in $\mathbb{N}$ such that $3n + 5m = 12$. In this case, $n = \frac{12-5m}{3}$, by rearrangement. Thus, $n =4- \frac{5m}{3}$. But, $\frac{5m}{3}\not\in \mathbb{N}$ for $m<3$. So, $m\geq3$, but in this case we have $5m\geq15$ and hence $n$ must be negative, i.e. $n\not\in \mathbb{N}$. This is a simple proof by contradiction.
JWeissman
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If $0\in\mathbb N$ then take $n=4$ and $m=0$.
Otherwise, $n\ge 1$ and $m\ge 1$, and so $3n+5m\ge 8$. The next possible values are $8+3=11$ and $8+5=13$. And so $12$ is out.
lhf
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