0

The lines given by $$f_1:\mathbb{W}\rightarrow \mathbb{N},~~f_1(n)=(6n+1)\cdot2^{k}$$ $$f_2:\mathbb{W}\rightarrow \mathbb{N},~~f_2(n)=(6n+3)\cdot2^{k}$$ $$f_3:\mathbb{W}\rightarrow \mathbb{N},~~f_3(n)=(6n+5)\cdot2^{k}$$ where $n,k$ are arbitrary whole numbers.

By ignoring the $2^{k}$ I can see that any given odd number will appear on one of the lines, and when multiplied by a factor of 2 will give any even number. How do you suppose I would go about and prove all of the natural numbers exist on the lines?

Alex Maslach
  • 77

1 Answers1

2

Your observation is pretty much correct, you just need to formalise it. You can start by proving that every odd $x$ is in one of the sequences, probably by assuming that there's an $x$ not in $f_1$ or $f_2$, and hence showing that it must be in $f_3$. After that, if you can then prove that any even number can be written as $2^k\cdot x$ for some odd $x$, you're done.

ConMan
  • 24,300