I am having trouble solving this equation, if someone can provide some help that would be much appreciated!
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The map $f:\mathbb{Z}^+=\{1,2,3,\dots\}\to \{4,6,\dots\}$ given by $f(n)=2n+2$ is a bijection, showing that the two sets have the same cardinality.
To see that $f$ is indeed a bijection, note that with $g:\{4,6,\dots\}\to \mathbb{Z}^+$ given by $g(n)=n/2-1$ we have $f\circ g$ and $g\circ f$ are identical maps on $\{4,6,\dots\}$ and $\mathbb{Z}^+,$ respectively.
nonuser
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Reiner Martin
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1@John Watson, thanks for the correction – Reiner Martin Nov 30 '17 at 21:22