# Bijection between $\mathbb N^2$ and $\mathbb N$ [duplicate]

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A question regarding a bijection between $\mathbb N^2$ and $\mathbb N$. I know about cantor pairing function but I wanted to ask about a bijection I have seen around the site which is $n=2^{u-1}(2v-1)$.

Can anyone explain why is this bijection surjective (I understand how it's injective) as it seems to give only values of $\mathbb N_{even}$ (power of 2 which is even times an odd number).

Thank you

## marked as duplicate by Guy Fsone, Asaf Karagila♦ elementary-set-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 6 '18 at 11:56

• If $u=1$ then $2^{u-1}=2^0=1$ and you get an odd number! – user491874 Feb 6 '18 at 11:48
It is surjective because if $n\in\mathbb N$ and you write $n$ as $2^ab$, with $a\in\mathbb{Z}_+$ and $b$ and odd natural, then the function that have in mind maps $\left(a+1,\frac{b+1}2\right)$ into $2^ab=n$.
• @GalushBalush When $a=0$. – José Carlos Santos Feb 6 '18 at 11:53