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I need help with this excercise:

Define $$f:\mathbb{N}\times\mathbb{N}\longrightarrow \mathbb{N}$$ Where, $$f(a,b)=2^{a-1}(2b-1).$$

Proof that $f$ is bijective function..

I try

$1-$ Injective:

Get $(a,b),(c,d)\in \mathbb{N}\text{x}\mathbb{N}$ and supose that $a>c$ , $b>d$ and $f(a,b)=f(c,d)$ then,

$$2^{a-1}(2b-1)=2^{c-1}(2d-1)$$ $$2^{a-c}(2b-1)=2d-1$$

If $a \neq c$ then $2d-1$ is even, $\bot$, therefore $a=c$, then,

$$2b-1=2d-1,$$ therefore $b=d$

Then, $(a,b)=(c,d)$

This part is ok??

How proof that $f$ is surjective?

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    f(0,0) is not an integer – 2017-01-15
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    I'm taking the natural numbers as the set $\{1,2,3,...\}$ – 2017-01-15
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    Tip: Use `\times` ($\times$) instead of `\text{x}` ($\text{x}$) for a multiplication sign. – 2017-01-15
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    That's good, but you don't assume but $a > c; b>d$ is only one way to have $(a,b) \ne (c,d)$. Don't state $(a,b) \ne (b,d)$ just note that as $2^{a-c}(2b -1) = 2d-1 \in \mathbb N$ is odd and $2b-1$ is also odd that $a > c$ makes an even and $a < c$ makes a fraction. So $a-c = 0$ and that means $a =c$ which means $b= d$. – 2017-01-15
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    Wow, that is a much cleaner bijection than the "diagonal" picture one which visually is intuitive but to actually write the arithmetic to do the bijection is nightmarish. – 2017-01-15

5 Answers 5

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Hint: $f$ surjective follows from the factoring theorem.

So take $n\in\mathbb N$, and you can write n=$2^{a-1}p_1^{\alpha_1}\ldots p_k^{\alpha_k}$ with $a,\alpha_1,\ldots,\alpha_k>0$ and $p_1,\ldots,p_k$ odd primes. Note that $a-1$ can be zero when $n$ is odd.

Then take $2b-1=p_1^{\alpha_1}\ldots p_k^{\alpha_k}$, which is guaranteed to be odd, since the product of two odd numbers is odd.

Then $f(a,b)=n$.

Also, your reasoning for $2^{a-c}(2b-1)=2d-1$ works only when $a>c$, but it can be easily fixed by symmetry.

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Every positive $n$ integer can be writen $2^au$ (to see this write $n$ as the product of power of primes, $n=2^a3^{n_3}...p_i^{n_i}$, where $2,3,..,p_i$ are prime numbers. $3^{n_3}...p_i^{n_i}=u$ is odd) where $u$ is odd $u=2b-1$, $f(a-1,b)=n$.

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Suppose $n$ is odd. Then there exists a $b$ such that $2b-1=n$ and hence $f(1,(n+1)/2)=n$.

Suppose $n$ is even. Then there exists $2^{a-1}$ such that $n=2^{a-1}m$ with $m$ odd. So first determine $a$ such that $n/2^{a-1}$ is odd and realize $$f(a-1,(n/2^{a-1}+1)/2)=2^{a-1}m=n $$

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A simpler way to prove injectivity: it follows from unique factorisation.

Indeed, if $2^{a-1}(2b-1)=2^{c-1}(2d-1)$, both sides have the same $2$-valuation (the exponent of $2$ in their decomposition into prime factors), and as $2b-1$ and $2d-1$ are odd, this means $a-1=c-1$, i.e. $a=c$.

Cancelling the powers of $2$, there results $2b-1=2d-1$, whence $c=d$.

Surjectivity:

If $n$ is a ntural number, let $r$ the greatest power of $2$ which divides $n$. Then $n=2^r m$, and $m$ is odd. It is easy to check that $$n=f\Bigl(r+1,\frac{m+1}2\Bigr).$$

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Let's fix two arbitrary integers $i,j\in \mathbb{N}-\{0\}$. Set the function $\varphi:\mathbb{R}^2\to \mathbb{R}^2$ by $$ \varphi(x,y)=\left( 2^{x-1}(2y-1)\, , 2^{[x+i]-1}(2[y+j]-1) \right). $$ Note that $\varphi(r,s)\in \mathbb{N}\times \mathbb{N}$ for all $r,s\in \mathbb{N}$. If $\varphi$ is globally invertible at $\mathbb{R}^2$ for all $i,j\in \mathbb{N}-\{0\}$ then, in particular, the restriction $\varphi|_{\mathbb{N}\times \mathbb{N} }$ is also globally invertible at $\mathbb{N}^2$ for any $i,j\in \mathbb{N}-\{0\}$.

The tip to prove what you want, the idea is to use the

Hadamard global inverse function theorem. A $C^1$-map $f:\mathbb{R}^n\to \mathbb{R}^n$ is a $C^1$-diffeomorphism if and only if the Jacobian $\det Df(z_1,\ldots,z _n)$ never vanishes and $|f(x_1,\ldots,x_n)|\to\infty$ whenever $|(x_1,\ldots,x_n)|\to\infty$.