Prove whether this function is an injection, surjection, bijection or neither.

Question

The function is written as follows $p:\mathbb{N}\times\mathbb{N} \to \mathbb{N}$ where $p(a,b)=(ab(b+1))/2$

So far I tried supposing that $p(a_1, b_1) = p(a_2,b_2)$ but I couldn't reach a conclusion when I subbed it back into the equation. Thanks for the help!

0 is not considered a natural.

Answer 1

Simce, for each $a\in\mathbb N$,$$p\left(\frac{a(a+1)}2,1\right)=p(1,a),$$your function is far from injective. It is surjective, though.

Answer 2

Note that$$ P(10,2)=P(2,5)=30$$

Thus it is not injective.

However, it is surjective because $$P(n,1)=n$$

for all natural numbers $ n.$

Answer 3

Sometimes intuition is just great. But what if you don't have it?

For surjective:

If $n = \frac {ab(b+1)}2$ then $a = \frac {2n}{b(b+1)}$ which can be solved be solved if we can find $b(b+1)|2n$. We could do $b(b+1) = m$ where $m|2n$ and solve with the quadratic formula that $b^2 + b -m = 0$ so $b =\frac {-1\pm\sqrt{1+4m}}{2}$ so $1+4m$ must be an odd perfect square greater than $1$ and $m|2n$ and... well the simplest such case considering we don't really know anything about $n$ would be if $m = 2$ (and $m$ must divide $2n$) and $b =\{-2, 1\}$ so $b= 1$. And $a = \frac {2n}{1(1+1)} = n$

So $f(n, 1) = n$ for all $n$ so surjective.

And then we kick ourselves for not seeing that that was obvious.

Now injective.

If we have intution and we just saw that that $f(n,1) = n$; so.... then $f(a,b) = f(f(a,b), 1)$. And so long as $b \ne 1$, we have $(f(a,b), 1) \ne (a,b)$. So $f$ is not injective.

[This assumes $f(a,b) \in \mathbb N$ for all $(a,b)$. In other words that $f$ actually IS a well-defined function. As either $b$ or $b+1$ is even then the $\frac {b(b+1)}2$ is a natural number. And so $f(a,b) = \frac{ab(b+1)}2$ is a natural number.]

But what if we didn't have the intuition?

If $\frac{ab(b+1)}2 = \frac {cd(d+1)}2$

$ab(b+1) = cd(d+1)$.

Surely there are examples where $a\ne c; b\ne d$.

Fix $b, d$ so $b\ne d$ then $a*M = c*N$ where $M = b(b+1)$ and $N=d(d+1)$. Solving $a*M = c*N$ should be easy.

For example, just let $a =N$ and $c = M$ so $a*M= c*N =N*M$. Or, more generally, for any multiple of $gcd(N,M)$ we can have $a*M = c*N = k*\gcd(N,M)$ by setting $a = \frac {kN}{\gcd(N,M)}$ and $c = \frac {kM}{\gcd(N,M)}$.

So $f$ is not injective.

Just to be thorough...

When we were trying to show surjectivity we consider $b =\frac {-1 \pm \sqrt{1 + 4m}}2$ where $m|2n$. If $1+ 4m$ is any odd perfect square and $m|2n$ we have a solution. So if we let $m = 2, 6, 12, 20,30,42,$ etc. we will have solutions. So if $n$ has two or more $1,3,6,10, 15, 21,$ etc. as factors, we will have two or more solutions.

Example: if $n = 60$ we can have $m|120; m = 2,6,12, 20, 30, b = 1,2, 3, 4, 5$ and $a= 60, 20, 10, 6, 4$ so $60 = f(60,1)=f(20,2)=f(10,3) = f(6,4)= f(4,5)$.