Bijective function problem

Question

Given function $$F(x,y)=\frac{(x+y-2)(x+y-1)}{2}+y$$ is bijective or not.

Attempt: I check $F(x_1,y_1)$ and $F(x_2,y_2)$ and ends up in $$(x_1+y_1)^2-3x_1-y_1=(x_2+y_2)^2-3x_2-y_2$$ Now how to show $(x_1,y_1)=(x_2,y_2)$? How one can show it is surjective?

Find $x_1$ in terms of $y_1$, $x_2$, $y_2$. Replace all $x_1$ in the equation. Find $y_1$ in terms of $x_2$ and $y_2$. Find back $x_1$ on the same dependencies by replacing every occurrence of $y_1$. Does $x_1 = x_2$ and $y_1 = y_2$ ? — Lærne, Jun 04 '18 at 08:16
It is too cumbersome exercise to find x1 in terms of y1,x2 and y2 and next are also so please tell another approach — maths student, Jun 04 '18 at 09:13

Steven Alexis Gregory · Accepted Answer · 2018-06-04T14:29:10.333

See this Wikipedia article

This is a table of $M_{i,j} = F(i,j)$ where $1 \le i,j \le 6$.

$$M = \left[ \begin{array}{cccccc} 1 & 3 & 6 & 10 & 15 & 21 \\ 2 & 5 & 9 & 14 & 20 & 27 \\ 4 & 8 & 13 & 19 & 26 & 34 \\ 7 & 12 & 18 & 25 & 33 & 42 \\ 11 & 17 & 24 & 32 & 41 & 51\\ 16 & 23 & 31 & 40 & 50 & 61 \\ \end{array}\right]$$

We can show that the "up-one over-one" diagonal elements increase by $1$ by computing

$$F(i-1,j+1)-F(i,j) = 1$$

Then the sequence $F(n,1), F(n-1,2),F(n-2,3), \dots, F(1,n)$ is an arithmetic sequence with common difference $1$.

To continue this sequence starting at the next diagonal, we see that

$$F(n+1,1) - F(1,n) = 1 $$

Since $F(1,1)=1$, it follows that $F$ is bijective.

score 1 · Answer 2 · answered Jun 04 '18 at 08:17

1

Write it as

$(x_1+y_1)^2-(x_2+y_2)^2+3(x_2-x_1)+(y_2-y_1)=0$ Now suppose, $x_1 \neq x_2$ and $y_1 \neq y_2$.

What can you say about the expression i.e., can it be $0$? Remember $x_1,x_2,y_1,y_2 \in \mathbb{N}$.

answered Jun 04 '18 at 08:17

Topology

1,566
11
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how you can tell that it is zero if x1=x2 and y1=y2 only? there might be other solutions. – maths student Jun 04 '18 at 09:06
No, I made a mistake. It's maybe hard to conclude from this. – Topology Jun 04 '18 at 09:13

score 0 · Answer 3 · answered Jun 04 '18 at 08:22

0

Its not even injective, since $F(2,0) = 0 = F(1,0)$

answered Jun 04 '18 at 08:22

Mike

1,312

I'm assuming that $0$ is not supposed to be a natural number in this case and he is not aware that there is a confusion about that. – Steven Alexis Gregory Jun 04 '18 at 08:32
@stevengregory The OP forgot to mention it in the question, it is made clear only in the comments. I am personally unclear, what to do in this case. – peterh Jun 04 '18 at 12:00
Some people say that the set of natural numbers is ${0,1,2, \dots}$ and some say ${1,2,3,\dots}$. In the case of this problem, I am pretty sure that $\mathbb N = {1,2,3,\dots}$. So $F(2,0)$ and $F(1,0)$ are not included and their values are of no concern to your problem. – Steven Alexis Gregory Jun 04 '18 at 14:24

Bijective function problem

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