Show that a multivariate function is bijective

  • #1
Mr Davis 97
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Homework Statement


Show that ##f: \mathbb{Z}^{+} \times \mathbb{Z}^{+} \longrightarrow \mathbb{Z}^{+} ##where ##\displaystyle f(m,n) = \frac{(m+n-2)(m+n-1)}{2}+m## is bijective

Homework Equations

The Attempt at a Solution


First, we show that ##f(a,b) = f(c,d) \implies a=c \land b=d##.

##f(a,b) = f(c,d)##

##\displaystyle \frac{(a+b-2)(a+b-1)}{2}+a = \frac{(c+d-2)(c+d-1)}{2}+c##

After simplification, we find that

##(a+b)^2 + a - b = (c+d)^2 + c - d##

Comparing sides, then ##a + b = c+d, ~a-b = c-d##
Then ##a=c,~b=d##

Second, we show that for all positive integers a, there exist an m and an n such that f(m,n) = a.
We see that ##f(m,2-m) = m##, so the function maps to every element in the codomain

Does this show that the function is bijective?
 
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  • #2
For the second part 2-m might not necessarily be a positive integer which seems to be a problem since the function is defined for pairs of positive integers.

For first part this conclusion
Mr Davis 97 said:
Comparing sides, then a+b=c+d, a−b=c−da+b=c+d, a−b=c−da + b = c+d, ~a-b = c-d
doesn't necessarily follow from the previous line. It might be correct but needs some additional justification me thinks.
 
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