5.Consider the functions f → g∶ Z × Z → Z × Z deﬁned as f (m,n) = (3m − 4n,2m + n) and g(m,n) = (5m + n,m). Find the formulas for g o f and f o g.

Question

anonymous · Answer

??

anonymous · Answer

I'll do $g\circ f$:

Since $f(m,n) = (3m-4n,2m+n)$ and $g(m,n) = (5m+n,m)$, when we look at $g(f(m,n))$, we plug in the first coordinate of $f(m,n)$ wherever we had an $m$ in $g(m,n)$, and plug in the second coordinate of $f(m,n)$ wherever we had an $n$ in $g(m,n)$.  Thus, it follows that
$$\begin{aligned} g(f(m,n)) &= g(3m-4n,2m+n)\ & = (5(3m-4n)+(2m+n), (3m-4n))\ &= (15m-20n+2m+n,3m-4n)\ &= (17m-19n,3m-4n)\end{aligned}$$
Do you think that you can do $f\circ g$ following a similar process?

I hope this made sense! :-)

anonymous · Answer

thank you.. i will try first for fog

anonymous · Answer

@ChristopherToni can u check for me..
f(g(m,n)) = f(5m + n,m)
             = (3(5m+n) + (2m + n), (5m + n))
             = (15m + 3n + 2m +n , 5m + n )
             = (17m + 4n, 5m + n)

is tht true ?

anonymous · Answer

Unfortunately, you didn't evaluate $f(5+n,m)$ correctly. :/

When doing this, think of it as letting $s=5m+n$ and $t=m$ in $f(s,t) = (3s-4t,2s+t)$

Thus, you should have that $f(5m+n,m) = (3(5m+n) - 4(m), 2(5m+n)+(m))=\ldots$

Does this make sense?

anonymous · Answer

owh.. sorry.. im still weak in this topic.. i will do again as exercise.. I have one question that i still dont know how to answer.. here @ChristopherToni 	Let f: A → B be a function. Prove that  f is bijective if and only if the inverse relation f^(-1) is a function from B to A. Give a numerical example.