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I have the following:

$f(m,n) = (3m+7n, 2m+5n)$ and I want to know if it is a bijection and if so, fine the inverse as well.

Here's my approach:

Suppose $f(m_1,n_1)=f(m_2,n_2)$ then:

$$ (3m_1+7n_1,2m_1+5n_1)=(3m_2+7n_2,2m_2+5n_2)$$ $$3m_1+7n_1=3m_2+7n_2 $$ and $$2m_1+5n_1=2m_2+5n_2$$

Now my issue begins here and don't know what the next step should be... Perhaps adding both equations? I'm not sure.

tc1729
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Dimitri
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3 Answers3

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You can solve them simultaneously : Rewrite them as $$ 3(m_1-m_2) + 7(n_1-n_2) = 0, \text{ and } 2(m_1-m_2) + 5(n_1-n_2) = 0 $$ So set $m=m_1-m_2, n=n_1-n_2$ and see that $$ 6m + 14n = 0, \text{ and } 6m + 15n = 0 $$ Subtracting, gives $n=0$, and hence $m=0$, which is what you want.

As for the inverse, solve $$ 3m+7n=x, \text{ and }2m+5n = y $$ Multiplying gives $$ 6m+14n = 2x, \text{ and } 6m+15n = 3y $$ and so $$ n = 3y-2x $$ Can you solve for $m$ similarly? This will give you a formula $$ g(x,y) = (\ldots, 3y-2x) $$ which is $f^{-1}$

Prahlad Vaidyanathan
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  • Please explain how you got the formula for solving it simultaneously. –  Nov 02 '13 at 01:33
  • I have solved for m, which is $ m = y-2n $, But how does that fit into the function $g(x,y)$ ? – Dimitri Nov 02 '13 at 02:03
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Hint: The natural proof is to note that the matrix of the transformation has determinant $1$, so the transformation is non-singular and the inverse matrix has integer entries.

But you could use your two equations, and eliminate $m_1$. (Multiply the first by $3$, the second by $2$, and subtract.) That also eliminates $m_2$, and you get that $n_1=n_2$.

You will then still need to deal with the onto part. For that, write down the equations $3m+7n=a$, $2m+3n=b$, and solve for $m$ and $n$. You will get an explicit expression which is clearly an integer.

André Nicolas
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Solve the system to show that $m_1=m_2$ and $n_1=n_2$. Subtract two times the first equation from three times the second equation to get $n_1=n_2$. Subtract five times the second equation from seven times the first equation to get $m_1=m_2$. Then $(m_1,n_1)=(m_2,n_2)$.

Hence $f$ is an injection.

Next consider $(3m+7n, 2m+5n)$. Since $3m+7n$ is an integer, there exists $l_1\in\mathbb{Z}$ such that $l_1=3m+7n$. Since $2m+5n$ is an integer, there exists $l_2\in\mathbb{Z}$ such that $l_2=2m+5n$. Hence there exists an element in $\mathbb{Z}^2$, namely $(l_1,l_2)$ such that $f(l_1,l_2)=(3m+7n, 2m+5n)$.

Hence $f$ is a surjection.

Since $f$ is both an injection and a surjection, $f$ is a bijection.

Chuck Franklin
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