1
$\begingroup$

Is the function $f(x) = 7x + 11 \pmod{10}$ invertible for all positive integers?

I don't know how to go about it, mainly because I have never dealt with functions such as this one.

  • 1
    Did you try to graph it? Do you know a condition that a function must have to be invertible?2012-11-28

2 Answers 2

5

Working modulo $\,10\,$ all the time:

$y=7x+11\Longrightarrow 7x=y-11=y+9\Longrightarrow x=\frac{y+9}{7}=(y+9)\cdot 3=3y+7\Longrightarrow$

the inverse function is $\,g(x)=3x+7\,$

Notes:

$(1)\;\;\;\;\;\;\;\;-11=9\pmod{10}\Longleftrightarrow -11-9=-20=0\pmod{10}$

$(2)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{1}{7}=3\pmod{10}\Longleftrightarrow 3\cdot 7=21=1\pmod{10}$

1

This function cannot be invertible because it is not injective. Note that f(0) = f(10), for example.

Hope this helps!

  • 0
    @DonAntonio- I absolutely see your point. I wasn't sure when I posted the answer what the OP meant, since from the question description I thought the OP might be taking an algebra course rather than a course in number theory or group theory. I was hoping this answer would provide useful input in that case.2012-11-29