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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.

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Did you try to graph it? Do you know a condition that a function must have to be invertible? –  Rick Decker Nov 28 '12 at 20:18

2 Answers 2

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:

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

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

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

Hope this helps!

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This is wrong: $\,10=0\pmod {10}\,$. –  DonAntonio Nov 28 '12 at 19:43
    
@DonAntonio, that just depends on how you interpret the question. Is $f : \mathbb Z \to \mathbb Z/10\mathbb Z$ (your interpretation) or $f : \mathbb Z \to\mathbb Z$ (his interpretation)? It's unclear from the question asked so both answers could be right. –  user50336 Nov 28 '12 at 20:34
    
Well, if it were a function $\,\Bbb Z\to \Bbb Z/10\Bbb Z\,$ then it could hardly have an inverse, could it? Thus my interpretation is $\,\Bbb Z/10\Bbb Z\to\Bbb Z/10\Bbb Z\,$ , and not what you wrote. I'm guessing that by "...for all integers" the OP meant "..integers modulo $10$, otherwise the question is nonsensical. –  DonAntonio Nov 28 '12 at 22:39
    
@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. –  templatetypedef Nov 29 '12 at 0:06

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