# If it does what is x?

1. Jun 11, 2012

### limitkiller

51^(-51)$\equiv x (mod 8)$ .
Does it mean anything at all?
If it does what is x?

2. Jun 11, 2012

### Millennial

Re: Mod

I doubt it means anything with a negative exponent of an integer.

3. Jun 11, 2012

### Infinitum

4. Jun 11, 2012

### limitkiller

Re: Mod

Thanks.

5. Jun 11, 2012

### I like Serena

Re: Mod

Hi limitkiller!

Yes, it does have meaning.
The inverse of a number in modular arithmetic is the number by which you must multiply to get 1 modulo 8.

So for instance:
$$3^{-1} \equiv 3 \pmod 8$$
This is true since $3 \cdot 3 \equiv 9 \equiv 1 \pmod 8$.

Furthermore, according to Euler's theorem (skipping a few steps here) we have:
$$51^{-51} \equiv 51^{-51 \mod \phi(8)} \equiv 51^{-51 \mod 4} \equiv 51^1 \pmod 8$$

So what do you think x is?

6. Jun 11, 2012

### DonAntonio

Re: Mod

Of course it has a meaning: modulo 8, we have that $\,3^{-1}=\frac{1}{3}=3\,$ , since $\,3\cdot 3=1\pmod 8$ , so doing

arithmetic modulo 8 all the way in the following we get:$$51=3\pmod 8\Longrightarrow 51^{-51}=\left(3^2\right)^{25}\cdot 3=1\cdot 3 = 3$$

DonAntonio

7. Jun 12, 2012

### Infinitum

Re: Mod

Interesting. The number theory manual I used specifically defined it only for integers, and denied its existence for rational numbers.

So, does $\sqrt{17} \equiv x (mod3)$ mean something too?

Edit : Changed numbers.

8. Jun 12, 2012

### haruspex

Re: Mod

I think you dropped a minus sign in the middle there, but since $3^{-1}\equiv3 (mod 8)$ the answer is right.

9. Jun 12, 2012

### DonAntonio

Re: Mod

The first line already explained why $\,3^{-1}=3\pmod 8$...

DonAntonio

10. Jun 12, 2012

### DonAntonio

Re: Mod

Sure: $\,x\equiv\sqrt{17}\pmod 3\Longleftrightarrow x^2\equiv 17\pmod 3$ , and since $\,17\equiv 2\pmod 3$ , we

get the equation $\,x^2=2\pmod 3\,$, which has no solution in the prime field of integer residues modulo $\,3$.

And your number theory manual is right: the modulo arithmetic is defined for integers, but its algebraic structure is rich when a prime is

involved and thus we have a field, where we can divide by non-zero elements and etc.

DonAntonio

Last edited: Jun 12, 2012
11. Jun 12, 2012

### Infinitum

Re: Mod

I see. Thanks for the reply.

12. Jun 12, 2012

### HallsofIvy

Staff Emeritus
Re: Mod

Nobody said anything about "rational numbers". if a and n are integers, then a-1 (mod n) is an integer.

If it exists, yes. First 17= 3(5)+ 2 so $17\equiv 2$ (mod 3) and the equation reduces to $\sqrt{2}\equiv x$. That is, we must have $x^2= 3k+ 2$ for some k. Further, since only 0, 1, and 2 are in the set of integers modulo 3, we need only note that 0(0)= 0, 1(1)= 1, and 2(2)= 4= 1 (mod 3) so that while $x^2\equiv 0$ and $x^2\equiv 1$ have solution modulo 3, $x^2\equiv 2$ does not. That is, $\sqrt{17}\equiv x (mod 3)$ does NOT have a meaning but $\sqrt{18}\equiv x (mod 3)$ does have meaning (18= 6(3)+ 0 so 18 is equivalent to 0 and x= 0 (mod 3) is a solution) and $\sqrt{19}\equiv x (mod 3)$ also has meaning (19= 6(3)+ 1 so 19 is equivalent to 1 and x= 1 and x= 2 (mod 3) are both solutions).

13. Jun 12, 2012

### Infinitum

Re: Mod

I saw the OP's question that stated $51^{-51}$, which is rational and not of the form a-1... Though I must admit, I was prejudicial.

This explanation was solid, thanks a lot, HallsofIvy.

This led me wonder whether there is a modulo for complex numbers, too. Say,

$$\sqrt{17i} \equiv x(mod 3)$$

Which reduces to,

$$\sqrt{2i} = x$$

So, we would have,
$$x^2 = 3k - 2 = 3(k-1) + 3 - 2 = 3n + 1$$

And hence x is equal to 1 or 2 (mod 3) are the solutions.

Is this correct?

14. Jun 13, 2012

### Infinitum

Re: Mod

Oops, I included i(iota) in the square root by mistake. Can't edit it now, though.

15. Jun 13, 2012

### HallsofIvy

Staff Emeritus
Re: Mod

No, $51^{-51}= (51^{-1})^{51}$ which will be an integer (or equivalence class of integers depending upon how you define "modulo")

Yes, assuming you simply extend the standard definition to include complex remainders.

16. Jun 13, 2012

### Infinitum

Re: Mod

Aye. Thank you.