What is the value of x in the equation 51^(-51)≡ x (mod 8)?

[limitkiller]

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

 

[Millennial]

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

 

[Infinitum]

Modular arithmetic is defined only for integers. So your expression doesn't give any meaning.

http://en.wikipedia.org/wiki/Modular_arithmetic

 

[limitkiller]

Thanks.

 

[I like Serena]

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

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?

 

[DonAntonio]

Modular arithmetic is defined only for integers. So your expression doesn't give any meaning.

http://en.wikipedia.org/wiki/Modular_arithmetic

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

 

[Infinitum]

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

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.

 

[haruspex]

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

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

 

[DonAntonio]

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

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

DonAntonio

 

[DonAntonio]

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.

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

 

[Infinitum]

I see. Thanks for the reply.

 

[HallsofIvy]

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

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

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

Edit : Changed numbers.

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).

 

[Infinitum]

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

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.

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).

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?

 

[Infinitum]

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

 

[HallsofIvy]

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.

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

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

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

Which reduces to,

\sqrt{2}i = 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?

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

 

[Infinitum]

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

Aye. Thank you.