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limitkiller
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51^(-51)[itex] \equiv x (mod 8)[/itex] .
Does it mean anything at all?
If it does what is x?
Does it mean anything at all?
If it does what is x?
limitkiller said:51^(-51)[itex] \equiv x (mod 8)[/itex] .
Does it mean anything at all?
If it does what is x?
Infinitum said:Modular arithmetic is defined only for integers. So your expression doesn't give any meaning.
http://en.wikipedia.org/wiki/Modular_arithmetic
DonAntonio said:Of course it has a meaning: modulo 8, we have that [itex]\,3^{-1}=\frac{1}{3}=3\,[/itex] , since [itex]\,3\cdot 3=1\pmod 8[/itex] , so doing
arithmetic modulo 8 all the way in the following we get:[tex]51=3\pmod 8\Longrightarrow 51^{-51}=\left(3^2\right)^{25}\cdot 3=1\cdot 3 = 3[/tex]
DonAntonio
I think you dropped a minus sign in the middle there, but since [itex]3^{-1}\equiv3 (mod 8)[/itex] the answer is right.DonAntonio said:Of course it has a meaning: modulo 8, we have that [itex]\,3^{-1}=\frac{1}{3}=3\,[/itex] , since [itex]\,3\cdot 3=1\pmod 8[/itex] , so doing
arithmetic modulo 8 all the way in the following we get:[tex]51=3\pmod 8\Longrightarrow 51^{-51}=\left(3^2\right)^{25}\cdot 3=1\cdot 3 = 3[/tex]
DonAntonio
haruspex said:I think you dropped a minus sign in the middle there, but since [itex]3^{-1}\equiv3 (mod 8)[/itex] the answer is right.
Infinitum said:Interesting. The number theory manual I used specifically defined it only for integers, and denied its existence for rational numbers.
So, does [itex]\sqrt{17} \equiv x (mod3)[/itex] mean something too?
Edit : Changed numbers.
Nobody said anything about "rational numbers". if a and n are integers, then a-1 (mod n) is an integer.Infinitum said:Interesting. The number theory manual I used specifically defined it only for integers, and denied its existence for rational numbers.
If it exists, yes. First 17= 3(5)+ 2 so [itex]17\equiv 2[/itex] (mod 3) and the equation reduces to [itex]\sqrt{2}\equiv x[/itex]. That is, we must have [itex]x^2= 3k+ 2[/itex] 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 [itex]x^2\equiv 0[/itex] and [itex]x^2\equiv 1[/itex] have solution modulo 3, [itex]x^2\equiv 2[/itex] does not. That is, [itex]\sqrt{17}\equiv x (mod 3)[/itex] does NOT have a meaning but [itex]\sqrt{18}\equiv x (mod 3)[/itex] does have meaning (18= 6(3)+ 0 so 18 is equivalent to 0 and x= 0 (mod 3) is a solution) and [itex]\sqrt{19}\equiv x (mod 3)[/itex] also has meaning (19= 6(3)+ 1 so 19 is equivalent to 1 and x= 1 and x= 2 (mod 3) are both solutions).So, does [itex]\sqrt{17} \equiv x (mod3)[/itex] mean something too?
Edit : Changed numbers.
HallsofIvy said: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 [itex]17\equiv 2[/itex] (mod 3) and the equation reduces to [itex]\sqrt{2}\equiv x[/itex]. That is, we must have [itex]x^2= 3k+ 2[/itex] 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 [itex]x^2\equiv 0[/itex] and [itex]x^2\equiv 1[/itex] have solution modulo 3, [itex]x^2\equiv 2[/itex] does not. That is, [itex]\sqrt{17}\equiv x (mod 3)[/itex] does NOT have a meaning but [itex]\sqrt{18}\equiv x (mod 3)[/itex] does have meaning (18= 6(3)+ 0 so 18 is equivalent to 0 and x= 0 (mod 3) is a solution) and [itex]\sqrt{19}\equiv x (mod 3)[/itex] also has meaning (19= 6(3)+ 1 so 19 is equivalent to 1 and x= 1 and x= 2 (mod 3) are both solutions).
No, [itex]51^{-51}= (51^{-1})^{51}[/itex] which will be an integer (or equivalence class of integers depending upon how you define "modulo")Infinitum said:I saw the OP's question that stated [itex]51^{-51}[/itex], which is rational and not of the form a-1... Though I must admit, I was prejudicial.
Yes, assuming you simply extend the standard definition to include complex remainders.This led me wonder whether there is a modulo for complex numbers, too. Say,
[tex]\sqrt{17}i \equiv x(mod 3)[/tex]
Which reduces to,
[tex]\sqrt{2}i = x[/tex]
So, we would have,
[tex]x^2 = 3k - 2 = 3(k-1) + 3 - 2 = 3n + 1[/tex]
And hence x is equal to 1 or 2 (mod 3) are the solutions.
Is this correct?
HallsofIvy said:Yes, assuming you simply extend the standard definition to include complex remainders.
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