Multiplicative inverse for Z31

[WVteacher]

Prove Multiplicative inverse for Z31

Can someone tell me the proof of multiplicative inverse for the ring Z31 using [x] and [y]. It is to prove an integral domain is a field.

 

[pka]

Can someone tell me the proof of multiplicative inverse for the ring Z31 using [x] and [y]. It is to prove an integral domain is a field.

Have a look at these calculations. Do you see the point?
Think about Euler & Fermat's work in the number theory of integers.

 

[WVteacher]

I understand what the multiplicative inverse is but I am having a hard time proving it. This is what I did but I don't know if I'm right:

[x]*[y]=1
[x*y]=1
[x]=1*[y^-1]
[x]=[y^-1]

where x and y are elements of Z₃₁ so y is the multiplicative inverse of x. Is this right?

 

[pka]

I understand what the multiplicative inverse is but I am having a hard time proving it. This is what I did but I don't know if I'm right:
[x]*[y]=1
[x*y]=1
[x]=1*[y^-1]
[x]=[y^-1]
where x and y are elements of Z₃₁ so y is the multiplicative inverse of x. Is this right?

Frankly, I have absolutely no idea what any of that means. Now I have taught this material and even contributed text material on this. I understood your to be asking about how to find the multiplicative inverse. But now you seem to have meant "what is an inverse?" If that is it, I find it to be very naive. Anyone working at this level should know what a multiplicative inverse is; we all know the definition. It is quite another matter is ask if a collection has that property. Does \(\displaystyle \mathbb{Z}_k\) have the multiplicative inverse property? The answer is yes if \(\displaystyle \bf{k}\) is prime. That is a number theory result from Euler & Fermat theorems.

 

[WVteacher]

I used Euler and formats work to prove this the first time and was told I was incorrect and that it did not prove that that the integral domain is a field. I was also told I need to prove this in terms of x and y that is why I wrote to you what I did in x and y to prove the multiplicative inverse to prove the field.

 

[pka]

I used Euler and formats work to prove this the first time and was told I was incorrect and that it did not prove that that the integral domain is a field. I was also told I need to prove this in terms of x and y that is why I wrote to you what I did in x and y to prove the multiplicative inverse to prove the field.

It appears that you & I are speaking two different languages. Because you asked a question about rings, I assumed that you knew basic group theory (i.e. about operations on equivalent classes, co-sets, factor groups etc.). So you needed to know that each element of Z_{31} had an inverse. Thus you need to have a sit-down with whomever you are talking to.

 

[HallsofIvy]

Do you know what "\(\displaystyle Z_{31}\)" is? That isn't clear from what you say. Two numbers, x and y, are "multiplicative inverses" in \(\displaystyle Z_{31}\) if and only if xy= 1 (mod 31) which means that xy is some multiple of 31 plus 1: xy= 31n+ 1. So one way to show that every member of \(\displaystyle Z_{31}\), except 0, has a multiplicative inverse is to go through all such 31 members and use that formula to show what the multiplicative inverse is.

For example, a multiplicative inverse of 7, x, must satisfy 7x= 31m+ 1 for so integer m. One way to do that is to try m= 1, 2, 3, ... until we get to a multiple of 7:
31*1+ 1= 32 is not a multiple of 7.
31*2+ 1= 63 is a multiple of 7: 7(9)= 63= 1 (mod 31).
The multiplicative inverse of 7 is 9 and the multiplicative inverse of 9 is 7.

Obviously the multiplicative inverse of 1 is 1 and we get two multiplicative inverse each time so we need to do this 15 times to determine all multiplicative inverses.