Modulo Arithmetic, set of all integers div 7

**Timsworth** · May 8th 2011, 04:49 AM

Hi there.

I've looked over my notes for my last lecture only to realise I don't understand what's going on with them at all and would like to know if anyone else can explain how the following works:

in $\mathbb{Z}$ /11:
8+7 = 10 mod 11

We all know that 8+7 does not equal 10 mod 11, so what's going on?

It then goes on to prove that
7+8 = ?
7+8 = 15, too big so /11, remainder 4
7+8 = 4 mod 11

**Timsworth** · May 8th 2011, 05:31 AM

Okay, the problem was that I had typo'd my notes -.- It should be 7+3 = 10 mod 11 which is much more logical.

My problem now is that I need to use the answer to "Find a number in $\mathbb{Z}$ /7 which gives 1 mod 7 when multiplied by 3" (the answe of which is 5)
to solve the equation "3n = 4 in $\mathbb{Z}$ /7" and don't know where to start.

Any ideas/insights?

**HappyJoe** · May 8th 2011, 05:31 AM

It certainly looks like a mistake!

If you quote it in its context, then we can probably tell for sure whether it's a mistake or not.

**HappyJoe** · May 8th 2011, 05:35 AM

Ah, never mind my other post. I posted it before you found the typo.

Anyway, you know that 3*5 = 1 (mod 7). Thus 4*(3*5) = 4 (mod 7), or rearranging the left hand side 3*(4*5) = 4 (mod 7). But 4*5 = 20 = 6 (mod 7), hence 3*6 = 4 (mod 7).

**abhishekkgp** · May 8th 2011, 05:40 AM

Originally Posted by Timsworth

Okay, the problem was that I had typo'd my notes -.- It should be 7+3 = 10 mod 11 which is much more logical.

My problem now is that I need to use the answer to "Find a number in $\mathbb{Z}$ /7 which gives 1 mod 7 when multiplied by 3" (the answe of which is 5)
to solve the equation "3n = 4 in $\mathbb{Z}$ /7" and don't know where to start.

Any ideas/insights?

what do you mean by $\mathbb{Z}_{/7}$ ?? do you mean $\mathbb{Z}_7$ .
if that's the case then you need to solve $3n=1 \, in\, \mathbb{Z}_7$ , not the one you typed. for this you need to find the inverse of $3 \, in \, \mathbb{Z}_7$ using euclidean algorithm.

**Timsworth** · May 8th 2011, 05:44 AM

Cheers Joe. I was hoping that was the answer! It's the notation that threw me off, I think. There was no $\equiv_7$ or $\pmod{7}$ etc that I'm used to seeing.

**Timsworth** · May 8th 2011, 05:52 AM

Originally Posted by abhishekkgp

what do you mean by $\mathbb{Z}_{/7}$ ?? do you mean $\mathbb{Z}_7$ .
if that's the case then you need to solve $3n=1 \, in\, \mathbb{Z}_7$ , not the one you typed. for this you need to find the inverse of $3 \, in \, \mathbb{Z}_7$ using euclidean algorithm.

I'm just getting used to using TeX, what I meant was $\mathbb{Z}/7$ rather than $\mathbb{Z}_/_7$

**Timsworth** · May 8th 2011, 06:45 AM

What's the difference between $\mathbb{Z}/7$ and $\mathbb{Z}/7^*$ ? Are there any online resources I'm missing about it? I can't find anything good.

**HappyJoe** · May 8th 2011, 09:57 AM

The difference between $\mathbb{Z}/7$ and $(\mathbb{Z}/7)^*$ is that the former consists of the numbers $\{0,1,2,3,4,5,6\}$ , while the latter consists of $\{1,2,3,4,5,6\}$ .

In general, if n is some integer, then $\mathbb{Z}/n$ consists of the numbers $\{0,1,2,\ldots,n-1\}$ , while $(\mathbb{Z}/n)^*$ consists of all numbers that are relatively prime to n (that is, all numbers a with gcd(a,n)=1, so that the greatest common divisor of a and n is 1).

So for instance, we have that

$(\mathbb{Z}/8)^* = \{1,3,5,7\},$

and

$(\mathbb{Z}/15)^* = \{1,2,4,7,8,11,13,14\}.$

Importantly, whenever p is a prime, we have that

$(\mathbb{Z}/p)^* = \{1,2,3,\ldots,p-1\},$

like it happened in the case p=7 above.

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