(Elementary number theory) Understanding congruence and modulus

[Leo Liu]

TL;DR Summary

N/A

In this question, how does the step marked with 1 become the step marked with 2? I can see that the transitivity property of congruence is used, but I don’t know what exactly is going on here. Can someone please explain? Also at which step is Congruence Add and Multiply used?

Thanks.

Preposition used:

 

[jedishrfu]

Isn’t 9 mod 7 the value 2?

Similarly for 3^3 is 27 which is one less than 28.

what rule would they be using to do this?

 

[Leo Liu]

Isn’t 9 mod 7 the value 2?

Similarly for 3^3 is 27 which is one less than 28.

Yes. But I just don't understand how the theorem is applied here.

 

[jedishrfu]

The theorem says the modulo N of a sum or product of a list of numbers is the same as the sum or product of the modulos of those numbers, right?

 

[Leo Liu]

The theorem says the modulo N of a sum or product of a list of numbers is the same as the sum or product of the modulos of those numbers, right?

Right. But I am afraid that I still cannot see why as the congruence equations are not in the same form as the preposition.

 

[PeroK]

Right. But I am afraid that I still cannot see why as the congruence equations are not in the same form as the preposition.

What does that mean?

We have $9 = 2 + 7$. Therefore $9 \equiv 2 \ (mod \ 7)$. Therefore we may replace $9$ by $2$ in any equation modulo $7$. That's the idea.

 

[martinbn]

The proposition says that if $a_1\equiv a_2 \; (mod\; m)$ and $b_1\equiv b_2 \; (mod\; m)$, then you have $a_1b_1\equiv a_2b_2 \; (mod\; m)$.

In the example it is applied to $9 \equiv 2 \; (mod\; 7)$ and $3^3 \equiv -1 \; (mod\; 7)$.

 

[Leo Liu]

What does that mean?

We have $9 = 2 + 7$. Therefore $9 \equiv 2 \ (mod \ 7)$. Therefore we may replace $9$ by $2$ in any equation modulo $7$. That's the idea.

Thanks. You answer helped me understand it intuitively. Yet I still don't know how the congruence add and multiply is applied to this step.

 

[Leo Liu]

The proposition says that if $a_1\equiv a_2 \; (mod\; m)$ and $b_1\equiv b_2 \; (mod\; m)$, then you have $a_1b_1\equiv a_2b_2 \; (mod\; m)$.

In the example it is applied to $9 \equiv 2 \; (mod\; 7)$ and $3^3 \equiv -1 \; (mod\; 7)$.

This makes sense, but could you explain why this expression still holds when a1 and b1 is followed by (mod m)?

 

[PeroK]

Thanks. You answer helped me understand it intuitively. Yet I still don't know how the congruence add and multiply is applied to this step.

Proposition 3 that you quoted in your OP is a formal way to say you can replace a number by any other number of modular equivalence in products and sums. That's what it means - and Proposition 3 is that idea written out formally.