## anonymous one year ago given x ≡ 1 (mod 3). Is x ≡ 4 (mod 3) true?

1. ganeshie8

Below is not required but it helps in relating congruences and divisibility $x\equiv 1 \pmod 3 \implies 3 \mid (x-1) \implies 3\mid [(x-1) - 3] \\\implies 3\mid (x-4)\implies x\equiv 4\pmod{3}$

2. UsukiDoll

modulo 3 looks like [0,1,2] though?

3. UsukiDoll

like on x ≡ 1 (mod 3). it's x is congruent to the remainder 1 in modulo 3. I can see that part... it's x ≡ 4 (mod 3) that's giving me a hard time... because that's x is congruent to the remainder 4 in modulo 3.

4. ganeshie8

Notice that $$3\mid (3k)$$. This is same as saying $$3k\equiv 0 \pmod{3}$$. Therefore $$x\equiv x+0\equiv x+3k\pmod{3}$$

5. anonymous

so basically I can add multiple of 3 on either sides, and it still holds?

6. anonymous

like x + 3 ≡ 1 (mod 3) holds?

7. ganeshie8

Yes, congruence is just a neat notation that hides the messy divisibility rules

8. ganeshie8

$$x\equiv y\pmod{n}$$ and $$y\equiv z\pmod{n}$$ implies $$x\equiv z\pmod{n}$$ transitive property holds, so you can replace $$3$$ by $$0$$ in modulo 3

9. Empty

Whenever I use mods I end up just subtracting off the number until I have some number less than what I want, like this: $9 \equiv 5 \equiv 1 \mod 4$ So here I just kept subtracting 4 off, rather than trying to think "ok what's the remainder if I divide by 4?" There are a bunch of cute tricks you can use too, like: $100^{2523672} \mod 3$ just rewrite 100 as 3*33+1 and we have: $(3*33+1)^{klfaljag} \mod 3$ $(0*33+1)^{lsdjf0aug} \mod 3$ $1 \mod 3$

10. anonymous

ok. I know it seems kinda trivial but I just want to make sure I really understand it. x + 3k ≡ 1 + 3s (mod 3), for some integer k,s, holds ? right? ^^

11. anonymous

actually for *all* integer s,k

12. ganeshie8

x + 3k ≡ 1 + 3s (mod 3) thats not so important, all you need to do is to refer back to the definition : $a\equiv b\pmod{n}\iff n\mid (a-b)$

13. anonymous

@Empty i'm not quite there yet but I might find that useful one day ^^

14. ganeshie8

the moment you see 3k, and if you're in modulo 3, you see that 3k is divisible by 3 : $3\mid (3k) \iff 3k\equiv 0 \pmod{3}$ so you replace $$3k$$ by $$0$$

15. Empty

A way to help think of it is all even and all odd numbers can be represented as 2n or 2n+1. So you can see in mod 2, all even numbers are 0 and all odd numbers are 1.

16. anonymous

@ganeshie8 ah i see. x ≡ 4 (mod 3) x ≡ 1 + 3 (mod 3) and since 3 is a multiple of 3, replace 3 with 0. x - 99 ≡ 1 (mod 3). And since 99 is also a multiple of 3, replace it with 0. and x + 3k ≡ 1 + 3s (mod 3), replace 3k and 3s with 0.

17. ganeshie8

Yes, but careful, congruence relations don't work in both directions always

18. ganeshie8

 given x ≡ 1 (mod 3). Is x ≡ 4 (mod 3) true?  you should start with whats given, not what you want to prove

19. anonymous

ah yes. I just need to see what you can validly add to either side (or both). So in general, a ≡ b (mod c) implies a ≡ b + kc (mod c) implies a + kc ≡ b (mod c) implies a + kc ≡ b + sc (mod c)

20. ganeshie8

Yep, adding/subtracting a multiple of "c" to a side(s) wont change the congruence because : $c\mid (a-b) \implies c\mid [(a-b) + kc]$

21. IrishBoy123

"."

22. anonymous

yep! I understand everything now. Thank you!

23. ganeshie8

np