## anonymous one year ago if a ≡ b (mod c), then is it true that ad ≡ bd (mod c) for all integer d?

1. ganeshie8

$$a\equiv b\pmod{c} \implies c\mid (a-b) \implies c\mid d(a-b)\\~\\ \implies d(a-b)\equiv 0 \pmod{c} \implies ad\equiv bd\pmod{c}$$

2. ganeshie8

do you see any restrictions on $$d$$ ?

3. anonymous

No, I don't see any restriction on d. So it's true?

4. ganeshie8

Neither do I. So yes it is true for all $$d\in\mathbb Z$$

5. anonymous

how did you get d(a-b) ≡ 0 (mod c) though?

6. ganeshie8

That is the definition of congruence : $n\mid m \iff m\equiv 0 \pmod{n}$

7. anonymous

oh I see. Could you have done this though? c | d(a-b) is the same as c | d(a-b) - 0, which by definition d(a-b) ≡ 0 (mod c)

8. ganeshie8

I don't see why you want to have that intermediate step of subtracting 0

9. ganeshie8

unless you're taking the pattern seriously : $x\equiv y\pmod{n} \iff n\mid(x-y)$

10. anonymous

I was thinking about the definition in terns of two things. x ≡ y (mod z) means z | (x - y) so by subtracting 0, i.e c | d(a-b) - 0, I was comparing d(a-b) to x and 0 to y

11. anonymous

Yeah, that pattern ^^

12. ganeshie8

if it helps, then yes you may do that :)

13. ganeshie8

btw the converse of the conditional in main question does not hold : $ad\equiv bd\pmod{c} \implies a\equiv b\pmod{c}$ is false in general

14. anonymous

right. It's true if gcd(c,d) = 1. The proof was provided in the book :)

15. ganeshie8

Yes, more generally, below holds : $ad\equiv bd\pmod{c} \implies a\equiv b\pmod{\frac{c}{\gcd(d,c)}}$

16. anonymous

:O I didn't know that. Thank you for the info!

17. ganeshie8

np