Question

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

Answer 1

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

Answer 2

do you see any restrictions on \(d\) ?

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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)

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

Yeah, that pattern ^^

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

Yes, more generally, below holds :

\[ad\equiv bd\pmod{c} \implies a\equiv b\pmod{\frac{c}{\gcd(d,c)}}\]

Answer 16

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

Answer 17

np