Princeton Review Error : Congruences Page 225

[Legendre]

Princeton Review, rule 3 :

For any positive integer c, the statement

'a is congruent to b (mod n)'
<=>
'a is congruent to b (mod cn)'


Counter-example :

5 congruent 2 (mod 3)

Let c = 2.

5 congruent 2 (mod 9) is FALSE. Since (5-2) = 3 is not divisible by 9.

[hopeful_statguy]

What edition do you have? In the 3rd edition, rule 3 is:
If $$a_1 \equiv b_1 \pmod{n}$$ and $$a_2 \equiv b_2 \pmod{n}$$ , then
$$a_1 \pm a_2 \equiv b_1 \pm b_2 \pmod{n}$$
$$a_1 a_2 \equiv b_1 b_2 \pmod{n}$$

[kosuke]

Sorry to bring this post back from the dead, but I'm having trouble figuring out what LeDuc meant to say here... (I think Legendre meant to say rule #4, not 3 on page 225).

Any idea on what LeDuc really means?

[kosuke]

Sorry, nevermind.

I figured out what I was doing wrong.

[Legendre]

Sorry I meant rule 4, not 3.

I think he meant that a = b (mod n) then, a = b or b + n or b + 2n or ... or b + (c-1)n (mod cn).

This is clear because for b + cn, b + (c+1)n,... we can mod out the cn and obtain same expression in the list.