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)'
Counterexample :
5 congruent 2 (mod 3)
Let c = 2.
5 congruent 2 (mod 9) is FALSE. Since (52) = 3 is not divisible by 9.
Princeton Review Error : Congruences Page 225

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 Joined: Sun Sep 13, 2009 5:45 pm
Re: Princeton Review Error : Congruences Page 225
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
$$a_1 \pm a_2 \equiv b_1 \pm b_2 \pmod{n}$$
$$a_1 a_2 \equiv b_1 b_2 \pmod{n}$$
If $$a_1 \equiv b_1 \pmod{n}$$ and $$a_2 \equiv b_2 \pmod{n}$$ , then
$$a_1 \pm a_2 \equiv b_1 \pm b_2 \pmod{n}$$
$$a_1 a_2 \equiv b_1 b_2 \pmod{n}$$
Re: Princeton Review Error : Congruences Page 225
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?
Any idea on what LeDuc really means?
Re: Princeton Review Error : Congruences Page 225
Sorry, nevermind.
I figured out what I was doing wrong.
I figured out what I was doing wrong.
Re: Princeton Review Error : Congruences Page 225
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 + (c1)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.
I think he meant that a = b (mod n) then, a = b or b + n or b + 2n or ... or b + (c1)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.