Congruence
True or False: (give reason)
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2 ∈ 18 (mod 8)
Can someone please help with this problem? I'm confused. . . Thanks for any helpful replies.
The definition of congruence is when
(b-a)/m is an integer, then we can write
a ≡ b (mod m).
For example, we can write
3 ≡ 15 (mod 4)
because (15-3)/4 = 3 is an integer.
Alternatively, 15/4 gives 3 as a remainder, but this latter case is not always true.
Would this be sufficient for you to solve the given problem?
So, this would read as :
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2 = 18 (mod 8)
Meaning this would be considered true, right?Because 18/8 = 2, and then 8 · 2 = 16, making the remainder 2.
This might sound silly, but what does the symbol at the top of the number mean?
_
2
Sorry, I do not know.
I have seen the overbar used to mean:
1. mean (average) value,
2. recurring decimals,
3. complement (of a set)
4. complex conjugate
but not in this case.
You may want to check with your teacher. If it is in a computer generated exercise, there is usually a legend page into which you can look.
2 ∈ 18 (mod 8)
It is true though, right?
2 ∈ 18 (mod 8)
is not true, because ∈ is an operator for sets. For example:
2∈ℝ is correct, because 2 is a member of the real numbers.
Since 18 is not a set which includes 2, the above statement is not correct.
However, if you meant
2 ≡ 18 (mod 8)
that would be correct.
The code for ≡ is & # 8 8 0 1 ; without the spaces between & and ;.
Hmmm...I kind of get what you are saying, but why is 18 not a set that does not include 2?
Here is an example in the book that is true:
_
55 ∈ 7 (mod 3)
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7 (the line goes over 7 in the above)
Why would this be considered true?
Could you tell me if I am correct in thinking:
• With respect to congruence mod 29, 17 ∩ 423 = ∅ (True)
• Let a, b, and n be integers with n > 1. Then a ≡ b (mod n) ⇔ a = b (False)
•If ac ≡ bc(mod n), and gcd(c, n) = 1, then a ≡ b(mod m) (True)
I'll get back to the last three proposition. But could you check in your book what the overbar stands for? It does not seem to be a typo at all.
I think I found something about the overbar
_
a <--- equivalence class of a
_
b <---equivalence class of b
Let a, b, and n be integers with n > 1. Then a ≡ b (mod n) ⇔ a = b (False)
Correct.
Note that:
a=b => a≡b(mod n) but not the other way round.
For the other part:
ac≡bc (mod n)
=> (bc-ac)=kn k∈ℤ
=> (b-a)c=kn
=> (b-a)=k'n (GCD(c,n)=1)
=> b≡a (mod n)
(true)