Question

Proof that if a \equiv b (\bmod n), then a+c \equiv b+c(\bmod n) and ac \equiv bc (\bmod n).

Polynomial factorization
ANSWERED
asked 2021-02-26

Proof that if \(\displaystyle{a}\equiv{b}{\left({b}\mod{n}\right)}\), then \(\displaystyle{a}+{c}\equiv{b}+{c}{\left({b}\mod{n}\right)}\) and \(\displaystyle{a}{c}\equiv{b}{c}{\left({b}\mod{n}\right)}\).

Answers (1)

2021-02-28

Step 1
Apply the definition of congruence and divisibility, to express \(\displaystyle{a}\equiv{b}{\left({b}\mod{n}\right)}\).
From the definition of congruence, \(\displaystyle{a}\equiv{b}{\left({b}\mod{n}\right)}\) means that \(\displaystyle{n}{\mid}{\left({a}-{b}\right)}\).
By the definition of divisibility, \(\displaystyle{n}{\mid}{\left({a}-{b}\right)}\) implies that \((a-b)\) is a multiple of n.
That is, \(a-b=nk\), where k is an integer.
Step 2
To prove \(\displaystyle{a}+{c}\equiv{b}+{c}{\left({b}\mod {n}\right)}\), add and subtract c on the left hand side of the equation
\(a-b=nk\) and use the definition of congruence.
\(a-b=nk\)
\(a-b+c-c=nk\)
\((a+c)-(b+c)=nk\)
Since, \((a+c)-(b+c)\) is a multiple of n, implies that \(\displaystyle{n}{\mid}{\left[{\left({a}+{c}\right)}-{\left({b}+{c}\right)}\right]}\).
Hence, by the definition of congruence, \(\displaystyle{a}+{c}=\equiv{b}+{c}{\left({b}\mod{n}\right)}\).
Thus, if \(\displaystyle{a}\equiv{b}{\left({b}\mod{n}\right)}\) then \(\displaystyle{a}+{c}=\equiv{b}+{c}{\left({b}\mod{n}\right)}\).
Step 3
To prove \(\displaystyle{a}{c}\equiv{b}{c}{\left({b}\mod{n}\right)}\), use the definition of congruence and the properties of divisibility.
\(\displaystyle{a}\equiv{b}{\left({b}\mod{n}\right)}\) implies that \(\displaystyle{n}{\mid}{\left({a}-{b}\right)}\).
By the properties of divisibility, if n divides \((a-b)\) then n also divides the multiple of \((a-b)\).
That is, \(\displaystyle{n}{\left|{\left({a}-{b}\right)}{c}={n}\right|}{\left({a}{c}-{b}{c}\right)}\)
Hence, by the definition of congruence, \(\displaystyle{a}{c}\equiv{b}{c}{\left({b}\mod{n}\right)}\).
Therefore, if \(\displaystyle{a}\equiv{b}{\left({b}\mod{n}\right)}\) then \(\displaystyle{a}{c}\equiv{b}{c}{\left({b}\mod{n}\right)}\).

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