A Counterexample to the statement, If ac≡bc(modm), where a,b,c, and m are integers with m≥2, then a≡b(modm).

Here we need to just find one counter example that contradicts the validity of the given statement. Suppose we set values a=0, b=1, c=3, m=4 in ac≡bc(modm) We get, 0×1≡1×3(mod3) 0=3(mod3) 0=0 Thus, this congruence is satisfied. Next let us plug these values in a≡b(modm) We get, 0≡1(mod3), which is obviously incorrect. Thus a counterexample could be the set of value a=0, b=1, c=3, m=4

Expert Answer to "A Counterexample to the statement, If ac≡bc(modm), where..."

High School
Geometry
High school geometry
Congruence

Rivka Thorpe

Answered question

2021-04-30

A Counterexample to the statement, If $a c \equiv b c (\mod m)$ , where a,b,c, and m are integers with $m \geq 2$ , then $a \equiv b (\mod m)$ .

Answer & Explanation

pivonie8

Skilled2021-05-02Added 91 answers

Here we need to just find one counter example that contradicts the validity of the given statement.
Suppose we set values a=0, b=1, c=3, m=4 in $a c \equiv b c (\mod m)$
We get,
$0 \times 1 \equiv 1 \times 3 (\mod 3)$
$0 = 3 (\mod 3)$
0=0
Thus, this congruence is satisfied.
Next let us plug these values in $a \equiv b (\mod m)$
We get,
$0 \equiv 1 (\mod 3)$ , which is obviously incorrect.
Thus a counterexample could be the set of value
a=0, b=1, c=3, m=4

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