We then looked at the problem using a 24-hour clock. Dividing 43 by 24 leaves a remainder of 19, so
43 ≡ 19 (mod 24). This indicates we are 19 hours after noon. We also can see that 19 ≡ 7 (mod 12). This corresponds to being 7 a.m.
Since we have 43 ≡ 19 (mod 24) and 19 ≡ 7 (mod 12), does this mean that 43 ≡ 7 (mod 12)? I posed this question and asked students to think about the situation and not actually do any divisions to find remainders. A few students reasoned that it would be true because 19 and 7 leave the same remainder when dividing by 12. Since 24 is a multiple of 12, 19 would leave an equivalent remainder when divided by 24.
I shifted gears slightly at this point and asked what we know about the value (43-7) since 43 ≡ 7 (mod 12). A few students simply stated the value. But then a couple more chimed in that the difference is a multiple of 12.
I used this to transition to a formal definition of congruence:
If a and b are integers and m is a positive integer, then a is congruent to b modulo m if m divides a - b.
This is written as a ≡ b (mod m)
I reminded students that to say m divides a - b mean we can find an integer k such that km = a - b.
I went back to our example of 43 ≡ 7 (mod 12). This means that (43 - 7) must be a multiple of 12. However, for any other integer k, (43 - 7)k is also a multiple of 12. This means that 43k ≡ 7k (mod 12).
Congruence Property 1:
If a ≡ b (mod m) then ak ≡ bk (mod m) for any integer k.
This first congruence property may be an indication that congruence relations have similar properties to equality. I had the class explore this idea by looking at the additive property of equality.
In this case we have a ≡ b (mod m) and c ≡ d (mod m). The question is will a + c ≡ b + d (mod m)?
I asked students to create examples to see if this was the case. I reminded students to return to the definition of congruence to verify their results. It took some time for students to grasp what was going on and how things worked. As students came up with examples I put them on the board and used the definition of congruence to verify the results. As more examples went up on the board, almost every student was able to produce an example of this property.
I asked students how they could demonstrate that this property always held. Students thought about this and a couple of students came up with the idea that you were essentially subtracting off remainders and so the result would have to be divisible by m. I built upon this idea to produce a proof.
Congruence Property 2:
If a ≡ b (mod m) and c ≡ d (mod m) then a + c ≡ b + d (mod m) for integers a, b, c, d and positive
positive integer m.
Proof
We are told that a is congruent to b and c is congruent to d mod m, therefore we can find integers k and l such that (a - b) = km and (c - d) = lm. Now, look at the quantity (a + c) - (b + d). Is this a multiple of m?
First, (a + c) - (b + d) = a + c - b - d = a - b + c - d = (a - b) + (c - d), using basic operations and properties of arithmetic for integers. Substituting from above we have
(a - b) + (c - d) = km + lm = (k + l)m. But this means (a + c) - (b + d) = (k + l)m so m divides
(a + c) - (b + d). Therefore, by definition of congruence, a + c ≡ b + d (mod m). Q.E.D.
I pointed out that a, b, c, and d can be either positive or negative and provided an example to demonstrate this.
I next asked students what would have if we multiplied rather than added values. This corresponds to the multiplicative property of equality. I told students to use the examples they had started with for addition and multiply the values instead.
Students were seeing that if a ≡ b (mod m) and c ≡ d (mod m) then ac ≡ bd (mod m). The questions was how could they demonstrate that this was always true? The class was running out of time at this point, so it was left for homework that students were to attempt to demonstrate this result based upon the definition of a congruence relation and arithmetic properties.
Visit the class summary for a student's perspective and to view the lesson slides.
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