1     1     1
a     b     c
Replace the variables a, b, and c with numbers to complete this equation.

In this open ended problem we were asked to find how many numbers we could find to make this equation true.

Process Description:
When I first read the problem and started to think about it, I immediately thought about negative numbers. I decided to start out simple so I tried to figure out what numbers could be subtracted to equal a negative number. I came up with 1/1 – 1/0 = 1/(-1), however being an immediate guess I didn't realize that the numbers where in the denominators place and quickly started to come up with different numbers. The next set of numbers that I came up with were a = 20, b = 30, c = 60. The way I came up with these is I first started with 20 and 30, the equation was 1/20 – 1/30 = 1/c. Then I found the lcd, or least common denominator, for 20 and 30 which is 60. After that I put in the numerators which was 3 and 2 because 20*3=60 and 30*2=60. Then the equation looked like this: 3/60 – 2/60 = 1/60. This equation works because 3 – 2 = 1, and there is still a 1 above the c. After completing this example I continued to solve the problem using varying sets of numbers to replace the variables, including a=5 b=6 c=30. I found c in this equation by multiplying the numbers in the a and b places together, but this does not work in every situation.
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​A pattern that we also learned was: a = b/2. For example b = 4, (4/2=2) a = 2, c = 4. 1/2 – 1/4 = 1/4, the least common denominator is 4, 2/4 – 1/4 = 1/4. And the equation works 2 – 1 = 1. This works with almost any set of numbers.
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My solutions to the crazy fraction problem looked like this:
1/5 – 1/6 = 1/c
5*6=30
C = 30
1/5 – 1/6 = 1/30
Lcd = 30
30/5 = 6  30/6 = 5
6/30 – 5/30 = 1/30
6 – 5 = 1
1/30
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As well as:
1/20 – 1/30 = 1/60
60/20 = 3  60/30 = 2
3/60 – 2/60 = 1/60
3 – 2 = 1
1/60
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And:
½ – ⅓ = 1/c
2*3=6
C = 6
½ – ⅓ = 1/6
Lcd = 6
6/2 = 3  6/3 = 2
3/6 – 2/6 = 1/6
3 – 2 = 1
1/6
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I did multiple sets of numbers to test the equation in different situations. How I solved the problem was like this:
(using 2, and 3)
First I took the number and multiplied it to get the c value and the least common denominator. 2 multiplied by 3 equals 6.
Using the number six a c the equation stated that ½ subtracted by ⅓ equals 1/6. However the denominators are not the same.
Using 6 as the lcd we have 3/6 and 2/6. Now the equation looks like this: 3/6 – 2/6 = 1/6
If you solve this equation it is true. 3/6 – 2/6 = 1/6 (3 - 2 = 1), and it shows that is is correct.
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While doing this problem I reviewed subtracting fractions but I/we really went more in-depth looking at problems and trying different ways to solve them as well as collaborating with other students to see how they solved the problem to help improve our understanding. I thought that the problem was very interesting, yet very simple, and could be solved using an infinite combination of numbers. I would give my self a 10/10 on this assignment because I understood the problem very well, I found several number sets that work in the equation, collaborated with other students, and helped some students to solve the problem.
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Habits of Mind:
The habits of mind I choose are evidence and supposition. I used evidence to prove that 1/a – 1/b = 1/c was a true equation because I found several different ways to demonstrate that my answers to the equation are correct. Using supposition I showed that I could use different sets of numbers in different circumstances to keep the equation true.