Project 7: Clock Arithmetic

The main difference between clock arithmetic and “regular” arithmetic is that the numbers stop getting bigger after passing 12 o’clock. Instead, they jump back down to one before continuing. The numbers for the hours on a typical clock face, with the addition operation, make up something in mathematics called a group.

A set of objects, S, together with an operation ⊕, is a group if all the following properties hold:

Closure: If a and b are members of S, then a ⊕ b is also a member of S.
Associativity: If a, b, and c are members of S, then (a ⊕ b) ⊕ c = a ⊕ (b ⊕ c).
Identity: There is a member of S, call it e, such that, for any member a in S, e ⊕ a = a = a ⊕ e.
Inverses: If a is a member of S, there must be a member of S, call it b, such that a ⊕ b = e.

You may recognize the associativity property as the associative law from a typical algebra class, and that is because real numbers, with the operation of addition, form a group. The identity element is zero, and the inverse of any element is its value multiplied by −1, so 47 and −47 are inverses because (47) + (−47) = 0.

The real numbers, with the operation of multiplication, do not form a group. While properties A, B, and C are satisfied (note the identity would be one, not zero), property D is not satisfied, since zero has no inverse. In other words, there is no real number a such that 0 × a = 1. Note that if we only consider strictly positive real numbers, then combining these with multiplication does form a group. The inverse of any real number a would be its reciprocal, ¹/_a, since a × (¹/_a) = 1.

These two examples are infinite groups since the groups contain infinitely many items. In this project, you will demonstrate how the hours of a clock, using the symbol ⊕ for “clock addition”, form a finite group, by showing that the four properties hold.

Part 1 – Clock Arithmetic

Define the operation ⊕ by stating that, for any two numbers on a clock face, a and b, a ⊕ b will be the resulting time that happens b hours after time a. So, 3 ⊕ 5 = 8, since five hours after 3 o’clock is 8 o’clock, and 6 ⊕ 9 = 3 since nine hours after 6 o’clock is 3 o’clock. Notice that times, in some sense, “wrap around” after passing 12 o’clock.

Explain why this group satisfies the closure property (A).
Calculate (6 ⊕ 8) ⊕ 11 and 6 ⊕ (8 ⊕ 11) separately (showing your work). Can you explain why this group satisfies the associativity property (B) in all cases?
Which hour on the clock would be the identity member described in the identity property (C)? Why is this the case?
Which hour on the clock would be the inverse of 4? Describe a rule for calculating the inverse, mentioned in the inverse property (D), of any hour on the clock other than the identity member (the identity member will always be its own inverse).

A group is called an ablelian group if it also satisfies the following property:

E. Commutativity: If a and b are members of S, then a ⊕ b = b ⊕ a.

What time is 10 hours after 5 o’clock and what time is 5 hours after 10 o’clock? Can you explain why this group satisfies the commutativity property?.

Part 2 – Modulo Arithmetic

The rules outlined in this project are similar to the rules for modulo division by 12 (replace the 12 on the clock face with a zero in this case) and can be generalized to a clock with any positive number of hours on it. A clock with 60 numbers (from zero to 59, representing the minutes) would follow the same rules.

Where would the minute hand be 49 minutes after it was on the 35^th minute?
Even though there is not “98^th minute” on a clock with 60 minutes, you can still calculate where the minute hand would be 98 minutes after it was on the 35^th minute.

Notice that on the clock with hours, moving the time forward 13 hours is no different than moving the time forward one hour, or moving the time forward 25 hours, for that matter. This can be characterized by saying: 13 ≡ 1 mod 12 and 25 ≡ 1 mod 12, and we even get 25 ≡ 13 mod 12.

Moving the minute hand forward 98 minutes would be equivalent to moving the minute hand forward minutes. (choose a number between 0 and 59)
By answering the previous question, you can now state that 98 ≡ mod 60.
[a little tricky] What is the simplest way to verify whether a ≡ b mod c is true? (hint: look at the difference between a and b)

Once completed, this page to a pdf document; then hand it in through your course's Learning Management System.

Neil Simonetti