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Notes for Session 9, Part D

Note 7

One of the interesting extensions of these algebraic systems is the connection to equation solving. Many of the techniques for solving equations become automatic with experience, and solving equations in modular systems (mod 10 is the modular system for units digit arithmetic) draws attention to the properties of numbers that are sometimes assumed or taken for granted. For example, in solving 3x = 8, we need to determine whether or not 3 has an inverse in the system we're dealing with.

Groups: Write 3x = 8 on an overhead, and discuss in a full group how you might solve this equation in units digit arithmetic. Some may say you should divide by 3; others may say you should multiply by 1/3. Remember that the only operations we have in our system are addition and multiplication, and 1/3 is not in the domain of units digit arithmetic. In fact, you need to consider what number is the reciprocal of 3 in our new system. If you have trouble, you can look at the table to see what number multiplied by 3 yields 1. Then go through the steps to solve the equation.

Groups: Next, write x + 4 = 2 on an overhead, and discuss how to solve this equation by adding the opposite of 4 to both sides.

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Note 8

Groups: Work on Problems D1-D4 in small groups. You may notice that in mod 10, there are certain numbers that don't have reciprocals: all of the numbers that share a factor with 10. (Another way to say this is: The only numbers that do have reciprocals are relatively prime to 10.) You may also want to take time to discuss how solving equations in units arithmetic differs from solving equations in the real number system.

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Note 9

Working in different algebraic systems has applications not just inside mathematics, but in other fields as well. In this part, we'll explore an applied use of modular systems. We'll see how modular systems are helpful when you want to keep the answers to calculations within a certain range (in this case, the alphabet).

Note the meaning of the terms "plaintext" (the original message; it's just the English words) and "ciphertext" (the encoded message; it looks like nonsense). Caesar reportedly used a very simple cipher to send military secrets: He assigned each letter a number from 0-25, then added 3 to the letter in the plaintext to get the ciphertext.

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Note 10

Use the "letter tables" to work on Problems D6 and D7.

Groups: After thinking about how to encipher a "Y" with Caesar's rule, have a discussion about how modular systems keep the results of calculations within a certain range. In the case of mod 10, all the answers were 0-9 (that is, units digits). If you want the answers to be between 0-25, you should work modulo 26, or just mod 26.

Groups: Put Caesar's modified cipher on an overhead:
C = P + 3 (mod 26)

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Note 11

Groups: Work on Problem D8 alone or with a partner. You may find that even though you know the rule, you will work on this as a "cryptogram," making guesses and trying to see words. Explain your strategy to the group (usually looking at the letter table and subtracting 3 from each letter in the coded message).

Once you have a modular system, there's no reason to restrict yourself to the rule "add 3." You can perform any algebraic rule before using mod 26, and you'll still get a cipher. (Some ciphers are better than others, though. Working mod 26, if you multiply by an even number or by 13, you will have several plaintext letters map to the same ciphertext letters. This is no good, because the cipher can't be "undone," even by its intended recipient! The reasons for this are beyond the scope of this course. We have stayed with multiples that create good ciphers.)

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Note 12

Groups: Work on Problems D9-D11 with a partner. Problem D11 is quite challenging, and you may approach it any number of ways, either by using "data" from their secret words, or by trying to solve the equation for P.

Remember that there is no "divide by 3" rule in a mod 26 system. If you come up with the equation P = (C - 2) / 3 (mod 26), how would you decipher M? M = 12, so P = (12 - 2) / 3 = 10 / 3. How can you find 10 / 3 in this system?

Groups: Before wrapping up this part of the session, share your equations and solutions. If no one actually solved the equation for P, you can look at this solution on an overhead or on the board:
C = 3P + 2 (mod 26)
C - 2 (mod 26) = 3P (mod 26)

Notice that we need mod 26 on both sides at this point, because although C is between 0 and 25, C - 2 may not be (it may be -2, for example).

Now you need to multiply both sides of the equation by the reciprocal of 3. There is no 1/3 in a mod 26 system, but there is a number r, so that 3r = 1 (mod 26). Discuss what that number must be (the answer is 9, because 3 * 9 = 27 = 1 (mod 26)).

9 * (C - 2) (mod 26) = 9 * (3P) (mod 26)
P = 9 * (C - 2) (mod 26)

If you worked on this another way, you may have come up with different but equivalent equations:
P = 9 C - 18 (mod 26)
P = 9 C +12 (mod 26)

and so on.

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