CLOCK ARITHMETIC.

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Mod arithmetic.

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Presentation transcript:

CLOCK ARITHMETIC

... -4 -3 -2 -1 0 1 2 3 4 5 6 ... 2 + 3

... -4 -3 -2 -1 0 1 2 3 4 5 6 ... 3 - 7 We define ordinary arithmetic by hopping up and down a straight line. Let’s define a new kind of arithmetic by hopping around a circle!

4 1 3 2 2 + 4 = 1

4 1 3 2 In this system, 4 + 3 = 2

4 1 3 2 and 2 - 3 = 4

4 1 3 2 4  3 = 2

1 4 2 3  3 = 2 We evaluate 43 by taking 12 hops around the clock. Because every fifth hop is zero we land on the remainder when 12 is divided by 5.

1 4 2 3 5 3 8 4 + 4 = 3

1 4 2 3 10 2 12 4  3 = 2

1 4 2 3 5 4 9 3  3 = 4

15 10 16 etc 5 14 11 9 6 4 1 3 2 7 8 12 13 Numbers of the same color belong to the same modular class.

1 4 2 3 ARITHMETIC MOD 5 Solve for x: 2 x + 4 = 2 1 4 2 3 ARITHMETIC MOD 5 Solve for x: 2 x + 4 = 2 Additive inverse of 4 associativity 2 x + 4 = 2 ( ) + 1 3( ) 2 x + 0 = 3

1 4 2 3 ARITHMETIC MOD 5 Solve for x: 2 x + 4 = 2 2 x + 4 = 2 ( ) + 1 1 4 2 3 ARITHMETIC MOD 5 Solve for x: 2 x + 4 = 2 2 x + 4 = 2 ( ) + 1 Multiplicative inverse of 2 3( ) 2 x + 0 = 3

1 4 2 3 ARITHMETIC MOD 5 Solve for x: 2 x + 4 = 2 2 x + 4 = 2 ( ) + 1 1 4 2 3 ARITHMETIC MOD 5 Solve for x: 2 x + 4 = 2 2 x + 4 = 2 ( ) + 1 (3• ) 2 x + 0 = 3 3( ) 1 x = 4