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Upcoming MA 214 Schedule (2/21). The mid-term exam will be entirely take-home (with very specific “ground rules”). It will be handed out on Wednesday March 5 at class time and will be due at 10 am on Friday March 7. This will be a hard deadline – no late tests accepted .
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Upcoming MA 214 Schedule (2/21) • The mid-term exam will be entirely take-home (with very specific “ground rules”). • It will be handed out on Wednesday March 5 at class time and will be due at 10 am on Friday March 7. This will be a hard deadline – no late tests accepted. • There will be extended office hours on Thursday March 6 (2-5 pm). • There will be no class on Friday, March 7. (I’m sure you are heart-broken by this news.)
Reducing bases and exponents in congruences • In the linear congruence we considered in the last class (318x 42 (mod 186)), it would have been a little simpler to reduce the coefficient 318 down to its remainder modulo 186 right to start with. That is, an equivalent congruence is (318-186)x 42 (mod 186)), i.e.,132x 42 (mod 186)). • We are saying that if a = qm + r, then x satisfies a x c (mod m) if and only if it satisfies r x c (mod m). • We can also reduce c if we like. • Punch line: In any congruence, any number lying in the base (as opposed to being an exponent) can be replaced by its remainder modulo m.
But what about exponents? • Can we reduce exponents which are larger than the modulus? • The answer is yes, but not in the same way as with numbers in the base. • We’ll be able to pin this down for arbitrary moduli, but for the moment let’s assume that the modulus is a prime p. • Let’s look at some data for, say, the moduli being p = 3, 5, and 11, and using only the base 2 for starters: • p = 3: 22 1, 23 2, 24 1, etc. • p = 5: 22 4, 23 3, 24 1, 25 2, 26 4, etc. • p = 11: 22 4, 23 8, 24 5, 25 10, 26 9, 27 7,28 3, 29 6, 210 1, 211 2, 212 4, etc. • For a given prime modulus p, what exponent seems to get 1?
A Conjecture on Reducing Exponents • So in particular, we see that you cannot reduce exponents in exactly the way you can reduces bases. But what can we say? • Conjecture. If p is an odd prime, then 2p -1 1 (mod p). • Hence, at least for the base 2, we can reduce exponents modulo p – 1 (as opposed to modulo p) provided the modulus is an odd prime. • Let’s test this out for larger primes and for other bases in Mathematica. • Possible Generalized Conjecture: If a and p are relatively prime, then a p -1 1 (mod p). • This is in fact Fermat’s Little Theorem
Two Examples and Assignment for Monday • Use what we’ve learned to solve x 234167 (mod 5). • Use what we’ve learned to solve x82 38 (mod 11). • Read Chapter 9 and do Exercise 9.1.