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.

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Presentation transcript:

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 ( )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: 2 2  1, 2 3  2, 2 4  1, etc. p = 5: 2 2  4, 2 3  3, 2 4  1, 2 5  2, 2 6  4, etc. p = 11: 2 2  4, 2 3  8, 2 4  5, 2 5  10, 2 6  9, 2 7  7, 2 8  3, 2 9  6, 2 10  1, 2 11  2, 2 12  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 2 p -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  (mod 5). Use what we’ve learned to solve x 82  38 (mod 11). Read Chapter 9 and do Exercise 9.1.