MATH221Mathematics for Computer ScienceUnit 8Elementary Number Theory1
2OBJECTIVES•Understand the definition of divisibility.•Understand apply the Quotient-Remainder Theorem.•Understand and be able to find the Greatest Common Divisor.•Understand and apply the Euclidean Algorithm.•Understand the concept of the Fundamental Theorem of Arithmetic.
3DivisibilityIf nand dare integers and d0, then n is divisible by dif and only if n= dkfor some kℤ. We write d| nand say that d divides n (n is divisible by d).In logic notation, the definition of divisibility is writtend| nkℤ, n = dk.
4DivisibilityAlternatively, ifd| nkℤ, n = dk.We say thatnis a multiple of d, ordis a factor of n, ordis a divisor of n, orddivides n.
5Discussion:1.Is -16 a divisor of 32? Yes.32 = -16 -2. 2.If l ℤand l0, does l| 0? Yes.0 = l k for some kℤ.
6Discussion:3.Find all values of aℤsuch that a| 1.a| 1kℤ, 1= ak.For a= 1 and k= 1, ak= 1.For a= -1 and k= -1, ak= 1.Therefore, a= -1 or 1.
7Discussion:4.What is the relationship between aand bif a | b, and b| a, a, bℤ?Now, a| bkℤ, b = ak(1)b| alℤ, a = bl(2)Substitute (2) into (1),b = (bl)kDivide both sides by b,1 = lkSince l, kℤ, l= k= 1. or l= k= -1.Hence, a= b.
8Discussion:5.If a, bℤ, is 3a+ 3bdivisible by 3?Now,3a+ 3b= 3(a+ b)= 3swhere s= a+ b and s ℤ.Hence, 3 | 3a+ 3b, that is, 3a+ 3bis divisible by 3.
9Discussion:6.If a, b, c, x, yℤ. If b| aand b| c, does b| (ax+ cy)? Why?Now, b| akℤ, a = bk(1)b| clℤ, c = bl(2)(1) x,ax= bkx(3)(2) y,cy= bly(4)(3) + (4),ax+ cy =bkx+ bly= b(kx+ ly)= bmwhere m= kx+ ly, mℤ. Hence, b| (ax+ cy).
10Discussion:7.If a, b, is it true that ℕa| bimplies ab?Now,a| bkℤ, b = ak.Since a, b 1, sok1. Multiply both sides of the inequality by a,ka1a.Hence, ba.That is,abTherefore, if a, b, it is true that ℕa| bimplies ab.
11Theorem - Transitivity of DivisibilityFor all integers a, band c, if a| band b| c, then a| c.Note: After going through the details of the proof for this theorem, for the remaining theorems, lemma, etc., we shall focus on the application. For their proofs, please refer to the supplementary notes.
12Transitivity of Divisibility: ProofWe know that a| bkℤ, b = ak(1)b| clℤ, c = bl(2)Show that a| c, that is, find mℤsuch that c= ma.Now, c = blby(2)= (ak)lby (1)= (kl)aby associativity and commutativityLet m= kl, then mℤ(since is a closed operation on ℤ).