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More Number Theory Proofs Rosen 1.5, 3.1 Prove or Disprove • • • • If m and n are even integers, then mn is divisible by 4. The sum of two odd integers is odd. The sum of two odd integers is even. If n is a positive integer, then n is even iff 3n2+8 is even. • n2 + n + 1 is a prime number whenever n is a positive integer. • n2 + n + 1 is a prime number whenever n is a prime number. • |x| + |y| |x + y| when x,y R. • 3 is irrational. If m and n are even integers, then mn is divisible by 4. Proof: m and n are even means that there exists integers a and b such that m =2a and n = 2b Therefore mn = 4ab. Since ab is an integer, mn is 4 times an integer so it is divisible by 4. The sum of two odd integers is odd. This is false. A counter example is 1+3 = 4 The sum of two odd integers is even. Proof: If m and n are odd integers then there exists integers a,b such that m = 2a+1 and n = 2b+1. m + n = 2a+1+2b+1 = 2(a+b+1). Since (a+b+1) is an integer, m+n must be even. If n is a positive integer, then n is even iff 3n2+8 is even. Proof: We must show that n is even 3n2+8 is even, and that 3n2+8 is even n is even. First we will show if n is even, then 3n2+8 is even. n even means there exists integer a such that n = 2a. Then 3n2+8 = 3(2a)2 + 8 = 12a2 + 8 = 2(6a2 + 4) which is even since (6a2 + 4) is an integer. If n is a positive integer, then n is even iff 3n2+8 is even (cont.). Now we will show if 3n2+8 is even, then n is even using the contrapositive (indirect proof). Assume that n is odd, then we will show that 3n2+8 is odd. n odd means that there exists integer a such that n = 2a+1. 3n2+8 = 3(2a+1)2 + 8 = 3(4a2 + 4a + 1) + 8 = 2(6a2 + 6a + 5) + 1, which is odd. Therefore, by the contrapositive if 3n2+8 is even, then n is even. n2 + n + 1 is a prime number whenever n is a positive integer. Try some examples: n = 1, 1+1+1 = 3 is prime n = 2, 4+2+1 = 7 is prime n = 3, 9+3+1 = 13 is prime n = 4, 16+4+1 = 21 is not prime and is a counter example. Not true. n2 + n + 1 is a prime number whenever n is a prime number. Try some examples: n = 1, 1+1+1 = 3 is prime n = 2, 4+2+1 = 7 is prime n = 3, 9+3+1 = 13 is prime n = 5, 25+5+1 = 31 is prime n = 7, 49+7+1 = 57 is not prime (19*3). Not true. Prove |x| + |y| |x + y| when x,y R. Note: |z| is equal to z if z0 and equal to -z if z<0 There are four cases x y 0 0 <0 0 0 <0 <0 <0 Prove |x| + |y| |x + y| when x,y R. Case 1 x,y are both 0 Then |x| + |y| = x + y = |x+y| since both x and y are positive. Prove |x| + |y| |x + y| when x,y R. Case 2 x < 0 and y 0 so |x| + |y| = -x + y If y -x, then x+y is nonnegative and |x+y| = x+y Since x is negative, -x > x, so that |x| + |y| = -x + y > x+y = |x+y| If y < -x, then |x+y| = -(x+y) = -x + -y. Since y 0, then y -y, so that |x| + |y| = -x + y -x + -y = |x+y| Prove |x| + |y| |x + y| when x,y R. Case 3 x0 and y <0 so |x| + |y| = x + -y If x -y, then x+y is nonnegative and |x+y| = x+y Since y is negative, -y > y, so that |x| + |y| = x + -y > x+y = |x+y| If x < -y, then |x+y| = -(x+y) = -x + -y. Since x0, then x-x , so that |x| + |y| = x + -y -x + -y = |x+y| Prove |x| + |y| |x + y| when x,y R. Case 4 x,y are both < 0 Then |x| + |y| = -x + - y = -(x+y) = |x+y| Therefore the theorem is true. This is know in mathematics as the Lipschitz condition. Prove that 33 is irrational. Proof (by contradiction): Assume that 33 is rational, i.e. that 33 = a/b for a,bZ and b0. Since any fraction can be reduced until there are no common factors in the numerator and denominator, we can further assume that a and b have no common factors. Prove that 33 is irrational. (cont.) Then 3 = a3/b3 which means that 3b3 = a3 so a3 is divisible by 3. Lemma: When m is a positive integer, then if m3 is divisible by 3, then m is divisible by 3. (Left as an exercise). By the lemma since a3 is divisible by 3, then a is divisible by 3. Thus kZ a = 3k. Prove that 33 is irrational. (cont.) Now, we will show that b is divisible by 3. From before, a3/b3 = 3 3b3 = a3 = (3k)3. Dividing by 3 gives b3 = 9k3 = 3(3k3). Therefore b3 is divisible by 3 and from the Lemma , b is divisible by 3. Prove that 33 is irrational. (cont.) But, if a is divisible by 3 and b is divisible by 3, then they have a common factor of 3. This contradicts our assumption that our a/b has been reduced to have no common factors. Therefore 33 a/b for some a,bZ, b0. Therefore 33 is irrational. Some Other Proof Strategies Rosen 3.1 Backward Reasoning We have used mostly forward reasoning strategies up to now. However, sometimes it is unclear how to proceed from the initial assumptions. Bacward reasoning may help-Motto: If you can’t prove the original proposition, equate it to one you can prove. Prove (a,b Z+, a≠b)[(a+b)/2> ab] (i.e., the arithmetic mean is always greater than the geometric mean for this universe of discourse.) Backward reasoning proof (a+b)/2> ab (a+b)2/4> ab (a+b)2> 4ab a2+2ab+b2 > 4ab a2-2ab+b2 > 0 (a-b)2> 0 Original Assumption Why? Why? Why? Why? Why? Prove (a,b Z+, a≠b)[(a+b)/2> ab] Backward reasoning proof (cont.) But, a,b Z+, (a-b)2> 0 implies a≠b. We can now easily start from a≠b and go backwards to reconstruct the path to prove the original proposition. Conjecture and Proof A conjecture is a plausible statement that has not been proved. A conjecture may result from recognizing that there are multiple examples for which it is true. For some conjectures, counterexamples are eventually found. However, even if a conjecture is valid for very many examples, this does not usually constitute a valid proof. (Why?) Conjecture and Proof Sometimes a complex proof is constructed as a series of conjectures that are then proved. Sometimes a proof is found by referring to the proof of a similar problem or class of problem. There are many famous conjectures that are still not proved (or only recently proved). Fermat’s Last Theorem xn + yn = zn has no solution for x,y,z,n Z, x,y,z≠0, n>2 This is a conjecture made by Pierre de Fermat (1601-1665), the French mathematician. He wrote in the margin of his copy of the works of Diophantus, an ancient Greek mathematician, that he had a “wondrous proof”, but that it wouldn’t fit in the margin. He then died, leaving no record of the proof! Fermat’s Last Theorem Attempts at proof over the years led to new fields, such as algebraic number theory. Finally, in 1994, Andrew Wiles provided a correct proof that required hundred of pages of advanced mathematics.