s Not Based On POMI James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University February 12, 2018
Outline 1 Non POMI Based s 2 Some Contradiction s 3 4 Homework
Some Contradiction s 2 is not a rational number We will prove this technque using a technique called contradiction. Let s assume we can find positive integers p and q so that 2 = (p/q) 2 with p and q having no common factors. When this happens we say p and q are relatively prime. This tells us p 2 = 2q 2 which also tells us p 2 is divisible by 2. Thus, p 2 is even. Does this mean p itself is even? Well, if p was odd, we could write p = 2l + 1 for some integer l. Then, we would know p 2 = (2l + 1) 2 = 4l 2 + 4l + 1. The first two terms, 4l 2 and 4l are even, so this implies p 2 would be odd. So we see p odd implies p 2 is odd. Thus, we see p must be even when p 2 is even. So we now know p = 2k for some integer k as it is even. But since p 2 = 2q 2, we must have 4k 2 = 2q 2. But this says q 2 must be even.
Some Contradiction s The same reasoning we just used to show p odd implies p 2 is odd, then tells us q odd implies q 2 is odd. Thus q is even too. Now here is the contradiction. We assumed p and q were relatively prime; i.e. they had no common factors. But if they are both even, they share the factor 2. This is the contradiction we seek. Hence, our original assumption must be incorrect and we can not find positive integers p and q so that 2 = (p/q) 2.
Some Contradiction s 3 is not a rational number Let s assume we can find positive integers p and q so that 3 = (p/q) 2 with p and q being relatively prime. This tells us p 2 = 3q 2 which also tells us p 2 is divisible by 3. Does this mean p itself is divisible by 3? Well, if p was not divisible by 3, we could write p = 3l + 1 or 3l + 2 for some integer l. Then, we would know or p 2 = (3l + 1) 2 = 9l 2 + 6l + 1. p 2 = (3l + 2) 2 = 9l 2 + 12l + 4. The first two terms in both choices are divisible by 3 and the last terms are not. So we see p 2 is not divisible by 3 in both cases. Thus, we see p must be divisible by 3 when p 2 is divisible by 3.
Some Contradiction s So we now know p = 3k for some integer k as it is divisible by 3. But since p 2 = 3q 2, we must have 9k 2 = 3q 2. But this says q 2 must be divisible by 3. The same reasoning we just used to show p 2 divisible by 3 implies p is divisible by 3, then tells us q 2 divisible by 3 implies q is divisible by 3. Now here is the contradiction. We assumed p and q were relatively prime; i.e. they had no common factors. But if they are both divisible by 3, they share the factor 3. This is the contradiction we seek. Hence, our original assumption must be incorrect and we can not find positive integers p and q so that 3 = (p/q) 2.
Some Contradiction s Let s introduce some notation: 1 if p and q are relatively prime integers, we say (p, q) = 1. 2 if the integer k divides p, we say k p.
Some Contradiction s Let s introduce some notation: 1 if p and q are relatively prime integers, we say (p, q) = 1. 2 if the integer k divides p, we say k p. Now let s modify the two proofs we have seen to attack the more general problem of showing n is not rational if n is prime.
Some Contradiction s n is not a rational number when n is a prime number Let s assume there are integers u and v with (u, v) = 1 so the n = (u/v) 2 which implies nv 2 = u 2. This tells us n u 2. Now we will invoke a theorem from number theory or abstract albebra which tells us every integer u has a prime factor decomposition: u = (p 1 ) r 1 (p 2 ) r2 (p s ) rs for some prime numbers p 1 to p s and positive integers r 1 to r s. For example, here are two such prime decompositions. 66 = 2 3 11 80 = 2 4 5 It is easy to see what the integers p 1 to p s and r 1 to r s are in each of these two examples and we leave that to you!
Some Contradiction s Thus, we can say u 2 = (p 1 ) 2r1 (p 2 ) 2r2 (p s ) 2rs Since n u 2, n must divide one of the terms in the prime factor decomposition: i.e. we can say n divides the term p r i i. Now the term p r i i is a prime number to a positive integer power r i. The only number that can divide into that evenly are appropriate powers of p i. But, we know n is a prime number too, so n must divide p i itself. Hence, we can conclude n = p i. But this tells us immediately that n divides u too.
Some Contradiction s Hence, we know now u = nw for some integer w. This tells us (nw) 2 = nv 2 or nw 2 = v 2. Thus, n v 2. The previous argument applied to v then tells us n v too. Hence u and v share the factor n which is a contradiction. Thus we conclude n is irrational.
Definition Absolute Values Let x be any real number. We define the absolute value of x, denoted by x, by x = { x, if x 0 x, if x < 0. For example, 3 = 3 and 4 = 4.
Theorem Triangle Inequality Let x and y be any two real numbers. Then and for any number z. x + y x + y x y x + y x y x z + z y
We know ( x + y ) 2 = (x + y) 2 which implies But 2xy 2 x y impyling ( x + y ) 2 = x 2 + 2xy + y 2. ( x + y ) 2 x 2 + 2 x y + y 2 = x 2 + 2 x y + y 2 = ( x + y ) 2. Taking square roots, we find x + y x + y. Of course, the argument for x y is similar as x y = x + ( y). To do the next part, we know a + b a + b for any a and b. Let a = x z and b = z y. Then (x z) + (z y) x z + z y.
Comment The technique where we do x y = (x z) + (z y) is called the Add and Subtract Trick and we will use it a lot! Comment Also note x c is the same as c x c and we use this other way of saying it a lot too. Theorem Backwards Triangle Inequality Let x and y be any two real numbers. Then x y x y y x x y x y x y
Let x and y be any real numbers. Then by the Triangle Inequality x = (x y) + y x y + y x y x y Similary, y = (y x) + x y x + x y x x y Combining these two cases we see x y x y x y But this is the same as saying x y x y.
Lemma Proving a Number x is Zero via Estimates Let x be a real number that satisfies x < ɛ, ɛ > 0. Then, x = 0. We will prove this by contradiction. Let s assume x is not zero. Then x > 0 and x /2 is a valid choice for ɛ. The assumption then tells us that x < x /2 or x /2 < 0 which is not possible. So our assumption is wrong and x = 0.
Theorem Extended Triangle Inequality Let x 1 to x n be a finite collection of real numbers with n 1. Then x 1 + x n x 1 + + x n or using summation notation n x i n x i. BASIS : P(1) is the statement x 1 x 1 ; so the basis step is verified. INDUCTIVE : We assume P(k) is true for an arbitrary k > 1. Hence, we know k k x i x i.
Now look at P(k + 1). We note by the triangle inequality applied to a = k x i and b = x k+1, we have a + b a + b or ( k+1 ) k x i x i + x k+1 Now apply the induction hypothesis to see k+1 k x i x i + x k+1 = k+1 This shows P(k + 1) is true and by the POMI, P(n) is true for all n 1. x i
Theorem l 2 Norm Inequality Let {a 1,..., a n } and {b 1,..., b n } be finite collections of real numbers with n 1. Then n 2 ( n ) ( n ) a i b i ai 2 bi 2 BASIS : P(1) is the statement a 1 b 1 2 a1 2b2 1 ; the basis step is true. INDUCTIVE : We assume P(k) is true for k > 1. Hence, we know k 2 ( k a i b i a 2 i ) ( k b 2 i )
Now look at P(k + 1). k+1 2 a i b i = k 2 a i b i + a k+1 b k+1 Let A denote the first piece; i.e. A = k a ib i. Then expanding the term A + a k+1 b k+1 2, we have k+1 2 a i b i = = ( k+1 a i b i ) 2 = A 2 + 2Aa k+1 b k+1 + a 2 k+1 b 2 k+1 k 2 ( k ) a i b i + 2 a i b i a k+1 b k+1 + ak+1 2 bk+1 2
or k+1 a i b i 2 k 2 k a i b i + 2 a i b i a k+1 b k+1 + ak+1 2 bk+1 2 Now use the induction hypothesis to see k+1 2 k k a i b i ai 2 bi 2 + 2 k a 2 i k bi 2 a k+1 b k+1 (1) +a 2 k+1 b 2 k+1
k k Now let α = a2 i b k+1 and β = b2 i a k+1. We know for any real numbers α and β that (α β) 2 0. Thus, α 2 + β 2 2α β. We can use this in our complicated sum above. We have 2α β = 2 k k bi 2 a k+1 b k+1 α 2 + β 2 = ( k a 2 i a 2 i ) b 2 k+1 + ( k b 2 i ) a 2 k+1
Hence, the middle part of Equation 1 can be replaced by the 2αβ α 2 + β 2 inequality above to get k+1 2 a i b i = k + a 2 i ( k ( k k bi 2 + b 2 i ) a 2 i ) a 2 k+1 + a 2 k+1 b 2 k+1 b 2 k+1 (2) ( k ) ( k ) ai 2 + ak+1 2 bi 2 + bk+1 2
But this says k+1 2 a i b i ( k+1 a 2 i ) ( k+1 b 2 i ) This shows P(k + 1) is true and by the POMI, P(n) is true for all n 1.
Comment This is a famous type of theorem. For two vectors [ ] [ ] a1 b1 V = and W = a 2 the inner product of V and W, < V, W > is a 1 b 1 + a 2 b 2 and the norm or length of the vectors is V = a 21 + a22 and W = b1 2 + b2 2 The Theorem above then says < V, W > V W. This is called the Cauchy - Schwartz Inequality also. b 2 Comment This works for three and n dimensional vectors too.
Homework Homework 3 Prove the following propositions. 3.1 5 is not a rational number using the same sort of argument we used in the proof of 3 is not rational. 3.2 7 is not a rational number using the same sort of argument we used in the proof of 3 is not rational. 3.3 On the interval [1, 10], use factoring and the triangle inequality to prove x y 2 10 x y.