A Problem With The Rational Numbers
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- Lorraine Martin
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1
2 Solvability of Equations
3 Solvability of Equations 1. In fields, linear equations ax + b = 0 are solvable.
4 Solvability of Equations 1. In fields, linear equations ax + b = 0 are solvable. 2. Quadratic equations ax 2 + bx + c = 0 are a natural next target.
5 Solvability of Equations 1. In fields, linear equations ax + b = 0 are solvable. 2. Quadratic equations ax 2 + bx + c = 0 are a natural next target. 3. We run into problems rather quickly.
6 Proposition.
7 Proposition. There is no rational number r such that r 2 = 2.
8 Proposition. There is no rational number r such that r 2 = 2. Proof.
9 Proposition. There is no rational number r such that r 2 = 2. Proof. First let n N be so that n 2 = 2z for some z N.
10 Proposition. There is no rational number r such that r 2 = 2. Proof. First let n N be so that n 2 = 2z for some z N. Then 2 is a factor in the prime factorization of n 2.
11 Proposition. There is no rational number r such that r 2 = 2. Proof. First let n N be so that n 2 = 2z for some z N. Then 2 is a factor in the prime factorization of n 2. But every factor in the prime factorization of n 2 occurs with an even exponent.
12 Proposition. There is no rational number r such that r 2 = 2. Proof. First let n N be so that n 2 = 2z for some z N. Then 2 is a factor in the prime factorization of n 2. But every factor in the prime factorization of n 2 occurs with an even exponent. Thus n 2 = 2 2 k 2 for some k.
13 Proposition. There is no rational number r such that r 2 = 2. Proof. First let n N be so that n 2 = 2z for some z N. Then 2 is a factor in the prime factorization of n 2. But every factor in the prime factorization of n 2 occurs with an even exponent. Thus n 2 = 2 2 k 2 for some k. Hence n = 2k.
14 Proposition. There is no rational number r such that r 2 = 2. Proof. First let n N be so that n 2 = 2z for some z N. Then 2 is a factor in the prime factorization of n 2. But every factor in the prime factorization of n 2 occurs with an even exponent. Thus n 2 = 2 2 k 2 for some k. Hence n = 2k. That is, if n 2 is divisible by 2, then so is n.
15 Proposition. There is no rational number r such that r 2 = 2. Proof. First let n N be so that n 2 = 2z for some z N. Then 2 is a factor in the prime factorization of n 2. But every factor in the prime factorization of n 2 occurs with an even exponent. Thus n 2 = 2 2 k 2 for some k. Hence n = 2k. That is, if n 2 is divisible by 2, then so is n. Now suppose for a contradiction that there are n Z and d N ( n ) 2 so that = 2. d
16 Proposition. There is no rational number r such that r 2 = 2. Proof. First let n N be so that n 2 = 2z for some z N. Then 2 is a factor in the prime factorization of n 2. But every factor in the prime factorization of n 2 occurs with an even exponent. Thus n 2 = 2 2 k 2 for some k. Hence n = 2k. That is, if n 2 is divisible by 2, then so is n. Now suppose for a contradiction that there are n Z and d N ( n ) 2 so that = 2. WLOG, we can assume that n and d have no d common factors and that n N.
17 Proposition. There is no rational number r such that r 2 = 2. Proof. First let n N be so that n 2 = 2z for some z N. Then 2 is a factor in the prime factorization of n 2. But every factor in the prime factorization of n 2 occurs with an even exponent. Thus n 2 = 2 2 k 2 for some k. Hence n = 2k. That is, if n 2 is divisible by 2, then so is n. Now suppose for a contradiction that there are n Z and d N ( n ) 2 so that = 2. WLOG, we can assume that n and d have no d common factors and that n N. But n 2 = 2d 2 implies n = n 2 2.
18 Proposition. There is no rational number r such that r 2 = 2. Proof. First let n N be so that n 2 = 2z for some z N. Then 2 is a factor in the prime factorization of n 2. But every factor in the prime factorization of n 2 occurs with an even exponent. Thus n 2 = 2 2 k 2 for some k. Hence n = 2k. That is, if n 2 is divisible by 2, then so is n. Now suppose for a contradiction that there are n Z and d N ( n ) 2 so that = 2. WLOG, we can assume that n and d have no d common factors and that n N. But n 2 = 2d 2 implies n = n 2 2. Consequently, 2d 2 = (n 2 2) 2
19 Proposition. There is no rational number r such that r 2 = 2. Proof. First let n N be so that n 2 = 2z for some z N. Then 2 is a factor in the prime factorization of n 2. But every factor in the prime factorization of n 2 occurs with an even exponent. Thus n 2 = 2 2 k 2 for some k. Hence n = 2k. That is, if n 2 is divisible by 2, then so is n. Now suppose for a contradiction that there are n Z and d N ( n ) 2 so that = 2. WLOG, we can assume that n and d have no d common factors and that n N. But n 2 = 2d 2 implies n = n 2 2. Consequently, 2d 2 = (n 2 2) 2, that is, d 2 = n 2 2 2
20 Proposition. There is no rational number r such that r 2 = 2. Proof. First let n N be so that n 2 = 2z for some z N. Then 2 is a factor in the prime factorization of n 2. But every factor in the prime factorization of n 2 occurs with an even exponent. Thus n 2 = 2 2 k 2 for some k. Hence n = 2k. That is, if n 2 is divisible by 2, then so is n. Now suppose for a contradiction that there are n Z and d N ( n ) 2 so that = 2. WLOG, we can assume that n and d have no d common factors and that n N. But n 2 = 2d 2 implies n = n 2 2. Consequently, 2d 2 = (n 2 2) 2, that is, d 2 = n 2 2 2, which implies d = d 2 2.
21 Proposition. There is no rational number r such that r 2 = 2. Proof. First let n N be so that n 2 = 2z for some z N. Then 2 is a factor in the prime factorization of n 2. But every factor in the prime factorization of n 2 occurs with an even exponent. Thus n 2 = 2 2 k 2 for some k. Hence n = 2k. That is, if n 2 is divisible by 2, then so is n. Now suppose for a contradiction that there are n Z and d N ( n ) 2 so that = 2. WLOG, we can assume that n and d have no d common factors and that n N. But n 2 = 2d 2 implies n = n 2 2. Consequently, 2d 2 = (n 2 2) 2, that is, d 2 = n 2 2 2, which implies d = d 2 2. But then 2 n and 2 d, contradiction.
22 Proposition. There is no rational number r such that r 2 = 2. Proof. First let n N be so that n 2 = 2z for some z N. Then 2 is a factor in the prime factorization of n 2. But every factor in the prime factorization of n 2 occurs with an even exponent. Thus n 2 = 2 2 k 2 for some k. Hence n = 2k. That is, if n 2 is divisible by 2, then so is n. Now suppose for a contradiction that there are n Z and d N ( n ) 2 so that = 2. WLOG, we can assume that n and d have no d common factors and that n N. But n 2 = 2d 2 implies n = n 2 2. Consequently, 2d 2 = (n 2 2) 2, that is, d 2 = n 2 2 2, which implies d = d 2 2. But then 2 n and 2 d, contradiction.
23 Definition.
24 Definition. Let X be a set and let be an order relation on X.
25 Definition. Let X be a set and let be an order relation on X. Let A X.
26 Definition. Let X be a set and let be an order relation on X. Let A X. 1. The element u X is called an upper bound of A iff u a for all a A.
27 Definition. Let X be a set and let be an order relation on X. Let A X. 1. The element u X is called an upper bound of A iff u a for all a A. If A has an upper bound, it is also called bounded above.
28 Definition. Let X be a set and let be an order relation on X. Let A X. 1. The element u X is called an upper bound of A iff u a for all a A. If A has an upper bound, it is also called bounded above. 2. The element l X is called a lower bound of A iff l a for all a A.
29 Definition. Let X be a set and let be an order relation on X. Let A X. 1. The element u X is called an upper bound of A iff u a for all a A. If A has an upper bound, it is also called bounded above. 2. The element l X is called a lower bound of A iff l a for all a A. If A has a lower bound, it is also called bounded below.
30 Definition. Let X be a set and let be an order relation on X. Let A X. 1. The element u X is called an upper bound of A iff u a for all a A. If A has an upper bound, it is also called bounded above. 2. The element l X is called a lower bound of A iff l a for all a A. If A has a lower bound, it is also called bounded below. A subset A X that is bounded above and bounded below is also called bounded.
31 Definition.
32 Definition. Let X be a set and let be an order relation on X.
33 Definition. Let X be a set and let be an order relation on X. Let A X.
34 Definition. Let X be a set and let be an order relation on X. Let A X. 1. The element s X is called lowest upper bound of A or supremum of A, denoted sup(a)
35 Definition. Let X be a set and let be an order relation on X. Let A X. 1. The element s X is called lowest upper bound of A or supremum of A, denoted sup(a), iff s is an upper bound of A and for all upper bounds u of A we have that s u.
36 Definition. Let X be a set and let be an order relation on X. Let A X. 1. The element s X is called lowest upper bound of A or supremum of A, denoted sup(a), iff s is an upper bound of A and for all upper bounds u of A we have that s u. 2. The element i X is called greatest lower bound of A or infimum of A, denoted inf(a)
37 Definition. Let X be a set and let be an order relation on X. Let A X. 1. The element s X is called lowest upper bound of A or supremum of A, denoted sup(a), iff s is an upper bound of A and for all upper bounds u of A we have that s u. 2. The element i X is called greatest lower bound of A or infimum of A, denoted inf(a), iff i is a lower bound of A and for all lower bounds l of A we have that l i.
38 Definition. Let X be a set and let be an order relation on X. Let A X. 1. The element s X is called lowest upper bound of A or supremum of A, denoted sup(a), iff s is an upper bound of A and for all upper bounds u of A we have that s u. 2. The element i X is called greatest lower bound of A or infimum of A, denoted inf(a), iff i is a lower bound of A and for all lower bounds l of A we have that l i. A supremum of A that is an element of A is also called maximum of A, denoted max(a).
39 Definition. Let X be a set and let be an order relation on X. Let A X. 1. The element s X is called lowest upper bound of A or supremum of A, denoted sup(a), iff s is an upper bound of A and for all upper bounds u of A we have that s u. 2. The element i X is called greatest lower bound of A or infimum of A, denoted inf(a), iff i is a lower bound of A and for all lower bounds l of A we have that l i. A supremum of A that is an element of A is also called maximum of A, denoted max(a). An infimum of A that is an element of A is also called minimum of A, denoted min(a).
40 Theorem.
41 Theorem. Suprema are unique. Let X be a set and let be an order relation on X.
42 Theorem. Suprema are unique. Let X be a set and let be an order relation on X. If the set A X is bounded above and s,t X both are suprema of A, then s = t.
43 Theorem. Suprema are unique. Let X be a set and let be an order relation on X. If the set A X is bounded above and s,t X both are suprema of A, then s = t. Proof.
44 Theorem. Suprema are unique. Let X be a set and let be an order relation on X. If the set A X is bounded above and s,t X both are suprema of A, then s = t. Proof. Let A X and let s, t X be as indicated.
45 Theorem. Suprema are unique. Let X be a set and let be an order relation on X. If the set A X is bounded above and s,t X both are suprema of A, then s = t. Proof. Let A X and let s,t X be as indicated. Then s is an upper bound of A
46 Theorem. Suprema are unique. Let X be a set and let be an order relation on X. If the set A X is bounded above and s,t X both are suprema of A, then s = t. Proof. Let A X and let s,t X be as indicated. Then s is an upper bound of A and, because t is a supremum of A
47 Theorem. Suprema are unique. Let X be a set and let be an order relation on X. If the set A X is bounded above and s,t X both are suprema of A, then s = t. Proof. Let A X and let s,t X be as indicated. Then s is an upper bound of A and, because t is a supremum of A, we infer s t.
48 Theorem. Suprema are unique. Let X be a set and let be an order relation on X. If the set A X is bounded above and s,t X both are suprema of A, then s = t. Proof. Let A X and let s,t X be as indicated. Then s is an upper bound of A and, because t is a supremum of A, we infer s t. Similarly, t is an upper bound of A
49 Theorem. Suprema are unique. Let X be a set and let be an order relation on X. If the set A X is bounded above and s,t X both are suprema of A, then s = t. Proof. Let A X and let s,t X be as indicated. Then s is an upper bound of A and, because t is a supremum of A, we infer s t. Similarly, t is an upper bound of A and, because s is a supremum of A
50 Theorem. Suprema are unique. Let X be a set and let be an order relation on X. If the set A X is bounded above and s,t X both are suprema of A, then s = t. Proof. Let A X and let s,t X be as indicated. Then s is an upper bound of A and, because t is a supremum of A, we infer s t. Similarly, t is an upper bound of A and, because s is a supremum of A, we infer t s.
51 Theorem. Suprema are unique. Let X be a set and let be an order relation on X. If the set A X is bounded above and s,t X both are suprema of A, then s = t. Proof. Let A X and let s,t X be as indicated. Then s is an upper bound of A and, because t is a supremum of A, we infer s t. Similarly, t is an upper bound of A and, because s is a supremum of A, we infer t s. Hence s = t.
52 Theorem. Suprema are unique. Let X be a set and let be an order relation on X. If the set A X is bounded above and s,t X both are suprema of A, then s = t. Proof. Let A X and let s,t X be as indicated. Then s is an upper bound of A and, because t is a supremum of A, we infer s t. Similarly, t is an upper bound of A and, because s is a supremum of A, we infer t s. Hence s = t.
53 Example.
54 Example. The set S := supremum in Q. { x Q : x 2 2 } does not have a
55 Example. The set S := supremum in Q. { x Q : x 2 2 } does not have a Proof.
56 Example. The set S := supremum in Q. { x Q : x 2 2 } does not have a Proof. Suppose for a contradiction that s Q is the supremum of S.
57 Example. The set S := supremum in Q. { x Q : x 2 2 } does not have a Proof. Suppose for a contradiction that s Q is the supremum of S. Then we must have s 2 2:
58 Example. The set S := supremum in Q. { x Q : x 2 2 } does not have a Proof. Suppose for a contradiction that s Q is the supremum of S. Then we must have s 2 2: Indeed, otherwise s 2 > 2.
59 Example. The set S := supremum in Q. { x Q : x 2 2 } does not have a Proof. Suppose for a contradiction that s Q is the supremum of S. Then we must have s 2 2: Indeed, otherwise s 2 > 2. In particular, s 0
60 Example. The set S := supremum in Q. { x Q : x 2 2 } does not have a Proof. Suppose for a contradiction that s Q is the supremum of S. Then we must have s 2 2: Indeed, otherwise s 2 > 2. In particular, s 0 and because ( x) 2 = x 2 for all rational numbers x
61 Example. The set S := supremum in Q. { x Q : x 2 2 } does not have a Proof. Suppose for a contradiction that s Q is the supremum of S. Then we must have s 2 2: Indeed, otherwise s 2 > 2. In particular, s 0 and because ( x) 2 = x 2 for all rational numbers x, s > 0.
62 Example. The set S := supremum in Q. { x Q : x 2 2 } does not have a Proof. Suppose for a contradiction that s Q is the supremum of S. Then we must have s 2 2: Indeed, otherwise s 2 > 2. In particular, s 0 and because ( x) 2 = x 2 for all rational numbers x, s > 0. Moreover, there would be an ε > 0 so that s 2 > 2 + ε.
63 Example. The set S := supremum in Q. { x Q : x 2 2 } does not have a Proof. Suppose for a contradiction that s Q is the supremum of S. Then we must have s 2 2: Indeed, otherwise s 2 > 2. In particular, s 0 and because ( x) 2 = x 2 for all rational numbers x, s > 0. Moreover, there would be an ε > 0 so that s 2 > 2 + ε. Let δ := ε 2s.
64 Example. The set S := supremum in Q. { x Q : x 2 2 } does not have a Proof. Suppose for a contradiction that s Q is the supremum of S. Then we must have s 2 2: Indeed, otherwise s 2 > 2. In particular, s 0 and because ( x) 2 = x 2 for all rational numbers x, s > 0. Moreover, there would be an ε > 0 so that s 2 > 2 + ε. Let δ := ε. Then for all ν > 0 with ν < δ we have 2s 2sν ν 2 < 2s ε 2s = ε.
65 Example. The set S := supremum in Q. { x Q : x 2 2 } does not have a Proof. Suppose for a contradiction that s Q is the supremum of S. Then we must have s 2 2: Indeed, otherwise s 2 > 2. In particular, s 0 and because ( x) 2 = x 2 for all rational numbers x, s > 0. Moreover, there would be an ε > 0 so that s 2 > 2 + ε. Let δ := ε. Then for all ν > 0 with ν < δ we have 2s 2sν ν 2 < 2s ε = ε. Consequently, for all ν with 0 < ν < δ 2s we would have
66 Example. The set S := supremum in Q. { x Q : x 2 2 } does not have a Proof. Suppose for a contradiction that s Q is the supremum of S. Then we must have s 2 2: Indeed, otherwise s 2 > 2. In particular, s 0 and because ( x) 2 = x 2 for all rational numbers x, s > 0. Moreover, there would be an ε > 0 so that s 2 > 2 + ε. Let δ := ε. Then for all ν > 0 with ν < δ we have 2s 2sν ν 2 < 2s ε = ε. Consequently, for all ν with 0 < ν < δ 2s we would have (s ν) 2
67 Example. The set S := supremum in Q. { x Q : x 2 2 } does not have a Proof. Suppose for a contradiction that s Q is the supremum of S. Then we must have s 2 2: Indeed, otherwise s 2 > 2. In particular, s 0 and because ( x) 2 = x 2 for all rational numbers x, s > 0. Moreover, there would be an ε > 0 so that s 2 > 2 + ε. Let δ := ε. Then for all ν > 0 with ν < δ we have 2s 2sν ν 2 < 2s ε = ε. Consequently, for all ν with 0 < ν < δ 2s we would have (s ν) 2 = s 2 2sν + ν 2
68 Example. The set S := supremum in Q. { x Q : x 2 2 } does not have a Proof. Suppose for a contradiction that s Q is the supremum of S. Then we must have s 2 2: Indeed, otherwise s 2 > 2. In particular, s 0 and because ( x) 2 = x 2 for all rational numbers x, s > 0. Moreover, there would be an ε > 0 so that s 2 > 2 + ε. Let δ := ε. Then for all ν > 0 with ν < δ we have 2s 2sν ν 2 < 2s ε = ε. Consequently, for all ν with 0 < ν < δ 2s we would have (s ν) 2 = s 2 2sν + ν 2 > 2 + ε (2sν ν 2)
69 Example. The set S := supremum in Q. { x Q : x 2 2 } does not have a Proof. Suppose for a contradiction that s Q is the supremum of S. Then we must have s 2 2: Indeed, otherwise s 2 > 2. In particular, s 0 and because ( x) 2 = x 2 for all rational numbers x, s > 0. Moreover, there would be an ε > 0 so that s 2 > 2 + ε. Let δ := ε. Then for all ν > 0 with ν < δ we have 2s 2sν ν 2 < 2s ε = ε. Consequently, for all ν with 0 < ν < δ 2s we would have (s ν) 2 = s 2 2sν + ν 2 > 2 + ε (2sν ν 2) > 2 + ε ε
70 Example. The set S := supremum in Q. { x Q : x 2 2 } does not have a Proof. Suppose for a contradiction that s Q is the supremum of S. Then we must have s 2 2: Indeed, otherwise s 2 > 2. In particular, s 0 and because ( x) 2 = x 2 for all rational numbers x, s > 0. Moreover, there would be an ε > 0 so that s 2 > 2 + ε. Let δ := ε. Then for all ν > 0 with ν < δ we have 2s 2sν ν 2 < 2s ε = ε. Consequently, for all ν with 0 < ν < δ 2s we would have (s ν) 2 = s 2 2sν + ν 2 > 2 + ε (2sν ν 2) > 2 + ε ε = 2.
71 Proof (concl.).
72 Proof (concl.). Hence no rational number between s δ and s would be in S.
73 Proof (concl.). Hence no rational number between s δ and s would be in S. Thus s δ is an upper bound of S
74 Proof (concl.). Hence no rational number between s δ and s would be in S. Thus s δ is an upper bound of S, contradiction.
75 Proof (concl.). Hence no rational number between s δ and s would be in S. Thus s δ is an upper bound of S, contradiction. The proof that s 2 cannot be strictly smaller than 2 is similar.
76 Proof (concl.). Hence no rational number between s δ and s would be in S. Thus s δ is an upper bound of S, contradiction. The proof that s 2 cannot be strictly smaller than 2 is similar. But this means that s 2 = 2 and s Q, contradiction.
77 Proof (concl.). Hence no rational number between s δ and s would be in S. Thus s δ is an upper bound of S, contradiction. The proof that s 2 cannot be strictly smaller than 2 is similar. But this means that s 2 = 2 and s Q, contradiction.
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