Solvability of Equations
Solvability of Equations 1. In fields, linear equations ax + b = 0 are solvable.
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.
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.
Proposition.
Proposition. There is no rational number r such that r 2 = 2.
Proposition. There is no rational number r such that r 2 = 2. Proof.
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.
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.
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.
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.
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.
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.
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
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.
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.
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
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
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.
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.
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.
Definition.
Definition. Let X be a set and let be an order relation on X.
Definition. Let X be a set and let be an order relation on X. Let A X.
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.
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.
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.
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.
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.
Definition.
Definition. Let X be a set and let be an order relation on X.
Definition. Let X be a set and let be an order relation on X. Let A X.
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)
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.
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)
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.
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).
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).
Theorem.
Theorem. Suprema are unique. Let X be a set and let be an order relation on X.
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.
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.
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.
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
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
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.
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
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
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.
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.
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.
Example.
Example. The set S := supremum in Q. { x Q : x 2 2 } does not have a
Example. The set S := supremum in Q. { x Q : x 2 2 } does not have a Proof.
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.
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:
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.
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
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
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.
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.
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.
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.
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
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
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
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)
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
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.
Proof (concl.).
Proof (concl.). Hence no rational number between s δ and s would be in S.
Proof (concl.). Hence no rational number between s δ and s would be in S. Thus s δ is an upper bound of S
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.
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.
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.
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.