No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
A Problem With The Rational Numbers
Bernd Schröder
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Solvability of Equations
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Solvability of Equations
1. In fields, linear equations ax + b = 0 are solvable.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Solvability of Equations
1. In fields, linear equations ax + b = 0 are solvable.
2. Quadratic equations ax2 + bx + c = 0 are a natural next
target.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Solvability of Equations
1. In fields, linear equations ax + b = 0 are solvable.
2. Quadratic equations ax2 + bx + c = 0 are a natural next
target.
3. We run into problems rather quickly.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Proposition.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Proposition. There is no rational number r such that r2 = 2.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Proposition. There is no rational number r such that r2 = 2.
Proof.
Bernd Schröder
A Problem With The Rational Numbers
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Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Proposition. There is no rational number r such that r2 = 2.
Proof. First let n ∈ N be so that n2 = 2z for some z ∈ N.
Bernd Schröder
A Problem With The Rational Numbers
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Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Proposition. There is no rational number r such that r2 = 2.
Proof. First let n ∈ N be so that n2 = 2z for some z ∈ N. Then 2
is a factor in the prime factorization of n2 .
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Proposition. There is no rational number r such that r2 = 2.
Proof. First let n ∈ N be so that n2 = 2z for some z ∈ N. Then 2
is a factor in the prime factorization of n2 . But every factor in
the prime factorization of n2 occurs with an even exponent.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Proposition. There is no rational number r such that r2 = 2.
Proof. First let n ∈ N be so that n2 = 2z for some z ∈ N. Then 2
is a factor in the prime factorization of n2 . But every factor in
the prime factorization of n2 occurs with an even exponent.
Thus n2 = 22 k2 for some k.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Proposition. There is no rational number r such that r2 = 2.
Proof. First let n ∈ N be so that n2 = 2z for some z ∈ N. Then 2
is a factor in the prime factorization of n2 . But every factor in
the prime factorization of n2 occurs with an even exponent.
Thus n2 = 22 k2 for some k. Hence n = 2k.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Proposition. There is no rational number r such that r2 = 2.
Proof. First let n ∈ N be so that n2 = 2z for some z ∈ N. Then 2
is a factor in the prime factorization of n2 . But every factor in
the prime factorization of n2 occurs with an even exponent.
Thus n2 = 22 k2 for some k. Hence n = 2k. That is, if n2 is
divisible by 2, then so is n.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Proposition. There is no rational number r such that r2 = 2.
Proof. First let n ∈ N be so that n2 = 2z for some z ∈ N. Then 2
is a factor in the prime factorization of n2 . But every factor in
the prime factorization of n2 occurs with an even exponent.
Thus n2 = 22 k2 for some k. Hence n = 2k. That is, if n2 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
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Proposition. There is no rational number r such that r2 = 2.
Proof. First let n ∈ N be so that n2 = 2z for some z ∈ N. Then 2
is a factor in the prime factorization of n2 . But every factor in
the prime factorization of n2 occurs with an even exponent.
Thus n2 = 22 k2 for some k. Hence n = 2k. That is, if n2 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.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Proposition. There is no rational number r such that r2 = 2.
Proof. First let n ∈ N be so that n2 = 2z for some z ∈ N. Then 2
is a factor in the prime factorization of n2 . But every factor in
the prime factorization of n2 occurs with an even exponent.
Thus n2 = 22 k2 for some k. Hence n = 2k. That is, if n2 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 n2 = 2d2 implies n = n2 · 2.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Proposition. There is no rational number r such that r2 = 2.
Proof. First let n ∈ N be so that n2 = 2z for some z ∈ N. Then 2
is a factor in the prime factorization of n2 . But every factor in
the prime factorization of n2 occurs with an even exponent.
Thus n2 = 22 k2 for some k. Hence n = 2k. That is, if n2 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 n2 = 2d2 implies n = n2 · 2.
Consequently, 2d2 = (n2 · 2)2
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Proposition. There is no rational number r such that r2 = 2.
Proof. First let n ∈ N be so that n2 = 2z for some z ∈ N. Then 2
is a factor in the prime factorization of n2 . But every factor in
the prime factorization of n2 occurs with an even exponent.
Thus n2 = 22 k2 for some k. Hence n = 2k. That is, if n2 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 n2 = 2d2 implies n = n2 · 2.
Consequently, 2d2 = (n2 · 2)2 , that is, d2 = n22 · 2
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Proposition. There is no rational number r such that r2 = 2.
Proof. First let n ∈ N be so that n2 = 2z for some z ∈ N. Then 2
is a factor in the prime factorization of n2 . But every factor in
the prime factorization of n2 occurs with an even exponent.
Thus n2 = 22 k2 for some k. Hence n = 2k. That is, if n2 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 n2 = 2d2 implies n = n2 · 2.
Consequently, 2d2 = (n2 · 2)2 , that is, d2 = n22 · 2, which implies
d = d2 · 2.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Proposition. There is no rational number r such that r2 = 2.
Proof. First let n ∈ N be so that n2 = 2z for some z ∈ N. Then 2
is a factor in the prime factorization of n2 . But every factor in
the prime factorization of n2 occurs with an even exponent.
Thus n2 = 22 k2 for some k. Hence n = 2k. That is, if n2 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 n2 = 2d2 implies n = n2 · 2.
Consequently, 2d2 = (n2 · 2)2 , that is, d2 = n22 · 2, which implies
d = d2 · 2. But then 2|n and 2|d, contradiction.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Proposition. There is no rational number r such that r2 = 2.
Proof. First let n ∈ N be so that n2 = 2z for some z ∈ N. Then 2
is a factor in the prime factorization of n2 . But every factor in
the prime factorization of n2 occurs with an even exponent.
Thus n2 = 22 k2 for some k. Hence n = 2k. That is, if n2 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 n2 = 2d2 implies n = n2 · 2.
Consequently, 2d2 = (n2 · 2)2 , that is, d2 = n22 · 2, which implies
d = d2 · 2. But then 2|n and 2|d, contradiction.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Definition.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Definition. Let X be a set and let ≤ be an order relation on X.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Definition. Let X be a set and let ≤ be an order relation on X.
Let A ⊆ X.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
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.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
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.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
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.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
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.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
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.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Definition.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Definition. Let X be a set and let ≤ be an order relation on X.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Definition. Let X be a set and let ≤ be an order relation on X.
Let A ⊆ X.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
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)
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
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.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
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)
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
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.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
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).
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
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).
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Theorem.
Bernd Schröder
A Problem With The Rational Numbers
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Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Theorem. Suprema are unique. Let X be a set and let ≤ be an
order relation on X.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
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.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
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.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
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.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
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
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
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
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
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.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
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
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
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
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
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.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
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.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
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.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Example.
Bernd Schröder
A Problem With The Rational Numbers
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Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
n
o
Example. The set S := x ∈ Q : x2 ≤ 2 does not have a
supremum in Q.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
n
o
Example. The set S := x ∈ Q : x2 ≤ 2 does not have a
supremum in Q.
Proof.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
n
o
Example. The set S := x ∈ Q : x2 ≤ 2 does not have a
supremum in Q.
Proof. Suppose for a contradiction that s ∈ Q is the supremum
of S.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
n
o
Example. The set S := x ∈ Q : x2 ≤ 2 does not have a
supremum in Q.
Proof. Suppose for a contradiction that s ∈ Q is the supremum
of S.
Then we must have s2 ≤ 2:
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
n
o
Example. The set S := x ∈ Q : x2 ≤ 2 does not have a
supremum in Q.
Proof. Suppose for a contradiction that s ∈ Q is the supremum
of S.
Then we must have s2 ≤ 2: Indeed, otherwise s2 > 2.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
n
o
Example. The set S := x ∈ Q : x2 ≤ 2 does not have a
supremum in Q.
Proof. Suppose for a contradiction that s ∈ Q is the supremum
of S.
Then we must have s2 ≤ 2: Indeed, otherwise s2 > 2. In
particular, s 6= 0
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
n
o
Example. The set S := x ∈ Q : x2 ≤ 2 does not have a
supremum in Q.
Proof. Suppose for a contradiction that s ∈ Q is the supremum
of S.
Then we must have s2 ≤ 2: Indeed, otherwise s2 > 2. In
particular, s 6= 0 and because (−x)2 = x2 for all rational
numbers x
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
n
o
Example. The set S := x ∈ Q : x2 ≤ 2 does not have a
supremum in Q.
Proof. Suppose for a contradiction that s ∈ Q is the supremum
of S.
Then we must have s2 ≤ 2: Indeed, otherwise s2 > 2. In
particular, s 6= 0 and because (−x)2 = x2 for all rational
numbers x, s > 0.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
n
o
Example. The set S := x ∈ Q : x2 ≤ 2 does not have a
supremum in Q.
Proof. Suppose for a contradiction that s ∈ Q is the supremum
of S.
Then we must have s2 ≤ 2: Indeed, otherwise s2 > 2. In
particular, s 6= 0 and because (−x)2 = x2 for all rational
numbers x, s > 0. Moreover, there would be an ε > 0 so that
s2 > 2 + ε.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
n
o
Example. The set S := x ∈ Q : x2 ≤ 2 does not have a
supremum in Q.
Proof. Suppose for a contradiction that s ∈ Q is the supremum
of S.
Then we must have s2 ≤ 2: Indeed, otherwise s2 > 2. In
particular, s 6= 0 and because (−x)2 = x2 for all rational
numbers x, s > 0. Moreover, there would be an ε > 0 so that
ε
s2 > 2 + ε. Let δ := .
2s
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
n
o
Example. The set S := x ∈ Q : x2 ≤ 2 does not have a
supremum in Q.
Proof. Suppose for a contradiction that s ∈ Q is the supremum
of S.
Then we must have s2 ≤ 2: Indeed, otherwise s2 > 2. In
particular, s 6= 0 and because (−x)2 = x2 for all rational
numbers x, s > 0. Moreover, there would be an ε > 0 so that
ε
s2 > 2 + ε. Let δ := . Then for all ν > 0 with ν < δ we have
2s
ε
2sν − ν 2 < 2s = ε.
2s
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
n
o
Example. The set S := x ∈ Q : x2 ≤ 2 does not have a
supremum in Q.
Proof. Suppose for a contradiction that s ∈ Q is the supremum
of S.
Then we must have s2 ≤ 2: Indeed, otherwise s2 > 2. In
particular, s 6= 0 and because (−x)2 = x2 for all rational
numbers x, s > 0. Moreover, there would be an ε > 0 so that
ε
s2 > 2 + ε. Let δ := . Then for all ν > 0 with ν < δ we have
2s
ε
2sν − ν 2 < 2s = ε. Consequently, for all ν with 0 < ν < δ
2s
we would have
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
n
o
Example. The set S := x ∈ Q : x2 ≤ 2 does not have a
supremum in Q.
Proof. Suppose for a contradiction that s ∈ Q is the supremum
of S.
Then we must have s2 ≤ 2: Indeed, otherwise s2 > 2. In
particular, s 6= 0 and because (−x)2 = x2 for all rational
numbers x, s > 0. Moreover, there would be an ε > 0 so that
ε
s2 > 2 + ε. Let δ := . Then for all ν > 0 with ν < δ we have
2s
ε
2sν − ν 2 < 2s = ε. Consequently, for all ν with 0 < ν < δ
2s
we would have
(s − ν)2
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
n
o
Example. The set S := x ∈ Q : x2 ≤ 2 does not have a
supremum in Q.
Proof. Suppose for a contradiction that s ∈ Q is the supremum
of S.
Then we must have s2 ≤ 2: Indeed, otherwise s2 > 2. In
particular, s 6= 0 and because (−x)2 = x2 for all rational
numbers x, s > 0. Moreover, there would be an ε > 0 so that
ε
s2 > 2 + ε. Let δ := . Then for all ν > 0 with ν < δ we have
2s
ε
2sν − ν 2 < 2s = ε. Consequently, for all ν with 0 < ν < δ
2s
we would have
(s − ν)2 = s2 − 2sν + ν 2
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
n
o
Example. The set S := x ∈ Q : x2 ≤ 2 does not have a
supremum in Q.
Proof. Suppose for a contradiction that s ∈ Q is the supremum
of S.
Then we must have s2 ≤ 2: Indeed, otherwise s2 > 2. In
particular, s 6= 0 and because (−x)2 = x2 for all rational
numbers x, s > 0. Moreover, there would be an ε > 0 so that
ε
s2 > 2 + ε. Let δ := . Then for all ν > 0 with ν < δ we have
2s
ε
2sν − ν 2 < 2s = ε. Consequently, for all ν with 0 < ν < δ
2s
we would have
2
2
2
2
(s − ν) = s − 2sν + ν > 2 + ε − 2sν − ν
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
n
o
Example. The set S := x ∈ Q : x2 ≤ 2 does not have a
supremum in Q.
Proof. Suppose for a contradiction that s ∈ Q is the supremum
of S.
Then we must have s2 ≤ 2: Indeed, otherwise s2 > 2. In
particular, s 6= 0 and because (−x)2 = x2 for all rational
numbers x, s > 0. Moreover, there would be an ε > 0 so that
ε
s2 > 2 + ε. Let δ := . Then for all ν > 0 with ν < δ we have
2s
ε
2sν − ν 2 < 2s = ε. Consequently, for all ν with 0 < ν < δ
2s
we would have
2
2
2
2
(s − ν) = s − 2sν + ν > 2 + ε − 2sν − ν
> 2+ε −ε
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
n
o
Example. The set S := x ∈ Q : x2 ≤ 2 does not have a
supremum in Q.
Proof. Suppose for a contradiction that s ∈ Q is the supremum
of S.
Then we must have s2 ≤ 2: Indeed, otherwise s2 > 2. In
particular, s 6= 0 and because (−x)2 = x2 for all rational
numbers x, s > 0. Moreover, there would be an ε > 0 so that
ε
s2 > 2 + ε. Let δ := . Then for all ν > 0 with ν < δ we have
2s
ε
2sν − ν 2 < 2s = ε. Consequently, for all ν with 0 < ν < δ
2s
we would have
2
2
2
2
(s − ν) = s − 2sν + ν > 2 + ε − 2sν − ν
> 2 + ε − ε = 2.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Proof (concl.).
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Proof (concl.). Hence no rational number between s − δ and s
would be in S.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
Proof (concl.). Hence no rational number between s − δ and s
would be in S. Thus s − δ is an upper bound of S
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
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.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
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 s2 cannot be strictly smaller than 2 is similar.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
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 s2 cannot be strictly smaller than 2 is similar.
But this means that s2 = 2 and s ∈ Q, contradiction.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science
No Rational Square Root for 2
Upper and Lower Bounds
A Missing Supremum in Q
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 s2 cannot be strictly smaller than 2 is similar.
But this means that s2 = 2 and s ∈ Q, contradiction.
Bernd Schröder
A Problem With The Rational Numbers
logo1
Louisiana Tech University, College of Engineering and Science