Positive numbers
R - The set of real numbers. Think of it as all decimals that terminate to the left of the decimal point.
Real numbers
0
0.5
2.7
0.16666…
−45=−1.25
2=1.414…
Not real numbers
i=−1
…2222.1
∞
N={1,2,3,4,…} - The set of natural numbers.
Z={…,−3,−2,−1,0,1,2,3…} - The set of integers.
Q={ba:a∈Z, b∈N} - The set of rational numbers.
R∖Q - The set of irrational numbers.
Definition 1. The set P, called the positive real numbers, is the nonempty subset of R with the following three properties:
a∈P,a=0,−a∈P.
Property 1. says that P is closed under addition
Property 2. says that P is closed under multiplication.
Property 3. is called the Trichotomy Property.
Is 1∈P? What about 2? Are squares in P?
Proposition 1. 1∈P.
Proof. Assume toward a contradiction that 1∈/P.
Proof. Assume toward a contradiction that 1∈/P. By the Trichotomoy Property, either −1∈P, or 1=0. Obiously 1=0, hence, it must be the case that −1∈P.
Proof. Assume toward a contradiction that 1∈/P. By the Trichotomoy Property, either −1∈P, or 1=0. Obiously 1=0, hence, it must be the case that −1∈P. Since P is closed under multiplication, we have
1=(−1)(−1)∈P.
This is a contradiction by the Trichotomy Property. We conclude that 1∈P. □
Proof. Assume toward a contradiction that 1∈/P. By the Trichotomoy Property, either −1∈P, or 1=0. Obiously 1=0, hence, it must be the case that −1∈P. Since P is closed under multiplication, we have
1=(−1)(−1)∈P.
Proof. From Proposition 1 we see that 1∈P. Since P is closed under addtion, we have
2=1+1∈P. □
Proposition 2. 2∈P.
Is 3∈P?
What about 4?
How many examples would you need to check before you would KNOW that all natural numbers are positive?
Proposition 3. N⊂P
Equivalent statements to N⊂P:
This is the perfect type of statement to prove using mathematical induction.
Proof. We will use induction to prove n∈P for all n∈N. We have already proved 1∈P. This finishes the base case.
Now, assume the claim is true for some k∈N, that is k∈P. Since P is closed under addition, and 1∈P, we conclude k+1∈P. This completes the inductive step, and shows that n∈P for all n∈N. □
Proof. We will use induction to prove n∈P for all n∈N.
Proof. We will use induction to prove n∈P for all n∈N. We have already proved 1∈P. This finishes the base case.
Proof. We will use induction to prove n∈P for all n∈N. We have already proved 1∈P. This finishes the base case.
Now, assume the claim is true for some k∈N, that is k∈P.
Proof. We will use induction to prove n∈P for all n∈N. We have already proved 1∈P. This finishes the base case.
Now, assume the claim is true for some k∈N, that is k∈P. Since P is closed under addition, and 1∈P, we conclude k+1∈P.
Proof. We will use induction to prove n∈P for all n∈N. We have already proved 1∈P. This finishes the base case.
Now, assume the claim is true for some k∈N, that is k∈P. Since P is closed under addition, and 1∈P, we conclude k+1∈P. This completes the inductive step, and shows that n∈P for all n∈N. □
Proposition 3. N⊂P
Is 1∈P? What about 2? Are squares in P?
Proposition 4. If x∈R and x=0, then x2∈P
Proof. We will prove this in two cases. By the Trichotomy property, either x∈P, or −x∈P.
Case 1. Assume x∈P. Since P is closed under multiplication, we have x2=(x)(x)∈P.
Case 2. Assume −x∈P. Since P is closed under multiplication, we have x2=(−x)(−x)∈P.
In either case we have shown that x2∈P. □
Proof. We will prove this in two cases.
Proof. We will prove this in two cases. By the Trichotomy property, either x∈P, or −x∈P.
Proof. We will prove this in two cases. By the Trichotomy property, either x∈P, or −x∈P.
Case 1. Assume x∈P.
Proof. We will prove this in two cases. By the Trichotomy property, either x∈P, or −x∈P.
Case 1. Assume x∈P. Since P is closed under multiplication, we have x2=(x)(x)∈P.
Proof. We will prove this in two cases. By the Trichotomy property, either x∈P, or −x∈P.
Case 1. Assume x∈P. Since P is closed under multiplication, we have x2=(x)(x)∈P.
Case 2. Assume −x∈P. Since P is closed under multiplication, we have x2=(−x)(−x)∈P.
Proof. We will prove this in two cases. By the Trichotomy property, either x∈P, or −x∈P.
Case 1. Assume x∈P. Since P is closed under multiplication, we have x2=(x)(x)∈P.
Case 2. Assume −x∈P. Since P is closed under multiplication, we have x2=(−x)(−x)∈P.
In either case we have shown that x2∈P. □
Read the section "The Order Properties of R"