Question

All theorems in the REAL NUMBERS

Answer 1

Theorem 1: If a and b are real numbers such that a+b=a, then b=0.

• Proof: Suppose that a+b=a. We will manipulate both sides of this equation to arrive at the conclusion that b=0.

(1)
a+b=a
(a+b)+(−a)=a+(−a)
(b+a)+(−a)=a+(−a)
b+(a+(−a))=a+(−a)
b+0=0
b=0

Theorem 2: If a and b are real numbers such that a⋅b=a, then b=1.

• Proof: Suppose that a⋅b=a. We will manipulate both sides of this equation to arrive at the conclusion that b=1.

(2)
a⋅b=a
a^(−1)⋅(a⋅b)=a^(−1)⋅a
(a^(−1)⋅a)⋅b=a^(−1)⋅a
1⋅b=1b=1

Theorem 3: If a is a real number then a⋅0=0.

• Proof: Let a∈R.

(3)
a⋅0=0
a+a⋅0=a+0
a⋅1+a⋅0=a+0
a⋅(1+0)=a+0
a⋅1=a+0
a=a

• Since we have shown that a+(a⋅0)=a+0 implies that a=a, then we must have that a⋅0=0. ■

Theorem 4: If a and b are real numbers where b≠0, then if a⋅b=1, then b=a−1.

• Proof: Let a,b∈R be such that b≠0 and suppose that a⋅b=1.

(4)
b=b1⋅b=b
(a⋅a^−1)⋅b=b
a−1⋅(a⋅b)=b
a−1⋅1=b
a−1=b

Theorem 5: If a and b are real numbers and a⋅b=0 then a=0 or b=0 or both a,b=0.

• Proof: If both a,b=0 then 0⋅0=0. Now without loss of generality, assume that a≠0.

(5)
a⋅b=0
a−1⋅(a⋅b)=a−1⋅0
a−1⋅(a⋅b)=0
(a−1⋅a)⋅b=0
1⋅b=0
b=0

• So if a≠0 then b=0. ■

Theorem 6: If a and b are real numbers such that a+b=0 then b=−a.

• Proof: Let a,b∈R be such that a+b=0.

(6)
a+b=0
(−a)+(a+b)=(−a)+0
((−a)+a)+b=−a
0+b=−a
b=−a

Theorem 7: If a is a real number, then −1⋅a=−a.

• Proof: Let a∈R. To show that −1⋅a=−a we will show that both values are additive inverses of a, and then by theorem 6 (which says that additive inverses are unique), we can conclude that −1⋅a=−a.

• First a+−1⋅a=1⋅a+−1⋅a=(1−1)⋅a=0⋅a=Theorem30. So a is the additive inverse of −1⋅a.

• Furthermore a+(−a)=0. Since additive inverses are unique we have that −1⋅a=−a. ■

Theorem 8: If a is a real number then −(−a)=a.

• Proof: Let a∈R. Notice that (−a)−(−a)=[(−1)⋅a]−[(−1)⋅a]=(−1)⋅(a−a)=−1⋅0=0. So the additive inverse to −a is −(−a). But the additive inverse to −a is also a which implies that −(−a)=a. ■

Theorem 9: (−1)⋅(−1)=1.

• Proof: Notice that (−1)⋅(−1)=−(−1)=Theorem81. ■

Theorem 10: If a and b are real numbers then −(a+b)=(−a)+(−b).

• Proof: Let a,b∈R. Then −(a+b)=(−1)⋅(a+b)=(−1)⋅a+(−1)⋅b=(−a)+(−b). ■

Theorem 11: If a and b are real numbers then (−a)⋅(−b)=a⋅b.

• Proof: Let a,b∈R. Then (−a)⋅(−b)=[(−1)⋅a]⋅[(−1)⋅b]=[(−1)⋅(−1)]⋅[a⋅b]=Theorem91⋅[a⋅b]=a⋅b. ■

Theorem 12: If a and b are real numbers such that b≠0 then −(a⋅b−1)=−a⋅b−1.

• Proof: Let a,b∈R be such that b≠0. Then −(a⋅b−1)=(−1)⋅(a⋅b−1)=[(−1)⋅a]⋅b−1=[−a]⋅b−1=−a⋅b−1

Hope this helps :)