2. Write the following properties in words:
(1) m-0= m
(2) n=1 = n
Answers
Explanation:
Theorem 0.3.4 (Distributive Property). If l, m and n are any natural numbers,
then
l · (m + n) = l · m + l · n
(l + m) · n = l · n + m · n
Proof. We will prove that multiplication is left distributive which is the first
of the two equations. The proof that it is right distributive will be left as an exercise.
We use induction on n. The base case can be shown as
l · (m + 0) = l · m, Additive identity
= l · m + 0, Additive identity
= l · m + l · 0. Definition 0.2.2
Now assume the theorem is true for n. Then
l · (m + (n++)) = l · ((m + n)++), Definition 0.2.2
= l · (m + n) + l, Definition 0.2.2
= (l · m + l · n) + l, Induction hypothesis
= l · m + (l · n + l), Associativity of addition
= l · m + l · (n++). Definition 0.2.2
This concludes the proof of the induction step and the theorem.
Now let us establish the algebraic properties of multiplication in N.
Theorem 0.3.5 (Algebraic properties of (N, ·)). following properties hold.
Multiplicative identity: For any n ∈ N, 1 · n = n = n · 1.
Associativity of multiplication: For any three natural numbers l, m, n,
(0.7) (l · m) · n = l · (m · n)
Commutativity of multiplication: For any two natural numbers m, n,
(0.8) m · n = n · m
Cancellation rule for multiplication: For any natural number n, if
n · m1 = n · m2 6= 0
then m1 = m2.
Proof. We examine these in order.
Multiplicative identity: Let us prove this by induction on n. The base
case states 1 · 0 = 0 = 0 · 1. The left equation follows from Definition
0.2.2. The right follows from this definition as well and the definition that
1 = 0 + + via the equation 0 · 1 = 0 ·(0++) = 0 · 0 + 0 = 0 + 0 = 0. In the
last equation we used the additive identity. Now assume 1 · n = n = n · 1.
Then 1·(n++) = 1·n+ 1 = n+ 1 = n++ by the induction hypothesis and
Proposition 0.2.5. For the other side we have (n++)·1 = (n++)·(0++) =
[(n++) · 0] + n++ = 0 + n++ = n++.
Below are given the properties in words.
Explanation:
Theorem 0.3.4 (Distributive Property). If l, m and n are any natural numbers,
then
l · (m + n) = l · m + l · n
(l + m) · n = l · n + m · n
Proof. We will prove that multiplication is left distributive which is the first
of the two equations. The proof that it is right distributive will be left as an exercise.
We use induction on n. The base case can be shown as
l · (m + 0) = l · m, Additive identity
= l · m + 0, Additive identity
= l · m + l · 0. Definition 0.2.2
Now assume the theorem is true for n. Then
l · (m + (n++)) = l · ((m + n)++), Definition 0.2.2
= l · (m + n) + l, Definition 0.2.2
= (l · m + l · n) + l, Induction hypothesis
= l · m + (l · n + l), Associativity of addition
= l · m + l · (n++). Definition 0.2.2
This concludes the proof of the induction step and the theorem.
Now let us establish the algebraic properties of multiplication in N.
Theorem 0.3.5 (Algebraic properties of (N, ·)). following properties hold.
Multiplicative identity: For any n ∈ N, 1 · n = n = n · 1.
Associativity of multiplication: For any three natural numbers l, m, n,
(0.7) (l · m) · n = l · (m · n)
Commutativity of multiplication: For any two natural numbers m, n,
(0.8) m · n = n · m
Cancellation rule for multiplication: For any natural number n, if
n · m1 = n · m2 6= 0
Then m1 = m2.
Proof. We examine these in order.
Multiplicative identity: Let us prove this by induction on n. The base
Case states 1 · 0 = 0 = 0 · 1. The left equation follows from Definition
0.2.2. The right follows from this definition as well and the definition that
1 = 0 + + via the equation 0 · 1 = 0 ·(0++) = 0 · 0 + 0 = 0 + 0 = 0. In the
Last equation we used the additive identity. Now assume 1 · n = n = n · 1.
Then 1·(n++) = 1·n+ 1 = n+ 1 = n++ by the induction hypothesis and
Proposition 0.2.5. For the other side we have (n++)·1 = (n++)·(0++) =
[(n++) · 0] + n++ = 0 + n++ = n++