Chapter 4 - Elementary Number Theory and Methods of Proof

Prove that every integer is a rational number.

Proof: Suppose n is any [particular but arbitrarily chosen] integer. Then n = n·1, and so n = n/1 by dividing both sides by 1. Now n and 1 are both integers, and 1 ̸= 0. Hence n can be written as a quotient of integers with a nonzero denominator, and so n is rational.

To show that a real number is rational, we must show that we can write it as _____.

To show that a real number is rational, we must show that we can write it as a ratio of integers with a nonzero denominator.

Is 0 even?

Yes, 0=2·0.

If a and b are integers, is 10a+8b+1 odd?

Yes, 10a+8b+1=2(5a+4b)+1, and since a and b are integers, so is 5a+4b (being a sum of products of integers).

If a and b are integers, is 6a2b even?

Yes, 6a2b = 2(3a2b), and since a and b are integers, so is 3a2b (being a product of integers).

Is −301 odd?

Yes, −301 = 2(−151)+1.

If k and m are integers, is 10km divisible by 5?

Yes. By the associative law of algebra, 10km = 5 · (2km) and 2km is an integer because it is a product of three integers.

If a and b are integers, is 3a + 3b divisible by 3?

Yes. By the distributive law of algebra, 3a + 3b = 3(a + b) and a + b is an integer because it is a sum of two integers.

For each of the following values of n and d, find integers q and r such that n = dq + r and 0 ≤ r < d. a. n=54, d=4 b. n=−54 d=4 c. n=54, d=70

a. 54=4·13+2; hence q =13 and r =2. b. −54=4·(−14)+2; hence q =−14 and r =2. c. 54=70·0+54; hence q =0 and r =54.

Assume that k is a particular integer. a. Is −17 an odd integer? b. Is 0 an even integer? c. Is 2k−1 odd?

a. Yes because 2(-9)+1 b. Yes because 2(0) c. Yes because 2k-1 = 2(k-1)+1 and k -1 is an integer because it is a difference of integers.

Assume that m and n are particular integers. a. Is 6m+8n even? b. Is 10mn+7 odd? c. If m>n>0, is m^2−n^2 composite?

a. Yes because factored it is 2(3m+4m) or 2k b. Yes because 2(5mn+3)+1 or 2k+1 form. c. Not necessarily because 3^2-2^2=5

Zero is a rational number because _____.

0/1

What are the three theorems of rational numbers:

1. Every integer is a rational number. 2. The sum of any two rational numbers is rational. 3. The double of a rational number is rational.

Prove the statement is true or false: The difference of any two odd integers is odd.

False, counter example is 3-1 = 2, 2 is even.

Prove the statement is true or false: If a sum of two integers is even, then one of the summands is even. (In the expression a + b, a and b are called summands.)

False, counter example, 3+1 = 4 but 3 and 1 are odd.

What is the Zero Product Property?

If neither of two real numbers is zero, then their product is also not zero.

Suppose that r and s are integers. Prove the following: ∃ an integer k such that 22r + 18s = 2k.

Let k = 11r + 9s. Then k is an integer because it is a sum of products of integers; and by substitution, 2k = 2(11r + 9s), which equals 22r + 18s by the distributive law of algebra.

Prove the following: ∃ an even integer n that can be written in two ways as a sum of two prime numbers.

Let n=10. Then 10=5+5=3+7 and 3,5, and 7 are all prime numbers.

What is a composite number?

n is composite ⇔ ∃ positive integers r and s such that n = r s and 1 < r < n and 1 < s < n.

What is an even integer?

n is even ⇔ ∃ an integer k such that n=2k.

What is an odd integer?

n is odd ⇔ ∃ an integer k such that n=2k+1.

What is a prime number?

n is prime ⇔ ∀ positive integers r and s, if n=rs then either r = 1 and s = n or r = n and s = 1.

Suppose a is an integer. If a mod 7=4, what is 5a mod 7? In other words, if division of a by 7 gives a remainder of 4, what is the remainder when 5a is divided by 7?

5a=7q+6 or 5a mod 7 = 6

An irrational number is a _____ that is _____.

An irrational number is a real number that is not rational.

Assume that a and b are both integers and that a does not equal 0 and b does not equal 0. Explain why (b − a)/(ab2) must be a rational number.

Because a and b are integers, b − a and ab2 are both integers (since differences and products of integers are integers). Also, by the zero product property, ab^2 does not equal 0 because neither a nor b is zero. Hence (b − a)/ab2 is a quotient of two integers with nonzero denominator, and so it is rational.

Suppose d is a positive integer and n is any integer. If d | n, what is the remainder obtained when the quotient- remainder theorem is applied to n with divisor d?

Because d | n n=dq+0 for some integer q. Thus the remainder is 0.