Question

Proofs

For this assignment, know that:

An integer is any countable number. Examples are: -3, 0, 5, 1337, etc.

A rational number is any number that can be written in the form a/b, a and b are integers in lowest terms, and b cannot equal 0. Examples are 27, 22/7, -3921/2, etc.

A real number is any number that is not imaginary or infinity. Examples are 0, 4/3, square root of 2, pi, etc.

1. Prove that the product of two odd integers is an odd integer.

2. Prove that the quotient of a rational number and a non-zero rational number is a rational number.

3. Prove by contrapositive that for every integer n, if 9n + 11 is even, then n is odd.

4. Prove by contrapositive that for every non-zero real number n, if is irrational, then 1/n is also irrational.

5. Prove by contradiction that the average of three real numbers is greater than or equal to at least one of the numbers.

6. Prove by cases that for all real numbers | x + y | <= |x| + |y|

7. Prove or disprove: The sum of any five consecutive integers is divisible by 5.

8. Prove or disprove: There exists no integers a and b such that 8a - 12b = 1

Answer 1

1. Prove that the product of two odd integers is an odd integer.
Solution:---
Let a and b be two odd numbers....
If a and b are odd, then a and b can be written as:
a=2m+1 ,
b=2n+1
Where m and n are whole numbers.
So....
ab= (2m+1)(2n+1)
= 4mn+2m+2n+1
= 2(2mn+m+n)+1

Which is odd.
Bcz it leaves the remainder 1 after being divided by 2.

2. Prove that the quotient of a rational number and a non-zero rational number is a rational number.
Solution:----
the quotient of two rational numbers is always rational, and the reason for this lies in the fact that the product of two integers is always an integer. To divide two rational numbers,
a/b ÷ c/d, we use the following rule:
a/b ÷ c/d = a/b × d/c
= ad / bc
By this rule, the quotient of any two rational numbers can always be written as a fraction with a product of integers in the numerator and a product of integers in the denominator. Since the product of two integers is always an integer, we have that the quotient of two rational numbers can be written as a fraction with an integer in the numerator and an integer in the denominator. This is the exact definition of a rational number, so we have that the quotient of two rational numbers is always rational.

3. Prove by contrapositive that for every integer n, if 9n + 11 is even, then n is odd.
Solution:-----
Proof by contraposition:-----------
The contrapositive is “If n is even, then 9n + 11 is odd.”
Assume that n is even. We can now
write n = 2k for some integer k. Then
9n + 11 = (9 x 2k) + 11 = 18k + 11 =2 (9k+ 5) + 1.
Which is odd.
Bcz it leaves the remainder 1 after being divided by 2.

4. Prove by contrapositive that for every non-zero real number n, if is irrational, then 1/n is also irrational.
Solution:---
The contrapositive is “If 1/x is rational, then x is rational.”
Suppose that 1/x is rational and x != 0. Then there exists integers p and q such that 1/x = p/q and
q != 0.
1/x != 0 because 1 != x · 0, this would mean that p != 0.
Since p != 0, then x = 1/ (1/x) = 1/(p/q) = q/p.
Hence x can be written as a quotient of two integers with a nonzero denominator. Thus, x is
rational.

5. Prove by contradiction that the average of three real numbers is greater than or equal to at least one of the numbers. Proof. Proof by contradiction. We assume that there are three real numbers x1, x2, and x3 such that the average of the three numbers is less than each of the three numbers.
That is (x1+ x2+ x3) /3 < x1, (x1+ x2+ x3)/3 < x2, and (x1+ x2+ x3)/3 < x3.
Adding the three inequalities gives:
x1+x2+x33+x1+x2+x33+x1+x2+x33 < x1+x2+x3.x1+x2+x33+x1+x2+x33+x1+x2+x33 < x1+x2+x3.
The inequality contradicts the algebraic fact that
x1+x2+x33+x1+x2+x33+x1+x2+x33 = x1+x2+x3.x1+x2+x33+x1+x2+x33+x1+x2+x33 = x1+x2+x3

7. Prove or disprove: The sum of any five consecutive integers is divisible by 5.
Solution:---
Suppose n is the sum of five consecutive integers. Then there is an integer m so that
n is the sum of the integers m, m + 1, m + 2, m + 3 and m + 4, that is, we have
n = m + (m + 1) + (m + 2) + (m + 3) + (m + 4)
Removing the parentheses and rearranging the terms, this means,
n = m + m + m + m + m + 1 + 2 + 3 + 4 = 5m + 10
In particular, this gives n = 5m + 10 = 5(m + 2) and, since m + 2 is an integer, this shows n
is divisible by 5.
Conversely, suppose n is divisible by 5, say n = 5a. Let m = a − 2 so a = m + 2. Then
n = 5(m+2) = 5m+10 = m+m+m+m+m+1+2+3+4 = m+(m+1)+m+2)+(m+3)+(m+4)
which shows that n is sum of five consecutive integers.