Question

explain why the sum of a rational number and an irrational number is always an irrational number. give an example to make your case.

question 3

explain why the product of two rational numbers is closed under multiplication. give an example to make your case.

Explanation:

Define rational and irrational numbers

We use the Rational Numbers and Irrational Numbers concepts.
A rational number can be written as \(\frac{a}{b}\), where \(a, b\) are integers and \(b
eq 0\).
An irrational number cannot be written as a fraction of integers.

Prove sum is always irrational

We use proof by contradiction.
Let \(r\) be a rational number and \(i\) be an irrational number.
Assume their sum is a rational number \(s\):

$$r + i = s$$

Subtract \(r\) from both sides:

$$i = s - r$$

Since the difference of two rational numbers is always rational, \(s - r\) must be rational.
This contradicts the fact that \(i\) is irrational.
Thus, the sum \(r + i\) must be irrational.

Provide sum example

Let the rational number be \(r = 3\).
Let the irrational number be \(i = \sqrt{2}\).
Their sum is \(3 + \sqrt{2}\).
Since \(\sqrt{2} \approx 1.414213...\) is non-repeating and non-terminating, \(3 + \sqrt{2} \approx 4.414213...\) is also non-repeating and non-terminating, which is irrational.

Prove closure under multiplication

We use the Closure Property concept.
Let \(x\) and \(y\) be any two rational numbers:

$$x = \frac{a}{b}, \quad y = \frac{c}{d}$$

where \(a, b, c, d\) are integers, and \(b, d
eq 0\).
Multiply the two numbers:

$$x \cdot y = \frac{a \cdot c}{b \cdot d}$$

Since integers are closed under multiplication, \(a \cdot c\) and \(b \cdot d\) are integers.
Since \(b
eq 0\) and \(d
eq 0\), \(b \cdot d
eq 0\).
Therefore, the product is a ratio of integers, which is a rational number.

Provide product example

Let the rational numbers be \(\frac{2}{3}\) and \(\frac{5}{7}\).
Their product is:

$$\frac{2}{3} \cdot \frac{5}{7} = \frac{10}{21}$$

Since \(10\) and \(21\) are integers and \(21
eq 0\), the product \(\frac{10}{21}\) is rational.

Answer:

Question 1

To explain why the sum of a rational number and an irrational number is always irrational, we can use a proof by contradiction:

1. Proof: Let \(r\) be a rational number and \(i\) be an irrational number. Assume their sum is a rational number \(s\), so \(r + i = s\). Solving for \(i\) gives \(i = s - r\). Since the difference of two rational numbers is always rational, \(s - r\) must be rational. This contradicts our starting condition that \(i\) is irrational. Therefore, the sum \(r + i\) must be irrational.
2. Example: Let the rational number be \(3\) and the irrational number be \(\sqrt{2}\). Their sum is \(3 + \sqrt{2}\). Since the decimal expansion of \(\sqrt{2}\) is non-terminating and non-repeating (\(1.414213...\)), adding \(3\) results in \(4.414213...\), which remains non-terminating and non-repeating, making it irrational.

Question 2

To explain why the product of two rational numbers is closed under multiplication:

1. Proof: Let \(x\) and \(y\) be any two rational numbers, which can be written as \(x = \frac{a}{b}\) and \(y = \frac{c}{d}\) (where \(a, b, c, d\) are integers, and \(b, d

eq 0\)). Their product is:

$$x \cdot y = \frac{a \cdot c}{b \cdot d}$$

Since integers are closed under multiplication, \(a \cdot c\) and \(b \cdot d\) are integers. Because \(b
eq 0\) and \(d
eq 0\), their product \(b \cdot d
eq 0\). Since the product can be written as a ratio of two integers with a non-zero denominator, it is by definition a rational number.

1. Example: Let the two rational numbers be \(\frac{2}{3}\) and \(\frac{5}{7}\). Their product is:

$$\frac{2}{3} \cdot \frac{5}{7} = \frac{10}{21}$$

Since \(10\) and \(21\) are integers and \(21
eq 0\), the product \(\frac{10}{21}\) is a rational number.