1.3-Revisiting Irrational Numbers

1.3-Revisiting Irrational Numbers Important Formulae

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Grade 10 → Math → Real Numbers → 1.3-Revisiting Irrational Numbers

After successful completion of this topic, you should be able to:

  • Apply theorems of irrational number in order to prove whether a given number is irrational or not.

Irrational numbers are an essential component of the real number system. They are defined as numbers that cannot be expressed as a fraction of two integers. This subtopic revisits the properties, examples, and significance of irrational numbers in mathematics.

1. Definition of Irrational Numbers

An irrational number is a number that cannot be represented in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. The decimal expansion of irrational numbers is non-terminating and non-repeating.

2. Examples of Irrational Numbers

Some common examples of irrational numbers include:

  • Square Roots: Numbers like $\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{7}$, etc., are irrational because they cannot be expressed as exact fractions. For example, $\sqrt{2} \approx 1.41421356\ldots$.
  • Pi ($\pi$): The value of $\pi$ is approximately $3.14159\ldots$, and it represents the ratio of the circumference of a circle to its diameter. It is known to be an irrational number.
  • Euler's Number ($e$): Approximately equal to $2.71828\ldots$, $e$ is another famous irrational number that is the base of natural logarithms.
3. Properties of Irrational Numbers

Irrational numbers possess specific properties that differentiate them from rational numbers:

  • Non-terminating and Non-repeating: The decimal representation of irrational numbers goes on forever without repeating any pattern.
  • Sum and Product: The sum or product of a rational number and an irrational number is always irrational. For example, $2 + \sqrt{3}$ is irrational.
  • Sum of Two Irrational Numbers: The sum of two irrational numbers can be rational or irrational. For example, $\sqrt{2} + (2 - \sqrt{2}) = 2$ (rational), while $\sqrt{2} + \sqrt{3}$ remains irrational.
4. Visual Representation of Irrational Numbers

Irrational numbers can be represented on the number line, just like rational numbers. The density of irrational numbers means that between any two rational numbers, there exist infinitely many irrational numbers. For example, between $1$ and $2$, we can find numbers like $\sqrt{2}, \sqrt{3},$ and $\pi$.

5. Common Misconceptions

It is important to address some common misconceptions regarding irrational numbers:

  • Not All Non-integer Numbers are Irrational: Numbers like $0.5$ and $1.75$ are not irrational; they are rational numbers.
  • Irrational Numbers are Not Whole: While they can take any position on the number line, irrational numbers cannot be counted like integers.
6. Importance of Irrational Numbers

Irrational numbers play a crucial role in various fields of mathematics and science:

  • Geometry: The calculation of areas and circumferences of circles involves the use of $\pi$, an irrational number.
  • Algebra: Many algebraic equations yield irrational solutions, emphasizing the need to understand these numbers.
  • Real Analysis: The study of functions and limits often requires the consideration of both rational and irrational numbers.
7. Rationalizing Denominators

In many mathematical problems, it is common to encounter expressions with irrational numbers in the denominator. To simplify such expressions, we often "rationalize" the denominator. For example:

To simplify $\frac{1}{\sqrt{2}}$, we multiply the numerator and denominator by $\sqrt{2}$:

$$\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$

This process helps in simplifying expressions for easier calculations and interpretations.


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Prove that $\sqrt{5}$ is irrational.

Solution:

Prove that $\sqrt{5}$ is irrational

Assume, for the sake of contradiction, that $\sqrt{5}$ is rational. This means that $\sqrt{5}$ can be expressed as a fraction of two integers, say $\sqrt{5} = \frac{p}{q}$, where $p$ and $q$ are integers with no common factors (i.e., $\frac{p}{q}$ is in its lowest terms), and $q \neq 0$.

Now, squaring both sides of the equation:

$\sqrt{5} = \frac{p}{q}$

Squaring both sides:

5 = $\frac{p^2}{q^2}$

Multiplying both sides by $q^2$, we get:

5$q^2 = p^2$

This implies that $p^2$ is divisible by 5. Therefore, $p$ must also be divisible by 5, because if a square of a number is divisible by a prime, the number itself must be divisible by that prime. So, let $p = 5k$, where $k$ is some integer.

Substitute $p = 5k$ into the equation $5q^2 = p^2$:

5$q^2 = (5k)^2$

5$q^2 = 25k^2$

Dividing both sides by 5:

$q^2 = 5k^2$

This shows that $q^2$ is also divisible by 5, so $q$ must be divisible by 5 as well.

Thus, both $p$ and $q$ are divisible by 5. However, this contradicts our original assumption that $p$ and $q$ have no common factors (since both are divisible by 5).

Therefore, our assumption that $\sqrt{5}$ is rational must be false. Hence, $\sqrt{5}$ is irrational.

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Prove that 3 + $2\sqrt{5}$ is irrational.

Solution:

Prove that $3 + 2\sqrt{5}$ is irrational

Let us assume, for the sake of contradiction, that $3 + 2\sqrt{5}$ is rational. This means that it can be expressed as a ratio of two integers, say:

$3 + 2\sqrt{5} = \frac{p}{q}$, where $p$ and $q$ are integers, and $q \neq 0$. Also, assume that the fraction is in its simplest form, i.e., $\gcd(p, q) = 1$.

Now, we will isolate $2\sqrt{5}$:

$2\sqrt{5} = \frac{p}{q} - 3$

Next, we rewrite the right-hand side to have a common denominator:

$2\sqrt{5} = \frac{p - 3q}{q}$

Now, divide both sides of the equation by 2:

$\sqrt{5} = \frac{p - 3q}{2q}$

Thus, $\sqrt{5}$ is expressed as a ratio of two integers, $\frac{p - 3q}{2q}$. This implies that $\sqrt{5}$ is rational.

However, this is a contradiction, because we know that $\sqrt{5}$ is an irrational number. Therefore, our assumption that $3 + 2\sqrt{5}$ is rational must be false.

Hence, $3 + 2\sqrt{5}$ is irrational.

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Prove that the following are irrationals:

(i) $\dfrac{1}{\sqrt{2}}$
(ii) $7\sqrt{5}$
(iii) $6 + \sqrt{2}$

Solution:

Proof that the following are irrationals:
(i) $\dfrac{1}{\sqrt{2}}$
Assume, for the sake of contradiction, that $\dfrac{1}{\sqrt{2}}$ is rational. This means that it can be expressed as $\dfrac{1}{\sqrt{2}} = \dfrac{p}{q}$, where $p$ and $q$ are integers with $q \neq 0$, and $\gcd(p, q) = 1$ (i.e., $p$ and $q$ are coprime). Squaring both sides of the equation: $$ \dfrac{1}{2} = \dfrac{p^2}{q^2} $$ Multiplying both sides by $2q^2$: $$ q^2 = 2p^2 $$ This implies that $q^2$ is even, and therefore $q$ must be even. Let $q = 2k$ for some integer $k$. Substituting this into the equation: $$ (2k)^2 = 2p^2 $$ $$ 4k^2 = 2p^2 $$ $$ 2k^2 = p^2 $$ This implies that $p^2$ is even, and therefore $p$ must also be even. Let $p = 2m$ for some integer $m$. Substituting this into the equation: $$ 2k^2 = (2m)^2 $$ $$ 2k^2 = 4m^2 $$ $$ k^2 = 2m^2 $$ This shows that $k^2$ is even, and so $k$ must also be even. However, this contradicts the assumption that $p$ and $q$ are coprime, as both $p$ and $q$ are divisible by 2. Hence, $\dfrac{1}{\sqrt{2}}$ cannot be rational. Therefore, $\dfrac{1}{\sqrt{2}}$ is irrational.
(ii) $7\sqrt{5}$
Assume, for the sake of contradiction, that $7\sqrt{5}$ is rational. This means that it can be expressed as $7\sqrt{5} = \dfrac{p}{q}$, where $p$ and $q$ are integers with $q \neq 0$, and $\gcd(p, q) = 1$. Dividing both sides by 7: $$ \sqrt{5} = \dfrac{p}{7q} $$ This implies that $\sqrt{5}$ is rational, which is a contradiction, because it is known that $\sqrt{5}$ is irrational. Hence, $7\sqrt{5}$ cannot be rational. Therefore, $7\sqrt{5}$ is irrational.
(iii) $6 + \sqrt{2}$
Assume, for the sake of contradiction, that $6 + \sqrt{2}$ is rational. This means that it can be expressed as $6 + \sqrt{2} = \dfrac{p}{q}$, where $p$ and $q$ are integers with $q \neq 0$, and $\gcd(p, q) = 1$. Subtract 6 from both sides: $$ \sqrt{2} = \dfrac{p}{q} - 6 $$ This implies that $\sqrt{2}$ is rational, which is a contradiction, because it is known that $\sqrt{2}$ is irrational. Hence, $6 + \sqrt{2}$ cannot be rational. Therefore, $6 + \sqrt{2}$ is irrational.

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