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Chapter 1: Real Numbers

    Introduction

    The Fundamental Theorem of Arithmetic

    Revisiting Irrational Numbers

    Exercise 1.1

    Exercise 1.2

    Summary

    Enhanced Curriculum Support

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Chapter 1: Real Numbers > Exercise 1.2

Exercise 1.2

Prove that 5 is irrational.

Solution

Asuumption(Contradictory method):

Let's assume 5 is rational. Thus, 3 = pq where p and q are two non-zero integers.

Suppose p and q have a common factor other than 1. If we divide by the common factor(s) to get ,5=ab where a and b are .

Squaring on both sides of the equation:

we get: 5 = .

Taking b2 on to the LHS.

5b2=a2

Implication:

Since 5b2=a2, a2 is divisible by 5.

From Theorem 1.2, If a prime p divides a2, then p must divide a.

This implies that since, 5 divides a2 which means divides a.

TIf 5 divides a then we can write a = 5c for some integer c.

⁡5b2 =

⁡5b2 = c2

b2 = c2

This means b2 is divided by 5 which means b is divided by .

Thus, a is divided by 5 and b is divided by 5. But this is not possible as a and b are .

This means that our assumption is .

Thus, 5 is .

Prove that 3 + 25 is irrational.

Solution

Asuumption(Contradictory method):

Let's assume that 3 + 25 is rational.

If 3 + 25 is rational that means it can be written in the form of ab where a and b are integers that have no common factor other than 1 and b ≠ 0.

3 + 25 = .

Rearranging to isolate 5 , we have:

(3 + 25) =

3b + 25b = a

25b = a -

5 = (a - 3b)2b

Analyze the Result:

Since a,b are integers, (a - 3b)2b is a number, then 5 is also a number.

But, we know that 5 is .

Therefore, our assumption was wrong that 3 + 25 is .

Hence, 3 + 25 is .

Prove that the following are irrationals :

(i)

(i) 12

Solution

Asuumption(Contradictory method):

Let us assume that 12 is a number.

Then, 12 = ab, where a and b have no common factors other than .

Rearranging to isolate 2 , we have:

√2 × a =

2 = .

Since b and a are integers, ba is a number and so, √2 is .

But we know that √2 is .

So, our assumption was wrong.

Therefore, 12 is an irrational number.

(ii)

(ii) 75

Solution

Asuumption(Contradictory method):

Let us assume that 75 is a rational number.

Then, 75= ab, where a and b have no common factors other than .

Rearranging to isolate 5, we have:

75 b =

5=

Since, a, 7, and b are integers, so, a7b is a number.

This means 5 is . But this contradicts the fact that 5 is .

So, our assumption was wrong.

Therefore, 75 is an irrational number.

(iii)

(iii) 6+2

Solution

Asuumption(Contradictory method):

Let us assume that 6+2 is rational.

Then, 6+2 = ab, where a and b have no common factors other than .

Rearranging to isolate 2 , we have:

2 = ab -

Since, a, b, and 6 are integers, so, ab - 6 is a number.

This means 2 is also a number.

But this contradicts the fact that 2 is .

So, our assumption was wrong.

Therefore, 6+2 is an irrational number.