Revisiting Irrational Numbers: Proving Pure Roots

Use contradiction and prime divisibility theorems to formally prove that square roots of primes are irrational.

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The Core Logic of Contradiction

Introduce Theorem 1.2 and the proof approach.

In math, sometimes the easiest way to prove something is true is to assume it's false and watch the logic break down. This is called a proof by contradiction.

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Prime Divisibility Theorem

Formal statement of the prime divisibility theorem.

Let

p

be a prime number. If

p

divides

a^{2}

, then

p

divides

a

, where

a

is a positive integer.

Concrete Example If

3

divides

81

(which is

9^{2}

), then

3

divides

9

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worked example

Proof: √2 is Irrational

Theorem 1.3.

Prove that $\sqrt{2}$ is irrational.

Let us assume, to the contrary, that $\sqrt{2}$ is rational. That is, we can find coprime integers $a$ and $b$ ( $b \neq 0$ ) such that: $\sqrt{2} = \frac{a}{b}$

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Proof: √3 is Irrational

Example 5.

Prove that $\sqrt{3}$ is irrational.

Let us assume, to the contrary, that $\sqrt{3}$ is rational. That is, we can find integers $a$ and $b$ ( $b \neq 0$ ) such that: $\sqrt{3} = \frac{a}{b}$

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sequence

Sequence the Proof for √5

Order the logical steps for proving sqrt(5) is irrational.

Arrange the steps to construct the proof by contradiction that $\sqrt{5}$ is irrational.

Drag to reorder

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reading

Complete the √5 Proof

Exercise 1.2 Q1.

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To prove that $\sqrt{5}$ is irrational, let us assume the opposite: suppose it is rational.

This means we can find integers $a$ and $b$ , where $b \neq 0$ , such that $\sqrt{5}=\frac{a}{b}$ .

Squaring both sides gives $5b^2=a^2$ .

This implies that $5$ divides $a^2$ , and therefore $5$ divides $a$ .

So, we can write $a=$ for some integer $c$ .

Substituting this back into $5b^2=a^2$ , we get $5b^2=$ .

This simplifies to $b^2=5c^2$ .

So, $5$ also divides $b$ .

Since both $a$ and $b$ are divisible by $5$ , this contradicts our initial assumption that $a$ and $b$ have no common factor other than $1$ .

Hence, $\sqrt{5}$ is irrational.

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worked example

Proof: √2 is Irrational

Theorem 1.3.

Prove that $\sqrt{2}$ is irrational.

Let us assume, to the contrary, that $\sqrt{2}$ is rational. That is, we can find coprime integers $a$ and $b$ ( $b \neq 0$ ) such that: $\sqrt{2} = \frac{a}{b}$

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worked example

Proof: √3 is Irrational

Example 5.

Prove that $\sqrt{3}$ is irrational.

Let us assume, to the contrary, that $\sqrt{3}$ is rational. That is, we can find integers $a$ and $b$ ( $b \neq 0$ ) such that: $\sqrt{3} = \frac{a}{b}$

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reading

Complete the √5 Proof

Exercise 1.2 Q1.

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To prove that $\sqrt{5}$ is irrational, let us assume the opposite: suppose it is rational.

This means we can find integers $a$ and $b$ , where $b \neq 0$ , such that $\sqrt{5}=\frac{a}{b}$ .

Squaring both sides gives $5b^2=a^2$ .

This implies that $5$ divides $a^2$ , and therefore $5$ divides $a$ .

So, we can write $a=$ for some integer $c$ .

Substituting this back into $5b^2=a^2$ , we get $5b^2=$ .

This simplifies to $b^2=5c^2$ .

So, $5$ also divides $b$ .

Since both $a$ and $b$ are divisible by $5$ , this contradicts our initial assumption that $a$ and $b$ have no common factor other than $1$ .

Hence, $\sqrt{5}$ is irrational.