Revisiting Irrational Numbers: Composite Expressions

Prove irrationality of expressions like a + b√c by isolating the root and showing a contradiction involving rational closures.

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Operations on Irrationals

Recap Class IX rules on sum and product.

In Class IX, you explored how rational and irrational numbers interact. A quick recap:

The sum or difference of a rational and an irrational number is always irrational.
The product and quotient of a non-zero rational and an irrational number is always irrational.

Now, instead of just stating these rules, we are going to prove them for particular cases.

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Pure Root vs Composite Expressions

A comparison showing the difference between pure root proofs (divisibility) and composite proofs (isolation).

$Clean side-by-side comparison diagram showing two math proof strategies. Left side titled 'Pure Root Proof' showing squaring both sides and using divisibility logic. Right side titled 'Composite Expression Proof' showing the algebraic isolation of the square root and using rational closure logic. Consistent visual treatment for both sides, same scale and style, pastel color coding to distinguish categories, elegant typography, white background, generous whitespace.$

Click to zoom

Notice how composite expressions do not require squaring both sides. Instead, we isolate the root to reveal a contradiction.

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The Isolation Strategy

Explain that composite proofs don't use divisibility; they isolate the root.

For expressions like $5 - 3$ , do not use the same method as pure roots like $2$ .

If we assume $5 - 3$ is rational, write:

5 - 3 = \frac{a}{b}

Now move the integer part to the other side:

3 = 5 - \frac{a}{b}

The right side, $5 - \frac{a}{b}$ , is rational because integers and fractions stay rational under subtraction.

So this would mean $3$ is rational, which is impossible.

Key idea: isolate the square root first, then show it would become rational.

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Proving a Difference is Irrational

Example 6: 5 - sqrt(3).

Problem. Show that $5 - \sqrt{3}$ is irrational.

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Proving a Product is Irrational

Example 7: 3 sqrt(2).

Problem. Show that $3\sqrt{2}$ is irrational.

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Sequence the Isolation Steps

Order the steps for proving 3 + 2sqrt(5) is irrational.

Arrange the logical steps to prove that $3 + 2\sqrt{5}$ is irrational.

Drag to reorder

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Proving a Quotient is Irrational

Exercise 1.2 Q3(i) 1/sqrt(2).

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Let us prove that $\frac{1}{\sqrt{2}}$ is irrational. First, assume to the contrary that $\frac{1}{\sqrt{2}}$ is rational.

This means we can find coprime integers $a$ and $b$ (where $a, b \neq 0$ ) such that $\frac{1}{\sqrt{2}} = \frac{a}{b}$ . By rearranging this equation and inverting both sides, we isolate the square root to get $\sqrt{2} =$ .

Since $a$ and $b$ are integers, the fraction $\frac{b}{a}$ is , which would imply that $\sqrt{2}$ is also rational. However, this the established fact that $\sqrt{2}$ is irrational.

Therefore, our initial assumption is incorrect, proving that $\frac{1}{\sqrt{2}}$ is indeed irrational.

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Proving a Difference is Irrational

Example 6: 5 - sqrt(3).

Problem. Show that $5 - \sqrt{3}$ is irrational.

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Proving a Product is Irrational

Example 7: 3 sqrt(2).

Problem. Show that $3\sqrt{2}$ is irrational.

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reading

Proving a Quotient is Irrational

Exercise 1.2 Q3(i) 1/sqrt(2).

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Let us prove that $\frac{1}{\sqrt{2}}$ is irrational. First, assume to the contrary that $\frac{1}{\sqrt{2}}$ is rational.

Since $a$ and $b$ are integers, the fraction $\frac{b}{a}$ is , which would imply that $\sqrt{2}$ is also rational. However, this the established fact that $\sqrt{2}$ is irrational.

Therefore, our initial assumption is incorrect, proving that $\frac{1}{\sqrt{2}}$ is indeed irrational.