$5 \times 11 \times 17 + 3 \times 11$

$= 11(5 \times 17 + 3)$

$= 11(85 + 3)$

$= 11 \times 88$

$= 11 \times 8 \times 11$

$= 8 \times 11^2$

Since the number can be expressed as a product of factors (other than 1 and itself), it is a composite number.

Source: Chapter 1, The Fundamental Theorem of Arithmetic

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• The key step is taking 11 as a common factor. This immediately shows the number is divisible by 11 (and also by 8), so it has factors other than 1 and itself — the definition of a composite number.
• Examiners expect you to show the factorisation step-by-step and explicitly state the conclusion. Do not just calculate the final value (968); show why it is composite by exhibiting its factors.
• This is similar to Exercise 1.1, Q.6 from Chapter 1.

Proof (by contradiction):

Assume $5 - 2\sqrt{3}$ is rational. Then we can write:

$$5 - 2\sqrt{3} = \frac{a}{b}, \quad \text{where } a, b \text{ are integers, } b \neq 0$$

Rearranging:

$$2\sqrt{3} = 5 - \frac{a}{b} = \frac{5b - a}{b}$$

$$\sqrt{3} = \frac{5b - a}{2b}$$

Since $a$ and $b$ are integers, $\dfrac{5b-a}{2b}$ is rational, which means $\sqrt{3}$ is rational.

But this contradicts the given fact that $\sqrt{3}$ is irrational.

Therefore, our assumption is wrong, and $5 - 2\sqrt{3}$ is irrational. $\blacksquare$

Source: Real Numbers, Section 1.3 – Revisiting Irrational Numbers

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• Use proof by contradiction: assume the number is rational, express it as $\frac{a}{b}$, and rearrange to isolate $\sqrt{3}$.
• The key step is showing this forces $\sqrt{3}$ to be rational — contradicting the given condition.
• Examiners award marks for: (1) correct assumption, (2) correct algebraic rearrangement to isolate $\sqrt{3}$, (3) stating the contradiction clearly.
• Don't forget to state the conclusion explicitly.

Assume, to the contrary, that $2 - 5\sqrt{3}$ is rational.

Then there exist coprime integers $a$ and $b$ ($b \neq 0$) such that:

$$2 - 5\sqrt{3} = \frac{a}{b}$$

$$\Rightarrow 5\sqrt{3} = 2 - \frac{a}{b} = \frac{2b - a}{b}$$

$$\Rightarrow \sqrt{3} = \frac{2b - a}{5b}$$

Since $a$ and $b$ are integers, $\dfrac{2b-a}{5b}$ is rational, which means $\sqrt{3}$ is rational.

But this contradicts the given fact that $\sqrt{3}$ is irrational.

Therefore, our assumption is wrong and $2 - 5\sqrt{3}$ is irrational.

Source: Chapter 1, Section 1.3 — Revisiting Irrational Numbers

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• This is a proof by contradiction question — always start by assuming the opposite of what you want to prove.
• The key step is isolating $\sqrt{3}$ on one side to show it would have to be rational — contradicting the given condition.
• Examiners award marks for: (1) correct assumption, (2) algebraic rearrangement to isolate $\sqrt{3}$, (3) stating the contradiction clearly. Don't skip the contradiction statement.

Option A — Two right triangles are always similar is false.

Right triangles have one angle = 90°, but the other angles can differ, so they need not be similar.

• Two figures are similar if all corresponding angles are equal and sides are proportional.
• Right triangles share one 90° angle, but the remaining two angles can vary (e.g., 30°-60°-90° vs. 45°-45°-90°), so similarity is not guaranteed.
• Squares, equilateral triangles, and circles are always similar because their shapes are fixed regardless of size.

Proof: Assume, to the contrary, that $\sqrt{2}$ is rational.

Then we can find integers $a$ and $b$ ($b \neq 0$) such that $\sqrt{2} = \dfrac{a}{b}$, where $a$ and $b$ are coprime (no common factor other than 1).

Squaring both sides: $2b^2 = a^2$

So 2 divides $a^2$. By Theorem 1.2, 2 divides $a$.

Let $a = 2c$ for some integer $c$. Substituting:

$$2b^2 = 4c^2 \implies b^2 = 2c^2$$

So 2 divides $b^2$, which means 2 divides $b$.

Thus 2 is a common factor of both $a$ and $b$, which contradicts the fact that $a$ and $b$ are coprime.

This contradiction arose because of our incorrect assumption. Hence, $\sqrt{2}$ is irrational.

Source: Chapter 1, Section 1.3 – Revisiting Irrational Numbers

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• This is a proof by contradiction — always start by assuming the opposite of what you want to prove.
• Key step: use Theorem 1.2 (if a prime $p$ divides $a^2$, then $p$ divides $a$) — examiners expect you to cite this.
• The contradiction must be clearly stated: $a$ and $b$ were assumed coprime, but both are divisible by 2.
• Write "Hence $\sqrt{2}$ is irrational" as the concluding line — never omit the conclusion.

(c) $\cos 20° > \cos 70°$

As the angle increases from 0° to 90°, cosine decreases. So $\cos 20° > \cos 70°$. (sin and tan increase, making options a, b, d false.)

Key rule: sin and tan are increasing functions on (0°, 90°), while cos is a decreasing function. So a smaller angle gives a larger cosine value — hence $\cos 20° > \cos 70°$ is the only true statement. Eliminate the others: sin 20° < sin 70° (sin increases), cos 20° > sin 20° only when angle < 45° but that's not what option (b) says — sin 20° < cos 20°, and tan 20° < tan 70° (tan increases). Examiners expect you to recall increasing/decreasing behaviour of trig ratios.

(a) a prime number

2 is a prime number because it has exactly two distinct factors: 1 and itself, and it is listed among the collection of prime numbers in the Fundamental Theorem of Arithmetic.

Source: Chapter 1, Section 1.2

2 is the smallest and the only even prime number. A prime number has exactly two factors — 1 and itself. 2 satisfies this condition. It is neither composite (which requires more than two factors) nor both. Examiners expect the correct option with a brief justification for MCQs.

(c) Assertion (A) is true, but Reason (R) is false.

Since OA = OB (radii), $\sqrt{(4-2)^2+(3-3)^2} = \sqrt{(x-2)^2+(5-3)^2}$ gives $2 = \sqrt{(x-2)^2+4}$, so $x = 2$. ✓
The centre is NOT the mid-point of every chord; it bisects a chord only when the line from centre is perpendicular to it.

• Assertion check: OA = $\sqrt{4+0}$ = 2; OB = $\sqrt{(x-2)^2+4}$ = 2 ⟹ $(x-2)^2$ = 0 ⟹ $x = 2$. True.
• Reason check: The centre is the mid-point of a chord only if it is the perpendicular bisector of that chord — it is not the mid-point of every chord. The statement as given is false.
• So the answer is (c): A is true, R is false.

Proof (by contradiction):

Assume $2 + \sqrt{3}$ is rational. Then we can write:

$$2 + \sqrt{3} = \frac{a}{b}, \quad \text{where } a, b \text{ are integers and } b \neq 0$$

Rearranging:

$$\sqrt{3} = \frac{a}{b} - 2 = \frac{a - 2b}{b}$$

Since $a$ and $b$ are integers, $\dfrac{a-2b}{b}$ is rational, which means $\sqrt{3}$ is rational.

But this contradicts the given fact that $\sqrt{3}$ is irrational.

Therefore, our assumption is wrong, and $2 + \sqrt{3}$ is irrational. $\blacksquare$

Source: Chapter 1, Section 1.3 – Revisiting Irrational Numbers

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• This is a standard "proof by contradiction" question. Always start by assuming the opposite of what you want to prove.
• The key step is isolating $\sqrt{3}$ on one side — showing it equals a rational expression, which contradicts the given condition.
• Examiners award marks for: (1) correct assumption, (2) correct rearrangement isolating $\sqrt{3}$, (3) stating the contradiction clearly, (4) conclusion. Don't skip the conclusion line.

Assume, to the contrary, that $\left(\sqrt{2} + \dfrac{\sqrt{3}}{2}\right)$ is rational.

Then there exist coprime integers $a$ and $b$ ($b \neq 0$) such that:

$$\sqrt{2} + \frac{\sqrt{3}}{2} = \frac{a}{b}$$

Rearranging:

$$\frac{\sqrt{3}}{2} = \frac{a}{b} - \sqrt{2}$$

$$\sqrt{3} = 2\left(\frac{a}{b} - \sqrt{2}\right) = \frac{2a}{b} - 2\sqrt{2}$$

$$2\sqrt{2} = \frac{2a}{b} - \sqrt{3}$$

$$\sqrt{6} = \frac{1}{2}\left(\frac{2a}{b} - \sqrt{3}\right) \cdot \sqrt{2}$$

A simpler path: squaring both sides of $\sqrt{2} + \dfrac{\sqrt{3}}{2} = \dfrac{a}{b}$:

$$2 + \sqrt{6} + \frac{3}{4} = \frac{a^2}{b^2}$$

$$\sqrt{6} = \frac{a^2}{b^2} - \frac{11}{4} = \frac{4a^2 - 11b^2}{4b^2}$$

Since $a$, $b$ are integers, the RHS $\dfrac{4a^2 - 11b^2}{4b^2}$ is rational.

This means $\sqrt{6}$ is rational — a contradiction, since $\sqrt{6}$ is given to be irrational.

Hence, our assumption is wrong, and $\left(\sqrt{2} + \dfrac{\sqrt{3}}{2}\right)$ is irrational. $\blacksquare$

Source: Chapter 1, Section 1.3 – Revisiting Irrational Numbers

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• The standard method is proof by contradiction: assume the expression is rational, then algebraically isolate a known irrational ($\sqrt{6}$ here) and show it would have to be rational — contradiction.
• The key step is squaring: $\left(\sqrt{2} + \dfrac{\sqrt{3}}{2}\right)^2 = 2 + \sqrt{6} + \dfrac{3}{4} = \dfrac{11}{4} + \sqrt{6}$, which lets you isolate $\sqrt{6}$.
• Always explicitly state "this contradicts the fact that $\sqrt{6}$ is irrational" — examiners award a mark specifically for this conclusion.
• The given condition "$\sqrt{6}$ is irrational" must be used; never leave it unused.

Proof by contradiction.

Assume, to the contrary, that $\dfrac{2-\sqrt{3}}{5}$ is rational.

Then we can find coprime integers $a$ and $b$ ($b \neq 0$) such that:

$$\frac{2-\sqrt{3}}{5} = \frac{a}{b}$$

Rearranging:

$$2 - \sqrt{3} = \frac{5a}{b}$$

$$\sqrt{3} = 2 - \frac{5a}{b} = \frac{2b - 5a}{b}$$

Since $a$ and $b$ are integers, $\dfrac{2b-5a}{b}$ is rational, which means $\sqrt{3}$ is rational.

But this contradicts the given fact that $\sqrt{3}$ is irrational.

This contradiction arose because of our incorrect assumption.

Therefore, $\dfrac{2-\sqrt{3}}{5}$ is irrational. $\blacksquare$

Source: Chapter 1, Section 1.3 Revisiting Irrational Numbers

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• The key technique is proof by contradiction: assume the number is rational, then algebraically isolate $\sqrt{3}$ and show it would have to be rational — contradicting the given condition.
• Always state that $a$ and $b$ are integers to justify that the rearranged expression is rational.
• Examiners award marks for: (1) correct assumption, (2) correct algebraic rearrangement to isolate $\sqrt{3}$, (3) citing the contradiction with the given fact, and (4) the conclusion. Don't skip any step.

Proof (by contradiction):

Assume, to the contrary, that $4 + 3\sqrt{2}$ is rational.

Then we can find coprime integers $a$ and $b$ ($b \neq 0$) such that:

$$4 + 3\sqrt{2} = \frac{a}{b}$$

Rearranging:

$$3\sqrt{2} = \frac{a}{b} - 4 = \frac{a - 4b}{b}$$

$$\sqrt{2} = \frac{a - 4b}{3b}$$

Since $a$ and $b$ are integers, $\dfrac{a - 4b}{3b}$ is rational, which means $\sqrt{2}$ is rational.

But this contradicts the given fact that $\sqrt{2}$ is irrational.

This contradiction arose because of our incorrect assumption.

∴ $4 + 3\sqrt{2}$ is irrational. $\hspace{1cm}\blacksquare$

Source: Chapter 1, Section 1.3 – Revisiting Irrational Numbers

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• Use proof by contradiction: assume the number is rational, express it as $\frac{a}{b}$ with $a$, $b$ coprime.
• Isolate $\sqrt{2}$ algebraically — it ends up equal to a rational expression, contradicting the given condition.
• Always state clearly what the contradiction is and why it arose.
• Examiners award marks for: correct assumption (1), correct algebraic rearrangement to isolate $\sqrt{2}$ (1), stating the contradiction and conclusion (1).

Proof: Assume, to the contrary, that $\sqrt{3}$ is rational.

Then we can find integers $a$ and $b$ ($b \neq 0$) such that $\sqrt{3} = \dfrac{a}{b}$, where $a$ and $b$ are coprime (no common factor other than 1).

So, $b\sqrt{3} = a$.

Squaring both sides: $3b^2 = a^2$

Therefore, $a^2$ is divisible by 3, and by Theorem 1.2, $a$ is also divisible by 3.

So, let $a = 3c$ for some integer $c$.

Substituting: $3b^2 = 9c^2 \Rightarrow b^2 = 3c^2$

This means $b^2$ is divisible by 3, so $b$ is also divisible by 3.

Therefore, $a$ and $b$ have at least 3 as a common factor. But this contradicts the fact that $a$ and $b$ are coprime.

This contradiction arose because of our incorrect assumption. Hence, $\sqrt{3}$ is irrational. $\blacksquare$

Source: Chapter 1, Section 1.3 – Revisiting Irrational Numbers

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• This is a proof by contradiction — always start by assuming the opposite of what you want to prove.
• The key tool used is Theorem 1.2: if a prime $p$ divides $a^2$, then $p$ divides $a$.
• The critical step students miss: after substituting $a = 3c$, show that 3 also divides $b$ — this creates the contradiction with "coprime."
• Write the conclusion clearly: state the contradiction and then state that $\sqrt{3}$ is irrational.
• Examiners award marks for: assumption (½), algebraic steps (1), applying Theorem 1.2 twice (½), contradiction + conclusion (1).

Answer: (B)

A square and a rhombus of the same area are not always similar, as their angles differ (a square has all 90° angles, a rhombus need not).

Similarity requires both equal corresponding angles AND proportional sides. A rhombus has equal sides but its angles can vary, so it is not always similar to a square. The other three statements are correct: congruent figures are always similar (ratio 1:1); two equilateral triangles always have 60°-60°-60°; similar triangles need not be congruent. Examiners expect you to identify the false statement and briefly justify.

Proof: Assume, to the contrary, that $\sqrt{5}$ is rational.

Then we can find integers $a$ and $b$ ($b \neq 0$) such that $\sqrt{5} = \dfrac{a}{b}$, where $a$ and $b$ are coprime.

Squaring both sides: $5b^2 = a^2$

So 5 divides $a^2$. By Theorem 1.2, 5 divides $a$.

Write $a = 5c$ for some integer $c$. Substituting:

$5b^2 = 25c^2 \Rightarrow b^2 = 5c^2$

So 5 divides $b^2$, and therefore 5 divides $b$.

Thus 5 is a common factor of $a$ and $b$, contradicting the fact that $a$ and $b$ are coprime.

This contradiction arose because we assumed $\sqrt{5}$ is rational. Hence, $\sqrt{5}$ is irrational. $\blacksquare$

Source: Chapter 1, Section 1.3 – Revisiting Irrational Numbers

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• This is a standard proof by contradiction. Always start by assuming the opposite of what you want to prove.
• The key tool is Theorem 1.2: if a prime $p$ divides $a^2$, then $p$ divides $a$. Examiners expect you to cite or use this explicitly.
• The contradiction must target the coprime condition — that is where the marks are.
• Write the substitution step ($a = 5c$) clearly; missing it loses a mark.