Irrational numbers

The governing concept is that the square-root function is increasing for non-negative numbers. Because 2 < 3, taking their principal square roots preserves the inequality, giving √2 < √3. This can also be checked by approximation: √2 is about 1.414 and √3 is about 1.732. Therefore option C is correct. The two numbers cannot be equal because equal non-negative square roots would have equal squares, whereas 2 and 3 are different. Option A reverses the valid inequality. Option D is also false because neither 2 nor 3 is a perfect square, so both √2 and √3 are irrational, although the question mainly asks about their order.

An irrational number has a non-terminating, non-repeating decimal expansion, such as \(\sqrt{2}\). Option C is repeating, so it is rational. Exam tip: a repeating decimal always represents a rational number.

In a right triangle, the hypotenuse is (\sqrt{1^2+1^2}=\sqrt{2}). Understand construction using Pythagoras theorem.

The number line represents real numbers, including both rational and irrational numbers. Irrationals can be constructed geometrically.

\(0.333...\) is a recurring decimal and \(0.333...=\frac{1}{3}\), so it is rational. An infinite decimal is not always irrational; in exams, check whether the digits repeat.

The positive number is (\sqrt{19}) because ((\sqrt{19})^2=19). Since (19) is not a perfect square, it is irrational.

(\sqrt{2}) is irrational so its decimal is non-terminating and non-repeating. Do not conclude from only a few decimal digits.

Every real number is not rational because irrational numbers are also real. The number system includes both types.

(\sqrt{2}) and (\sqrt{5}) are both irrational because (2) and (5) are not perfect squares. Check each number separately.

(\sqrt{28}=2\sqrt{7}) and (\sqrt{63}=3\sqrt{7}), so the sum is (5\sqrt{7}). Add like radicals.

(\sqrt{72}=6\sqrt{2}) and (\sqrt{8}=2\sqrt{2}), so the difference is (4\sqrt{2}). Simplify like radicals first.

The positive number is (\sqrt{37}) because ((\sqrt{37})^2=37). Since (37) is not a perfect square, it is irrational.

This decimal has no fixed repeating block. So it is a non-terminating non-repeating decimal and an irrational number.

Both \(\sqrt{2}\) and \(-\sqrt{2}\) are irrational, but their sum is \(0\), which is rational. Hence the claim is false. In contrast, \(3\sqrt{2}\) remains irrational. Exam tip: test an “always” statement using one counterexample.

(\sqrt{44}=2\sqrt{11}) and (\sqrt{99}=3\sqrt{11}), so the sum is (5\sqrt{11}). Simplify before adding like radicals.

An irrational number cannot be written as \(p/q\), so its decimal expansion is non-terminating and non-repeating. A repeating decimal is rational. Exam tip: look for both features together.

(\sqrt{18}=3\sqrt{2}) and (\sqrt{8}=2\sqrt{2}), so the sum is (5\sqrt{2}). Simplify radicals first.

An irrational number has a decimal expansion that never ends and never repeats a block of digits. Option C describes a rational number. Exam tip: non-terminating, non-repeating decimals are irrational.

Rationalising the denominator gives (\frac{1}{2+\sqrt{3}}=2-\sqrt{3}). Remember to multiply by the conjugate.

Writing a number in decimal form does not make it rational. The blocks of 1s grow as 1, 2, 3, ..., so no fixed repeating block exists. A non-terminating, non-repeating decimal is irrational. Exam tip: check for repetition.