Irrational numbers

Given \(x=\sqrt{7}+3\), we get \(x-3=\sqrt{7}+3-3=\sqrt{7}\). Since 7 is not a perfect square, \(\sqrt{7}\) cannot be expressed as the ratio of two integers, so it is irrational. Therefore, option B is correct. Exam tip: the square root of a non-perfect square is irrational.

(\sqrt{32}=4\sqrt{2}) and (\sqrt{18}=3\sqrt{2}), so the sum is (7\sqrt{2}). First convert radicals into like form.

Let \(r\ne0\) be rational and \(x\) be irrational. If \(rx\) were rational, then \(x=(rx)/r\) would be rational, a contradiction. In exams, always check the condition \(r\ne0\).

Both \(\sqrt{2}\) and \(-\sqrt{2}\) are irrational, but their sum is \(0\), which is rational. Hence the word “always” makes the claim false. Exam tip: one counterexample is enough to disprove a universal statement.

For natural \(n\), \(\sqrt{n}\) is rational only when \(n\) is a perfect square. Thus, a non-square has an irrational root. Since \(9=3^2\) but \(\sqrt{10}\) is irrational, check square status first.

For example, \(0.333\ldots = \frac{1}{3}\), so a non-terminating recurring decimal is rational. Only non-terminating non-recurring decimals are irrational. In exams, check for repetition.

(\sqrt{75}=5\sqrt{3}) and (\sqrt{12}=2\sqrt{3}), so the difference is (3\sqrt{3}). First take out perfect-square factors.

It has no fixed repeating block, so it is irrational. A non-terminating non-repeating decimal is irrational.

In A, \(\sqrt{7}+(-\sqrt{7})=0\). Both terms are irrational, but 0 is rational, so the claim fails. Exam tip: disprove “always” by finding one counterexample.

(\sqrt{80}=4\sqrt{5}) and (\sqrt{125}=5\sqrt{5}), so the sum is (9\sqrt{5}). Like radicals should be added.

Since (15<17<20), (\sqrt{17}) lies between them. Compare square roots using the numbers inside.

The number of zeros after successive 1s is 1, 2, 3, 4, ... , so no fixed repeating block exists. A non-terminating, non-repeating decimal is irrational. In exams, check for a repeating cycle, not merely repeated digits.

An irrational number cannot be expressed as a ratio of two integers and has a decimal expansion that is non-terminating and non-repeating. Option C, 0.4141141114…, is intended to show a decimal whose digits continue indefinitely without a fixed repeating block, so it is irrational. Option A displays a repeating block, 82, and is therefore rational; it can be written as a fraction. Option B is already a ratio of integers and is rational. Option D is terminating and can be expressed as 3625/1000, which reduces to a rational number. The slash-separated presentation in the options is slightly unusual, but the deciding property is the decimal pattern of the stated number.

The number of zeros between successive 1s keeps increasing, so no fixed repeating block occurs. Thus it is a non-terminating, non-recurring decimal and is irrational. Exam tip: check repetition, not the digits used.

Multiplying numerator and denominator by (\sqrt{7}) gives (\frac{3\sqrt{7}}{7}). Rationalisation keeps the value same.

If x is irrational and r is rational, assuming x+r is rational makes x=(x+r)−r rational, a contradiction. Thus A is correct. Exam tip: √2+(−√2)=0, so B is false.

\(6\) is rational because it can be expressed as a ratio of two integers. \(-\sqrt{10}\) is irrational because 10 is not a perfect square, and adding a negative sign does not change its irrational nature. Therefore, the irrational part is \(-\sqrt{10}\). Exam tip: In a sum or difference, identify the term containing the square root of an integer that is not a perfect square.

(\sqrt{45}=3\sqrt{5}), (\sqrt{125}=5\sqrt{5}) and (\sqrt{20}=2\sqrt{5}), so the answer is (6\sqrt{5}). Add and subtract coefficients of like radicals.

The number of zeros between successive 1s is 1, 2, 3, 4, ... , so no fixed block repeats periodically. Hence it is irrational. Exam tip: a rational decimal terminates or eventually repeats.

The square root of a perfect square is an integer, so it is rational. First check whether the number is a perfect square.