Irrational numbers and real numbers

Direct answer: Option A, \(\sqrt{2}\) is irrational. A rational number can be written as \(p/q\), where p and q are integers and \(q\ne0\). Fractions, integers, terminating decimals and repeating decimals are rational. The number \(\sqrt2\) cannot be expressed as such a fraction; 2 is not a perfect square, so its square root is irrational. Option A is correct. Option B, \(3/4\), is already a ratio of integers, so it is rational. Option C, 0.5, terminates and equals \(1/2\), so it is rational. Option D, 7, is an integer and can be written as \(7/1\), so it is rational. Do not confuse an irrational number with a non-real number: \(\sqrt2\) is real, just not rational. Memory cue: the square root of a positive non-perfect square is irrational.

(\frac{5}{8}) is a ratio of two integers. A rational number can be written as (\frac{p}{q}).

Real numbers include both rational and irrational numbers. These are placed on the number line.

0.333... is a repeating decimal, so it is a rational number. To show this, let x = 0.333...; then 10x = 3.333... and subtracting gives 9x = 3, so x = 3/9 = \(\frac{1}{3}\). Thus 0.333... can be written as a fraction and is rational. Option B (irrational) is wrong because irrational decimals are non-terminating and non-repeating. Options C and D are also incorrect: 0.333... is a real number, and every real number is either rational or irrational. Exam tip: Any terminating or repeating decimal is rational—use the multiply-and-subtract method to convert to a fraction quickly.

A decimal that is non-terminating and non-repeating represents an irrational number. Rational numbers either terminate or repeat a fixed block in their decimal form, so option B is incorrect. Integers and whole numbers are special cases of rationals and therefore do not describe a non-terminating non-repeating decimal; C and D are incorrect. Exam tip: if a decimal neither ends nor shows a repeating pattern, classify it as irrational; try expressing the number as a fraction to test rationality.

\(\sqrt{49}=7\). Since 7 can be written as \(\frac{7}{1}\), it is a rational number. The square root of a perfect square is an integer (hence rational). Option B is incorrect because irrational numbers have non-terminating, non-repeating decimals (e.g. \(\sqrt{2}\)); here the result is a whole integer. Option C is wrong because non-real complex numbers have a nonzero imaginary part; 7 has zero imaginary part. Option D is wrong because the value is well defined. Exam tip: first check if the radicand is a perfect square — then the square root will be an integer/rational.

A square root of a positive integer is rational only when the integer is a perfect square. The nearby perfect squares are 9 and 16, and 11 is neither of them. Therefore √11 cannot be written as a ratio of integers and is irrational. It is nevertheless a real number, because it has a point on the real number line and a positive decimal value approximately 3.316. It is not an integer or a natural number, since squaring any integer does not give 11; it is certainly not zero because 0² = 0, not 11. Thus option A is the only correct statement. The key test is whether the radicand is a perfect square, not whether the root symbol appears in the expression.

A rational number can be written as a ratio of two integers, \(\frac{p}{q}\), with p and q integers and \(q\neq0\). Rational numbers have decimal expansions that are either terminating or repeating. Option B describes irrational numbers (non-terminating, non-repeating decimals). Option C is wrong because rationals include positive, negative and zero. Option D contradicts the definition. Exam tip: check the decimal expansion — terminating or repeating indicates a rational number.

An irrational number cannot be expressed as a ratio of two integers, so its decimal expansion neither terminates nor becomes periodic. Examples: √2 = 1.41421356… and π = 3.14159265… are non-terminating and non-repeating. Option C (non-terminating and repeating) describes typical rational decimals such as 1/3 = 0.333…, so C is incorrect. Exam tip: Check the decimal expansion — if it terminates or eventually repeats the number is rational; if it is non-terminating and non-repeating, it is irrational.

The decimal expansion of \(\pi\) is non‑terminating and non‑repeating, so it cannot be expressed as a ratio of two integers — this is the definition of an irrational number. Moreover, \(\pi\) is known to be transcendental (not a root of any nonzero polynomial with integer coefficients). Option D (terminating decimal) is incorrect because terminating decimals have a finite number of decimal places (e.g. 0.5, 2.75), whereas \(\pi\) does not. Options B and C are wrong because integers and whole numbers are finite, discrete values and do not describe \(\pi\). Exam tip: check the decimal expansion properties — non‑terminating and non‑repeating implies irrational.

0 can be written as \\(\frac{0}{1}\\), so it is rational (a ratio of two integers). All rational numbers are also real, therefore 0 is real. Option B is incorrect because irrational numbers cannot be expressed as a ratio of integers (they have non-terminating, non-repeating decimals). Options C and D are false—0 is not non-real and not only negative. Exam tip: To check if a number is rational, try expressing it as \\(\frac{p}{q}\\) with integers p,q (q≠0) or check whether its decimal expansion terminates/repeats.

Adding irrational (\sqrt{3}) to rational (2) gives an irrational number. Do not treat such sums as rational.

Add like surd terms by summing their coefficients: \(\sqrt{2}+\sqrt{2}=1\cdot\sqrt{2}+1\cdot\sqrt{2}=(1+1)\sqrt{2}=2\sqrt{2}\). Option B (\(\sqrt{4}\)) equals 2, which is not equal to \(2\sqrt{2}\); option C is just one copy of \(\sqrt{2}\), and option D is four times larger. Exam tip: when adding surds with the same radicand, add the numerical coefficients and keep the common root unchanged.

\(\sqrt{18}=\sqrt{9\times2}=\sqrt{9}\times\sqrt{2}=3\sqrt{2}\). Hence the simplified form is \(3\sqrt{2}\). Option B (\(2\sqrt{3}\)) is incorrect because it equals \(\sqrt{12}\), not \(\sqrt{18}\). Exam tip: when simplifying square roots, factor out the largest perfect square and take its root outside the radical.

In Class 10 real numbers the square root of a negative number is not real. Note that (\sqrt{7}) is real irrational.

1.25 is a terminating decimal and can be expressed as a fraction: \(1.25=\frac{5}{4}\). Rational numbers are numbers that can be written as a ratio of two integers, so 1.25 is rational. The closest distractor is 'irrational number' — irrational decimals are non‑terminating and non‑repeating (e.g. \(\sqrt{2}\)), which 1.25 is not. Exam tip: convert a terminating decimal to a fraction by multiplying to remove the decimal places or by using place value (e.g. 1.25 = 125/100 = 5/4).

2.454545... has the block “45” repeating, so it is a repeating (periodic) decimal and therefore rational. In fact 0.454545... = 45/99 = 5/11, so 2.454545... = 2 + 5/11 = \(\frac{27}{11}\). Option B (1.010010001...) has increasing block lengths and is non‑repeating; option C (3.14159265...) represents the non‑repeating decimal expansion of π; option D (0.1234567891011...) is the concatenation of natural numbers — both C and D are non‑repeating (irrational). Exam tip: terminating or repeating decimals are rational — convert a repeating part to a fraction to verify.

Reena ignores the difference between non-terminating and recurring decimals. \(0.333...=\frac13\), so it is rational. Only non-terminating, non-recurring decimals are irrational. In exams, check whether digits repeat.

The sum of a rational number and an irrational number is always irrational. To see why, suppose a + b were rational. Since a is rational, subtracting a from the rational number a + b would make b = (a + b) - a rational, contradicting the given fact that b is irrational. Therefore a + b must be irrational for every rational value of a, including 0, 1, negative rationals, and fractions. Option A is correct. Option B and option C mention special choices that are sufficient but unnecessarily restrictive; the result does not depend on a being one particular rational number. Option D is the exact opposite of the closure argument. This property concerns addition, not multiplication or division, where different conditions may be needed.

\(\sqrt{12}=\sqrt{4\times3}=\sqrt{4}\times\sqrt{3}=2\sqrt{3}\). So the simplified form is \(2\sqrt{3}\). Option B (\(3\sqrt{2}\)) is incorrect because it equals \(\sqrt{18}\), not \(\sqrt{12}\). Exam tip: always factor the radicand and pull out the largest perfect square (here 4) to simplify quickly.

Real numbers have no imaginary part. In option A, -3 (integer), 0 and \(\sqrt{2}\) (irrational) are all real. Option B contains \(\sqrt{-1}=i\), which is imaginary; option C has \(\sqrt{-9}=3i\), imaginary; option D contains \(\sqrt{3}\) (real) but also \(\sqrt{-16}=4i\), so it is not all real. Thus A is the only choice with only real numbers. Exam tip: if you see \(\sqrt{\text{negative}}\) immediately mark it as non-real (imaginary).

(\sqrt{7}) is irrational and its negative is also irrational. The negative sign does not stop it from being real.

The number √5 is irrational because 5 is not a perfect square. Multiplying an irrational number by a non-zero rational number remains irrational. Here 3 is a non-zero rational number. For a short proof, suppose 3√5 were rational; dividing it by 3, a non-zero rational, would imply √5 is rational, which is impossible. Hence 3√5 is irrational, so option A is correct. It is not rational or an integer, and it is not zero because both 3 and √5 are positive. The common error is to think that multiplying by an integer automatically removes irrationality; that happens only in special expressions involving cancellation, not when a non-zero rational multiplies √5. The classification depends on the irrational factor and the non-zero multiplier.

Dividing (\sqrt{2}) by non zero rational (2) keeps it irrational. Check the nature of the root in such expressions.

(\sqrt{2}+(-\sqrt{2})=0) is rational but (\sqrt{2}+\sqrt{3}) is irrational. So there is no fixed always rule.