Proof of irrationality of square root 2 and square root 3

Correct answer: B. 4/5

The governing rule is that for positive numbers, √(a/b) = √a/√b, together with the convention that the radical denotes the principal, non-negative square root. Therefore √(16/25) = √16/√25 = 4/5. This is confirmed by squaring: (4/5)² = 16/25, and 4/5 is non-negative, so it is the principal square root. Thus option B is correct. Option A, 8/5, has square 64/25 and is too large. Option C is the original number inside the radical, not its square root. Option D, 5/4, is the reciprocal of the correct value. Since both 16 and 25 are perfect squares, the result is rational, even though the question belongs to work with square roots.

Correct answer: B. √3

The governing concept is extraction of perfect-square factors from a radical, followed by ordinary division. Rewrite 27 as 9 × 3. Then √27 = √(9 × 3) = √9 × √3 = 3√3. Substituting this into the expression gives √27 ÷ 3 = (3√3) ÷ 3 = √3. Therefore option B is correct. Option A confuses the square root of 27 with a whole-number value, while option C is unrelated to the given division and is the square of 3. Option D is the reciprocal of √3, not √3 itself. Since 3 is not a perfect square, √3 cannot be simplified to an integer or terminating rational value and remains irrational in the final answer.

Correct answer: A. (\sqrt{2}) is rational

In contradiction method we first assume (\sqrt{2}) is rational. Then a contradiction is obtained from this assumption.

Correct answer: B. They are coprime

A rational number is written in lowest form as (\frac{p}{q}). So (p) and (q) are assumed coprime.

Correct answer: A. \(\sqrt{3}\)

Assume \(\sqrt{3}=p/q\) in lowest terms. Since \(3\mid p^2\), we get \(3\mid p\), and then \(3\mid q\), which is a contradiction. Exam tip: the square root of a non-perfect-square integer is irrational.

Correct answer: D. (p^2) is even

The direct answer is Option D: p² is even. From p² = 2q², the number p² is equal to 2 multiplied by the integer q². Any integer that can be written as 2 times another integer is even. Step by step: q is an integer, so q² is an integer; multiplying q² by 2 gives 2q²; therefore p² is divisible by 2; hence p² is even. Option D is correct. Option A is wrong because a number divisible by 2 cannot be odd. Option B is wrong because p² need not be prime; for example, it may have several factors. Option C is wrong because the equation contains squares of integers and does not make p² negative; a square is non-negative. In the irrationality proof, this evenness is important because it later implies that p itself is even. Exam cue: “factor 2” means “even.”

Correct answer: A. (p) is even

If the square of a number is even then the number is also even. This fact is used in the proof for (\sqrt{2}).

Correct answer: C. 1.732 is only an approximation; the decimal expansion of \(\sqrt{3}\) is non-terminating and non-repeating.

A calculator rounds values to limited digits. Since \(1.732^2=2.999824\ne3\), 1.732 is not exact. It is only an approximation. Exam tip: use exact forms, not calculator displays, to judge rationality.

Correct answer: A. \(q^2=2r^2\)

Given \(p=2r\), substitute it into \(p^2=2q^2\): \((2r)^2=2q^2\), so \(4r^2=2q^2\). Dividing both sides by 2 gives \(q^2=2r^2\). The result \(q^2=4r^2\) would come from incorrect division. Exam tip: after substitution, remember that \((2r)^2=4r^2\).

Correct answer: C. (q) is even

In (q^2=2r^2), the right-hand side is a multiple of 2, so (q^2) is even. The square of an integer is even only when the integer itself is even; hence, (q) is even. Being prime, negative, or zero does not necessarily follow from this equation. Exam tip: Remember: an even square implies an even integer.

Correct answer: A. Both p and q become even

The proof uses contradiction. Assume that √2 is rational and write √2 = p/q in lowest terms, where p and q are integers, q is non-zero, and p and q have no common factor. Squaring gives p² = 2q², so p² is even and p must be even. Let p = 2k. Substitution gives 4k² = 2q², or q² = 2k², which shows that q is also even. Thus both p and q are divisible by 2, contradicting the assumption that p/q was in lowest terms. Therefore option A states the final contradiction. The other choices do not follow from the parity argument and do not contradict the original lowest-form assumption.

Correct answer: B. Contradiction method

The proof uses the method of contradiction, also called proof by contradiction or reductio ad absurdum. First, the opposite of the desired statement is assumed: √2 is taken to be rational and written as p/q in lowest form. Algebraic manipulation then shows that p and q must both be even. This is impossible because a fraction in lowest form cannot have a common factor greater than 1. The impossible conclusion contradicts the original assumption, so the assumption that √2 is rational must be false. Therefore √2 is irrational, and option B correctly names the method. Measurement, guessing, and drawing are not the logical proof procedures used here.

Correct answer: A. If a prime number divides the square of an integer, it also divides that integer.

From \(p^2=3q^2\), we get \(3\mid p^2\). Since 3 is prime, it divides \(p\); then it also divides \(q\), contradicting coprimality. Exam tip: remember the prime-divisor property.

Correct answer: A. Both p and q are even

From \(2q^2=p^2\), \(p^2\) is even, so \(p\) is even. Put \(p=2k\); then \(q\) also becomes even. Thus both share factor 2, contradicting coprimality. Exam tip: identify the common factor as the contradiction.

Correct answer: A. \(a^2\) is divisible by 3

Given \(a^2=3b^2\). Since \(3b^2\) is a multiple of 3, the equal quantity \(a^2\) must also be divisible by 3. The equation does not imply that \(a^2\) must be divisible by 2. Exam tip: if a number can be written as \(3\times\) an integer, it is divisible by 3.

Correct answer: B. (a) is divisible by (3)

If the square of a number is divisible by (3), the number is also divisible by (3). This uses prime factor property.

Correct answer: A. A rational number has a terminating or repeating decimal expansion; \(\sqrt{2}\) is non-terminating and non-repeating.

1.414 is only an approximation of \(\sqrt{2}\), not proof of rationality. Its expansion \(1.414213...\) is non-terminating and non-repeating, so it is irrational. Exam tip: do not confuse an approximation with the actual decimal expansion.

Correct answer: A. \(b^2=3k^2\)

Substitute \(a=3k\) into \(a^2=3b^2\): \((3k)^2=3b^2\), so \(9k^2=3b^2\). Dividing both sides by 3 gives \(b^2=3k^2\). The option \(b^2=9k^2\) is incorrect because after division, the left side becomes \(3k^2\), not \(9k^2\). Exam tip: square the substituted value first, then simplify by dividing out common factors.

Correct answer: A. b is divisible by 3

The equation \(b^2=3k^2\) shows that \(b^2\) is divisible by 3. Since 3 is prime, if it divides the square of an integer, it must also divide the integer itself. Therefore, \(b\) is divisible by 3. Divisibility of \(b\) by 2 does not follow from this equation. Exam tip: Remember that for a prime \(p\), \(p\mid n^2\Rightarrow p\mid n\).

Correct answer: A. \(m\) and \(n\) are coprime

Taking \(\frac{m}{n}\) in lowest terms makes \(m\) and \(n\) coprime. From \(m^2=2n^2\), \(m\) is even and then \(n\) is even, giving a contradiction. Exam tip: begin with a reduced fraction.

Correct answer: A. They are first assumed rational

Both standard proofs use contradiction. To establish that √2 or √3 is irrational, the proof begins by assuming the opposite: the number is rational. A rational number can be represented as a/b, where a and b are integers, b is non-zero, and the fraction is in lowest form. For √2, this leads to a² = 2b²; for √3, it leads to a² = 3b². In each case, divisibility arguments eventually force both a and b to share a factor, contradicting their lowest-form condition. Hence option A is correct. The numbers are not initially assumed to be integers, zero, or negative; those descriptions do not express the required contrary assumption.

Correct answer: A. If \(4\sqrt{3}\) were rational, dividing it by 4 would make \(\sqrt{3}\) rational, which is impossible.

Assume \(4\sqrt{3}\) is rational. Since 4 is a non-zero rational number, \((4\sqrt{3})/4=\sqrt{3}\) would also be rational, contradicting the fact that \(\sqrt{3}\) is irrational. Exam tip: division by a non-zero rational preserves rationality.

Correct answer: A. Because the fraction is written in lowest form

If √3 were rational, it could be expressed as a/b, with a and b integers, b ≠ 0. Any rational fraction can be reduced by cancelling common factors, so we may choose a representation in lowest form. In that form, a and b are coprime: their greatest common divisor is 1. This condition is essential because the proof later shows that 3 divides a and then, from a² = 3b², also 3 divides b. That would give a and b a common factor 3, contradicting their assumed lowest form. Thus option A is correct. The other choices make unsupported claims and are not properties of every rational representation.

Correct answer: B. Even

The square of an even number is also even. In the proof of (\sqrt{2}), this idea is also used in reverse direction.

Correct answer: A. It will be divisible by (3)

If a number has factor (3), its square also has factor (3). This is useful in the proof of (\sqrt{3}).