Proof of irrationality of square root 2 and square root 3

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.

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.

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

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

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.

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.”

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

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.

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 \((2r)^2=4r^2\).

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.

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.

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.

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.

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.

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.

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

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.

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.

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 \(p\), \(p\mid n^2\Rightarrow p\mid n\).

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.

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.

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.

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.

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

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