Proof of irrationality of square root 2 and square root 3

The governing concept is extracting perfect-square factors from radicals and then combining like surds. Factor 242 as 121 × 2, so √242 = √121 × √2 = 11√2. Factor 128 as 64 × 2, so √128 = √64 × √2 = 8√2. Both simplified terms contain the same radical √2, so they are like surds and their coefficients can be subtracted: 11√2 − 8√2 = (11 − 8)√2 = 3√2. Therefore option C is correct. Option A would come from adding the coefficients instead of subtracting them. Option B reflects an incorrect factorisation or subtraction. Option D incorrectly applies √a − √b = √(a−b), an identity that is not generally valid. Each radical must first be simplified separately.

The governing concept is rationalisation of a denominator containing a square root. Because x = 3 − √5, its reciprocal is 1/(3 − √5). Multiply numerator and denominator by the conjugate 3 + √5. The denominator becomes (3 − √5)(3 + √5) = 3² − (√5)² = 9 − 5 = 4, while the numerator becomes 3 + √5. Hence 1/x = (3 + √5)/4, making option A correct. Option B omits the denominator 4. Option C retains the original sign and does not use the conjugate correctly. Option D reverses the required division and multiplies by 4 instead. Substitution also confirms that (3 − √5)(3 + √5)/4 = 1.

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.

Since (p^2) has factor (2), (p^2) is even. Recognising evenness is a key step in this proof.

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.

To prove that √2 is irrational, suppose the contrary: √2 = p/q, where p and q are integers with no common factor and q is non-zero. Squaring gives p² = 2q², so p² is even and therefore p is even. Write p = 2k. Substitution gives q² = 2k², so q is also even. Thus p and q have 2 as a common factor, which contradicts the initial statement that p/q was in lowest form. Hence the final contradiction is that both p and q are even, making option A correct. The other choices do not follow from the parity argument.

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.