Proof of irrationality of square root 2 and square root 3

If both \(a\) and \(b\) are even, each is divisible by 2, so they cannot be coprime. This contradicts the original assumption; hence \(\sqrt{2}\) is irrational. Exam tip: always use the lowest-terms condition to identify the contradiction.

If 3 divides both \(a\) and \(b\), then 3 is their common factor. This contradicts the initial condition that \(a/b\) is in lowest terms and \(a,b\) are coprime. Exam tip: link the contradiction to coprimality.

Since 3 is prime, an integer leaves remainder 0, 1, or 2 on division by 3. The squares of remainders 1 and 2 both leave remainder 1. Thus, for n² to leave remainder 0, n must leave remainder 0 and be divisible by 3. Exam tip: this result is central to the proof that √3 is irrational.

In lowest form, (a) and (b) were assumed coprime. If both are even, common factor (2) appears.

Only (p) being divisible by (3) is not the final contradiction. Later (q) must also be proved divisible by (3).

The proof begins by assuming √2=a/b, with a and b integers, b≠0, and gcd(a,b)=1. Squaring gives a²=2b². Since a² is even, a is even; write a=2k. Substitution gives 4k²=2b², so b²=2k², which makes b even as well. Thus both a and b have the common factor 2, contradicting the assumption that the fraction was in lowest form. Option A states this exact conclusion. The result does not make the denominator zero, does not make √2 an integer, and gives no reason that a and b are equal.

For a square, \(s^2=A\). When \(A=2\), the side is \(s=\sqrt{2}\), which is irrational. Hence, a rational area need not give a rational side. In exams, first take the square root of the area.

From \\(p^2=3q^2\\), \\(3\\mid p^2\\); since 3 is prime, \\(3\\mid p\\). Put \\(p=3k\\) to get \\(q^2=3k^2\\), so \\(3\\mid q\\) too. This contradicts coprimality. Exam tip: if a prime divides a square, it divides the number itself.

From \(3q^2=p^2\), \(p\) must be divisible by 3. Put \(p=3k\); then \(q^2=3k^2\), so \(q\) is also divisible by 3. This contradicts coprimality. Exam tip: check divisibility of both terms.

From \(a^2=3b^2\), 3 divides \(a^2\). Since 3 is prime, it must divide \(a\). On putting \(a=3k\), one also obtains that 3 divides \(b\), contradicting coprimality. Exam tip: if a prime \(p\mid n^2\), then \(p\mid n\).

Prime factor (3) appears in (p^2) only if it appears in (p). The equation forces this divisibility.

Option A contains every essential step of the proof by contradiction. Assume \(\sqrt{2}=\frac{a}{b}\) in lowest terms. Then \(a^2=2b^2\), so \(a^2\), and hence \(a\), is even. Writing \(a=2k\) gives \(b^2=2k^2\), so \(b\) is also even. Thus both \(a\) and \(b\) are divisible by 2, contradicting that they are coprime. Option D is wrong because an infinite decimal alone does not prove irrationality; rational numbers may have infinite recurring decimals. Exam tip: always begin this proof by taking the fraction in lowest terms.

In lowest terms, \(p\) and \(q\) cannot have a common factor. From \(3\mid p^2\), we get \(3\mid p\); substitution then gives \(3\mid q\). Thus 3 is a common factor. Exam tip: use the prime divisibility rule for squares.

A rational number can be represented by many equivalent fractions, so merely writing √2=a/b is not enough for a contradiction. We choose the representation in lowest terms, meaning gcd(a,b)=1. From a²=2b², parity arguments show that a is even and then b is even. This conclusion means that 2 divides both numbers, so their greatest common divisor is at least 2. That directly conflicts with gcd(a,b)=1, producing the genuine contradiction required by the proof. Hence option A is correct. Without the coprime assumption, a pair such as 2/4 could be even in both entries without contradiction because it was not reduced. The other options do not follow from the equations.

Although \((\sqrt{2})^2=2\) is rational, \(\sqrt{2}\) is irrational. Thus, a rational square does not prove that the original number is rational. Exam tip: always test the converse of a statement separately.

From a=2r, a is a multiple of 2, so a is even. From b^2=2r^2, b^2 is even. If the square of an integer is even, then the integer itself must be even; hence b is even. Therefore, both a and b are even. They need not both be zero, since non-zero even integers are also possible. Exam tip: In the irrationality proof, this result contradicts the assumption that a and b are coprime.

From p = 3k, p is divisible by 3. Also, q² = 3k² shows that q² is divisible by 3. Since 3 is prime, if it divides the square of an integer, it must divide that integer itself. Hence q is also divisible by 3. Therefore, both p and q are divisible by 3. Option A has no basis because divisibility by 2 is not implied. Exam tip: In contradiction proofs of irrationality, showing that two assumed coprime integers share a prime factor produces the contradiction.

The relevant parity theorem is: an odd integer has an odd square. Its contrapositive is that if an integer’s square is even, then the integer itself must be even. In the proof, a²=2b² is even, so this rule establishes that a is even. After substituting a=2k, the resulting equation similarly shows that b² is even, and hence b is even. Therefore option A gives the correct logical direction. Option B reverses the parity result and is false because an even integer has an even square. An even square need not be zero, and its parity says nothing about a denominator being zero, so options C and D are also invalid.

Since (3) is prime, factor (3) in the square comes from the original number. This is the key rule.

From \(a^2=3b^2\), \(a^2\), and hence \(a\), is divisible by 3. Putting \(a=3k\) gives \(b^2=3k^2\), so \(b\) is also divisible by 3. This contradicts coprimality. Exam tip: state the common factor explicitly.