Proof of irrationality of square root 2 and square root 3

A calculator shows only a rounded approximation. Since \(1^2<3<2^2\), \(\sqrt{3}\) lies between 1 and 2, but its decimal expansion is non-terminating and non-repeating, so it is irrational. Exam tip: never treat a finite display as the exact decimal expansion.

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.

The direct answer is C, both a and b are even. In the contradiction proof, assume
\(\sqrt{2}=a/b\), where a and b are integers, b is not zero, and the fraction is in lowest form, so a and b are coprime. Squaring gives \(2b^2=a^2\). Therefore \(a^2\) is even, and an integer whose square is even must itself be even, so write \(a=2k\). Substitution gives \(b^2=2k^2\), so b is also even. Thus both a and b have the common factor 2, contradicting their being coprime. Option A is only an intermediate result. Option B is another intermediate result, not the complete contradiction. Option C states the complete contradiction and is correct. Option D, \(\sqrt{2}>0\), is true but does not conflict with the lowest-form assumption. Memory cue: the contradiction is not merely “one number is even”; it is “both numerator and denominator are even.”

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

The governing concept is proof by contradiction together with the meaning of lowest terms. Assume √2 = a/b, where a and b are integers, b is nonzero, and gcd(a,b)=1. Squaring gives a² = 2b². Thus a² is even, so a is even; write a=2k. Substitution gives 4k²=2b², hence b²=2k², so b is also even. Therefore 2 is a common factor of a and b, which contradicts gcd(a,b)=1. The fraction cannot therefore have been in lowest form, and the original rational assumption fails. Option A states this exact contradiction. The result does not imply that the denominator is zero, that √2 is an integer, or 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.

The direct answer is A: if 3 divides p squared, then 3 divides p. The reason is the prime-factor property. A prime number appearing in the factorisation of a square must already appear in the factorisation of its base. For example, if p had no factor 3, then p squared would also have no factor 3. Option A is correct and is the key step in proving the irrationality of the square root of 3. Option B is wrong because divisibility by 3 does not imply divisibility by 2. Option C is wrong because 3 dividing p does not mean p equals zero; p could be 3, 6, 9, or another multiple of 3. Option D is wrong because the divisibility statement says nothing about p and q being equal. In the full proof, 3 divides p, then substitution shows 3 divides q, contradicting lowest form. Memory cue: prime divides a square, so it divides the original number.

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.

First (p) is proved divisible by (3) from (p^2). Then putting (p=3k) gives the conclusion for (q).

Every rational number can be written as a/b with integer a and non-zero integer b, but the irrationality proof needs a special, reduced representation. We choose a/b in lowest terms, expressed by gcd(a,b)=1. Assuming √2=a/b then gives a²=2b². The parity argument forces a and b both to be even, which contradicts the reduced-form condition because they would share the factor 2. Thus option A states the proof-ready form precisely. A zero denominator is not allowed in a fraction, and requiring both entries to be even would destroy the contradiction rather than create it. Equal numerator and denominator is also irrelevant and would describe only a restricted fraction, not every rational number.

If \(\sqrt{n}=p/q\), then \(nq^2=p^2\). Since every prime exponent in a square is even, \(n\) must be a perfect square. Thus \(\sqrt{2}\) is irrational. Exam tip: first check whether the integer is a perfect square.

Both proofs have the same structure, but the prime factors are (2) and (3) respectively. Mention this in comparison.