Proof of irrationality of square root 2 and square root 3

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

Both proofs start with contradiction method. A coprime fraction is assumed rational and leads to contradiction.

In the proof of (\sqrt{2}), (2) is central and in the proof of (\sqrt{3}), (3) is central. This is the key difference.

A rational square does not guarantee that the original number is rational. For example, \((\sqrt{3})^2=3\), yet \(\sqrt{3}\) is irrational. Option B makes this incorrect inference. Exam tip: check whether the number under a square root is a perfect square.

Using contradiction, assume \(\sqrt{2}=p/q\) with \(p\) and \(q\) coprime. From \(p^2=2q^2\), \(p\) is even, and then \(q\) is also even, contradicting coprimality. Exam tip: identify the conclusion that both numerator and denominator share 2.

Assume that \(\sqrt{2}=\frac{m}{n}\), where \(m\) and \(n\) are integers and \(n\ne0\). Squaring both sides gives \(2=\frac{m^2}{n^2}\). Multiplying by \(n^2\) gives \(m^2=2n^2\), so option A is correct. Option B, with \(3n^2\), is the corresponding relation for \(\sqrt{3}\), not for \(\sqrt{2}\). Exam tip: for a square-root fraction relation, square both sides first and then clear the denominator.

A non-terminating decimal is not automatically irrational. For example, \(1/3=0.333\ldots\) is non-terminating but recurring, so it is rational. The decimal expansion of \(\sqrt{3}\) is non-terminating and non-recurring; check both features in exams.

A proof by contradiction begins by assuming the opposite of the statement to be proved. Here the assumption is that √2 is rational. If valid reasoning from that assumption produces an impossibility, the assumption must be false. Therefore √2 is not rational; it is irrational, so option A is correct. The contradiction does not imply that √2 is zero, an integer, or a natural number. In fact, every integer and every natural number is rational, so options B and D conflict with the conclusion. Option C is also numerically false because the square of zero is 0, not 2. The logical structure is assumption, contradiction, rejection of assumption, and conclusion.

For a contradiction proof, suppose √3 = m/n in lowest terms. Squaring gives m² = 3n², so the prime-factor rule implies that 3 divides m. Substituting m = 3k then shows that 3 also divides n, contradicting that m and n are coprime. Hence the rational assumption is impossible and √3 is irrational, so option B is correct.

From \(m^2=3n^2\), \(m^2\) is divisible by 3, so \(m\) is divisible by 3. Substituting this shows that \(n\) is also divisible by 3, contradicting coprimality. Exam tip: identify the common factor that causes the contradiction.

Lowest terms means that \(p\) and \(q\) have no common factor. From \(p^2=2q^2\), \(p\) is even, and then \(q\) is also even; this contradicts coprimality. Exam tip: state this contradiction explicitly.

To prove that \(\sqrt{2}\) is irrational by contradiction, we first assume it is rational and write \(\sqrt{2}=\frac{m}{n}\), where \(m,n\) are coprime and \(n\ne0\). The expression \(\frac{m}{0}\) is undefined, so option C is an invalid start. In option A, the denominator is non-zero, so it is a valid assumption. Exam tip: whenever a rational number is written as \(\frac{p}{q}\), check that \(q\ne0\).

From \(2q^2=p^2\), \(p^2\) is even, so \(p\) is even. Put \(p=2k\); then \(q^2=2k^2\), making \(q\) even too. This contradicts coprimality. Exam tip: identify the common factor 2.

\(\sqrt{3}\) is irrational because 3 is not a perfect square, so it cannot be expressed as \(p/q\). In contrast, \(\sqrt{36}=6\) is rational. Exam tip: the square root of a perfect square is an integer.

An infinite decimal alone does not prove irrationality: \(1/3=0.333\ldots\) is rational. An irrational number has a non-terminating, non-recurring decimal. Exam tip: distinguish recurring decimals from non-recurring ones.

If both \(m\) and \(n\) are even, then we can write \(m=2p\) and \(n=2q\) for some integers \(p,q\). Thus, \(2\) is a common factor of both \(m\) and \(n\), so their highest common factor is at least \(2\). In the irrationality proof of \(\sqrt{2}\), this contradicts the assumption that \(m\) and \(n\) are coprime. Exam tip: if two integers are both even, their HCF cannot be \(1\).

In a perfect square, every prime factor has an even exponent. An odd exponent means the number is not a perfect square, so its square root is irrational. Exam tip: check whether all prime exponents are even.

The relevant parity rule is that the square of an odd integer is odd. If m is assumed odd, then m² must be odd. However, the equation m² = 2n² has an even right-hand side because it is two times an integer square. It therefore forces m² to be even. An integer cannot be both odd and even, so the assumption that m is odd creates the contradiction. This parity observation supports the usual proof: m² is even, hence m is even, and later the equation shows that n is even too. Option A accurately states the problem. The other options are unrelated conclusions and do not follow from the equation.

If (p) has no factor (3), then (p^2) also has none. But (p^2=3q^2) shows it divisible by (3).

The proof begins by assuming, for contradiction, that √2 can be written as a rational number p/q in lowest form, where p and q are integers, q ≠ 0, and gcd(p,q) = 1. Rearranging and comparing prime factors shows that both p and q must be divisible by 2. That contradicts the choice that the fraction was already in lowest form. In a proof by contradiction, the contradiction rejects the original assumption, not the valid algebraic steps or the fact that √2 is real and positive. Hence the correct conclusion is that √2 is irrational. Option A states the opposite, while C and D do not express the logical conclusion.