Proof of irrationality of square root 2 and square root 3

Not being a perfect square only shows that \(\sqrt{3}\) is not an integer. In \(m^2=3n^2\), the power of 3 is even on the left but odd on the right, giving a contradiction. Exam tip: always assume lowest terms.

The direct answer is option A. In a proof by contradiction, write \(\sqrt{2}=m/n\) in lowest terms. This requires \(m,n\) to be integers, \(n\neq0\), and \(m,n\) to be coprime. Squaring gives \(m^2=2n^2\). Thus \(m^2\) is even, so \(m\) is even; writing \(m=2k\) then shows \(n\) is even too. That contradicts the coprime condition. Option A is correct because both the nonzero denominator and lowest-form condition are necessary. Option B is wrong because assuming both even at the start destroys the lowest-form setup and already assumes the contradiction. Option C is impossible because division by zero is undefined. Option D is not required: numerator and denominator need not be equal, and equality would not prove irrationality. The lowest-form condition is what makes “both even” a contradiction. Memory cue: rational fraction means coprime numerator and denominator, with denominator nonzero.

If \(m=2k\), then \(4k^2=2n^2\), so \(n^2=2k^2\) and \(n\) is also even. Thus both numbers have a common factor 2, contradicting coprimality. Exam tip: state this contradiction explicitly.

Since \(p^2=3q^2\), \(p^2\) is divisible by 3, so \(p\) is divisible by 3. Put \(p=3k\); then \(q^2=3k^2\), making \(q\) divisible by 3 too. This contradicts coprimality. Exam tip: if a prime divides a square, it divides its root.

The governing fact is Euclid’s lemma for a prime: if a prime divides the square of an integer, it divides that integer itself. In the √3 proof, the equation p² = 3q² shows that 3 divides p². Because 3 is prime, it follows that 3 divides p, so p can be written as p = 3k. This is the required step and makes option B correct. The conclusion does not say that p is divisible by 2, equal to zero, or equal to q. After this step, substitution is used to show that q is also divisible by 3, producing the contradiction with lowest terms.

The final conclusion comes from contradiction with the original lowest-terms assumption. Suppose √2 = m/n, where m and n are coprime integers. Squaring gives m² = 2n², so m is even; writing m = 2r then leads to n² = 2r², making n even as well. Thus both numerator and denominator have the common factor 2, which contradicts their being coprime. The rational assumption must therefore be false, and √2 is irrational. Option A states the decisive basis. A terminating decimal is not the conclusion of this proof, m = n is never established, and n = 0 is excluded because a fraction cannot have a zero denominator.

The direct answer is option B. Assume \(\sqrt{3}=p/q\) in lowest terms, with \(q\neq0\) and \(p,q\) coprime. Squaring gives \(p^2=3q^2\). Therefore 3 divides \(p^2\), and because 3 is prime, 3 divides \(p\). Put \(p=3k\). Then \(9k^2=3q^2\), so \(q^2=3k^2\), which means 3 divides \(q\) as well. Thus both \(p\) and \(q\) are divisible by 3, contradicting the lowest-form assumption. Hence \(\sqrt{3}\) is not rational and is irrational. Option B states this exact basis. Option A is wrong because a terminating decimal would indicate rationality, not irrationality. Option C is wrong because the proof does not show \(p=q\). Option D is wrong because \(q=0\) is forbidden from the beginning. Memory cue: final contradiction means a common prime factor violates lowest form.

From (m^2=2n^2), first (m^2) and then (m) are even. Only after taking (m=2r), (n) is proved even.

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.

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.