Proof of irrationality of square root 2 and square root 3

The equation \(q^2=2m^2\) makes \(q^2\) even, so \(q\) must be even. Since \(p\) was already even, both share factor 2, contradicting coprimality. Exam tip: if a square is even, its integer root is even.

From \(p^2=3q^2\), \(p^2\) is divisible by 3, so \(p\) is divisible by 3. Put \(p=3k\); then \(q\) is also divisible by 3, contradicting lowest terms. Exam tip: use the prime-divisibility property of squares.

Since
\(p^2=3q^2\),
\(p^2\) is divisible by 3, so
\(p\) is divisible by 3. Put
\(p=3k\); then
\(q^2=3k^2\), so
\(q\) is also divisible by 3. This contradicts lowest terms. Exam tip: if a prime divides a square, it divides the number itself.

From h² = 3k², h² is divisible by 3. Since 3 is prime, if it divides the square of an integer, it must also divide that integer. Hence 3 divides h, contradicting the given condition that h is not divisible by 3. Option A is not the required contradiction; the contradiction at this step concerns h directly. Exam tip: For a prime p, use p | n² ⇒ p | n.

\(\sqrt{3}\) is irrational because 3 is not a perfect square: its prime factor 3 has exponent 1, which is odd. In contrast, \(\sqrt{9}=3\) is rational. Exam tip: a perfect square has only even prime exponents.

From \(p^2=2q^2\), \(p^2\), and hence \(p\), is even. Put \(p=2r\); then \(q^2=2r^2\), so \(q\) is also even. This contradicts coprimality. Exam tip: an even square has an even root.

Since 3 is prime, \(p^2=3q^2\) implies that 3 divides \(p\). Put \(p=3k\); then \(q^2=3k^2\), so 3 divides \(q\) as well. This contradicts that \(p\) and \(q\) are coprime. Exam tip: if a prime divides a square, it divides the number itself.

Since 3 is prime, \(3\mid p^2=p\times p\) implies \(3\mid p\) by the prime-divisor property. It does not imply that \(p\) is divisible by 9. Exam tip: apply this rule only when the divisor is prime.

A fraction \(c/d\) represents a number only when its denominator is non-zero, so \(d\neq0\) is required to make the expression defined. This condition does not say that the fraction is reduced, and it does not imply any equality between \(c\) and \(d\). The condition \(\gcd(c,d)=1\) has a different purpose: it says that numerator and denominator have no common factor and that the fraction is in lowest terms.

Assuming \(\sqrt{2}=c/d\), squaring gives \(c^2=2d^2\). The parity argument shows that \(c\) is even and then that \(d\) is even. Thus both have the common factor 2, contradicting \(\gcd(c,d)=1\). The contradiction rests on the lowest-terms condition, while the non-zero condition only keeps the fraction meaningful. Hence option B is correct.

Direct answer: Option C. In a proof of irrationality, assume that sqrt(3)=h/k, where h and k are integers, k is not zero, and the fraction is in lowest terms, so gcd(h,k)=1. The condition k≠0 has a basic definition-related role: division by zero is not allowed, so h/k must be a genuine fraction. After squaring, h^2=3k^2. This shows that 3 divides h^2, and hence 3 divides h; write h=3r. Substitution then shows that 3 divides k as well. Thus h and k have a common factor 3, contradicting gcd(h,k)=1. Option A is wrong because h=3r comes from the divisibility equation, not merely from k≠0. Option B is wrong because the conditions have different jobs. Option C is correct: k≠0 keeps the fraction defined, while gcd(h,k)=1 creates the final contradiction. Option D reverses the logic and is false. Memory cue: denominator nonzero makes a fraction valid; lowest terms make a common-factor contradiction possible.

Assume \(\sqrt{3}=\frac{p}{q}\), where \(p\) and \(q\) are coprime integers. Then \(p^2=3q^2\), so \(3\mid p^2\). Since 3 is prime, \(3\mid p^2\) implies \(3\mid p\). On writing \(p=3k\), we also obtain \(3\mid q\), which contradicts the fact that \(p\) and \(q\) are coprime. Option B is incorrect because divisibility of \(n^2\) by 3 does not mean that \(n=3\); it means that \(n\) is divisible by 3. Exam tip: In irrationality proofs, look for the prime-divisibility property that forces both numerator and denominator to have a common factor.

From \(p^2=3q^2\), 3 divides \(p^2\), so the prime-factor property gives 3 divides \(p\). Put \(p=3k\); then 3 also divides \(q\), contradicting coprimality. Exam tip: transfer prime divisibility from a square back to its base.

The governing idea is the role of lowest terms in a contradiction proof. Assume √2=c/d with d nonzero. Squaring gives c²=2d² whether or not c/d has been reduced, and the positive value √2>0 is also unaffected. The argument then shows that c and d are both even. This becomes a contradiction only if the initial representation was in lowest terms, meaning gcd(c,d)=1. Without that condition, a numerator and denominator may legitimately share a factor; for example, 2/4 has both entries even and is still a valid representation of 1/2. Therefore the decisive conclusion is the contradiction arising when both are even, so A is correct. The other steps remain valid independently of reduction.

From \(3p^2=q^2\), \(q^2\) is divisible by 3, so \(q=3r\). Substituting gives \(3p^2=9r^2\), hence \(p^2=3r^2\); therefore, \(p\) is divisible by 3. Exam tip: if a prime divides a square, it divides the number itself.

From \(3q^2=p^2\), \(p^2\) is divisible by 3. Since 3 is prime, \(p\) must be divisible by 3. Putting \(p=3k\) then shows that \(q\) is also divisible by 3, contradicting coprimality. Exam tip: use the prime-factor rule for square terms.

A contradiction proof must begin with the strongest legitimate assumption, not with the contradiction that it intends to derive. We assume √3 = h/k, where h and k are integers, k ≠ 0, and gcd(h,k) = 1. Squaring gives h² = 3k²; divisibility arguments then show that 3 divides h and, after substitution, also divides k. This final result contradicts gcd(h,k) = 1. If both numbers were assumed divisible by 3 at the beginning, the contradiction would be presupposed and the proof would become circular. Hence option C is correct. The other options either impose false conditions or use an irrelevant decimal representation.

Assume \(\sqrt{3}=a/b\) with coprime integers \(a,b\). Then \(a^2=3b^2\), so \(3\mid a^2\); the key property gives \(3\mid a\). It later gives \(3\mid b\) too, contradicting coprimality. Exam tip: for prime \(p\), remember \(p\mid a^2\Rightarrow p\mid a\).

From \(h^2=2k^2\), \(h^2\), and hence \(h\), is even. Put \(h=2m\); then \(k^2=2m^2\), so \(k\) is even too. A common factor 2 contradicts coprimality. Exam tip: establish evenness of both integers.

Let \(\sqrt{3}=p/q\) be in lowest terms. From \(p^2=3q^2\), 3 divides \(p^2\), so it divides p. This leads to the contradiction; B gives only the reverse implication. Exam tip: state that p and q are coprime.

The direct answer is A. A fraction is in lowest form when its numerator and denominator have no common factor greater than 1. If \\(c/d\\) can be changed to \\((c/2)/(d/2)\\), then both c and d are divisible by 2. The two 2s cancel, so the same value is represented by a smaller numerator and denominator. In the irrationality proof, the fraction was assumed to be in lowest form at the start. Finding that both terms are even contradicts that assumption. Therefore the assumed fraction was not actually in lowest form. Option A is correct. Option B is wrong because this reduction does not say \\(\\sqrt{2}=2\\); it only concerns a common factor. Option C is wrong because d is a denominator and the expression would not be a valid fraction if d were zero. Option D is wrong because no equality between c and d follows. The useful cue is: if numerator and denominator can both be divided by the same number, the fraction was not simplest.

Since \(p^2=2q^2\), \(p^2\) is even, so \(p\) is even. Put \(p=2r\): \(4r^2=2q^2\), hence \(q^2=2r^2\), making \(q\) even too. This contradicts coprimality. Exam tip: if a square is even, its integer root is even.

To prove (d) even, (c=2u) must be substituted first. In exams, do not write denominator even directly.

The governing concept is the definition of divisibility. The statement 3|h means that there exists an integer r for which h=3r. Writing h=3r therefore converts an abstract divisibility statement into an explicit algebraic form; it is a definitional substitution, not an approximation or a conclusion about a denominator. In the proof, if h²=3k², substituting h=3r gives (3r)²=3k², so 9r²=3k². Dividing by 3 yields k²=3r², which then shows that 3 divides k² and hence, by the prime-square rule, 3 divides k. Thus the substitution is an intermediate step that enables the next divisibility argument. Option A is correct.

If both have common factor (2), the fraction reduces to (\frac{u}{v}). So the old form was not lowest.

The square-root spiral repeatedly forms a right triangle. At each stage, the old hypotenuse becomes one leg of the next right triangle, and a new perpendicular leg of length 1 is drawn. If the old hypotenuse is \(\sqrt{n}\), the new one is \(\sqrt{n+1}\), because of the Pythagorean theorem.

If \(h=3r\) and \(k=3s\), then both the numerator and denominator contain the common factor 3. A fraction in lowest form must have numerator and denominator with no common factor greater than 1. Thus the lowest-fraction condition is contradicted; it does not become stronger, make the root an integer, or prove \(k=0\). Therefore option C correctly describes the effect.