Proof of irrationality of square root 2 and square root 3

Correct answer: A. p² should not be divisible by 3 but the equation makes it divisible

Explanation: The correct answer is A. If p is not divisible by 3, then its prime factorisation contains no factor 3. Squaring p does not introduce a new prime factor, so p² is also not divisible by 3. However, the equation p² = 3q² writes the left side as three times an integer square, so the right-hand expression is divisible by 3; consequently p² must be divisible by 3. This is a direct contradiction. In the standard irrationality proof, one first assumes √3 = p/q in lowest terms, with q nonzero and p, q coprime. The contradiction forces both numbers to be divisible by 3, violating the lowest-form assumption. The other options do not follow from the equation.

Correct answer: A. Rational assumption → a² = 2b² → a even → b even → contradiction

Explanation: The correct answer is A because it follows the contradiction proof in the required logical order. Assume √2 is rational and write it as a/b in lowest terms, where a and b are coprime integers and b is nonzero. Squaring gives a²/b² = 2, hence a² = 2b². Therefore a² is even, so a is even; write a = 2k. Substitution gives 4k² = 2b², or b² = 2k², so b is also even. Thus both a and b have a common factor 2, contradicting the assumption that the fraction was in lowest terms. The other sequences omit the essential algebra or replace proof with an unsupported decimal guess.

Correct answer: A. Rational assumption → p² = 3q² → p divisible by 3 → q divisible by 3 → contradiction

Explanation: The correct answer is A. Begin by assuming that √3 is rational and express it as p/q in lowest terms, with p and q coprime and q nonzero. Squaring produces p² = 3q², so p² is divisible by 3. Since 3 is prime, divisibility of p² by 3 implies that p itself is divisible by 3; put p = 3k. Substituting and cancelling gives q² = 3k², so q is also divisible by 3. This contradicts the assertion that p and q have no common factor. The proof therefore establishes irrationality. The other options do not derive the required divisibility chain and cannot produce a valid contradiction.

Correct answer: A. The left side is a perfect square but the right side can have an odd exponent of 2

Explanation: The correct answer is A. In the prime factorisation of any perfect square, every prime occurs with an even exponent. Thus the left side a² must have an even exponent for the prime 2, as well as for every other prime. On the right side, 2b² contains the factor 2 outside the square b². If the exponent of 2 in b² is 2r, then its exponent in 2b² is 2r + 1, which is odd. The same integer cannot simultaneously have an even and an odd exponent of 2 in its unique prime factorisation. This contradiction is another way to see why the assumed rational representation of √2 is impossible. The other options make false claims about zero, negativity, or decimals.

Correct answer: C. c² = 2d² implies c = 2d

Explanation: The governing concept is the parity argument used in the proof that √2 is irrational. Starting from c²=2d², the right side is even, so c² is even; an integer whose square is even must itself be even. Writing c=2u and substituting gives 4u²=2d², hence d²=2u², so d is also even. If c/d was assumed to be in lowest terms, both numerator and denominator being even contradicts that assumption. However, c²=2d² does not imply c=2d. Equality of squares with a factor of 2 does not permit taking such a direct linear conclusion. Therefore option C is the invalid shortcut; A, B, and D represent valid steps or conclusions in the proof.

Correct answer: A. A prime factor occurs in a square only if it occurs in the original number

Explanation: The governing principle is unique prime factorisation. If the prime 3 divides h², then the prime factor 3 must already occur in the factorisation of h; squaring only doubles its exponent and cannot create a new prime factor. Therefore 3 divides h. The other options are false general statements and do not support the irrationality proof.