Proof of irrationality of square root 2 and square root 3

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.

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.

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.

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.

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.

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.