Proof of irrationality of square root 2 and square root 3

Assume that \(\sqrt{2}=\frac{m}{n}\), where m and n have no common factor. Squaring gives \(m^2=2n^2\). Since the right side is even, \(m^2\) is even, and therefore m is even. Write \(m=2k\). Substitution gives \(4k^2=2n^2\), so \(n^2=2k^2\), which means n is also even.

Thus the proof’s important middle objective is to establish, in sequence, that m is even and then n is even. If both are even, they share the factor 2, contradicting the assumption that \(\frac{m}{n}\) was in lowest terms. The proof does not aim to show m equals n, either variable is zero, or \(\sqrt{2}=2\). Therefore option C correctly describes the central intermediate step.

The direct answer is D. Assume, for contradiction, that the square root of 3 is rational and write it as p/q in lowest form, with q not equal to zero and p and q coprime. Squaring gives p squared equals 3q squared, so 3 divides p squared. Since 3 is prime, 3 must divide p. Substituting p equals 3k shows that 3 also divides q. Thus both p and q are divisible by 3, contradicting that they were coprime. Option A, q=0, is not the middle objective and is forbidden because a denominator cannot be zero. Option B, p=q, does not follow. Option C, square root of 3 equals 3, is false. Option D correctly states the essential route: first prove p divisible by 3, then q divisible by 3. Exam cue: divisibility of both numerator and denominator creates the contradiction.

From \(a^2=2b^2\), the right side is even, so \(a^2\) is even. The valid standard conclusion is that a is even; write \(a=2k\). Substituting gives \(4k^2=2b^2\), and after dividing by 2 we get \(b^2=2k^2\). This shows that b is even as well. These parity conclusions are enough to produce the contradiction when a and b were initially assumed to have no common factor.

It is not valid to conclude directly that \(a=2b\). Taking square roots would give a relation involving \(\sqrt{2}\), not the equation \(a=2b\); moreover, the variables need not have that particular relationship. The correct step is to infer that a is divisible by 2, then use substitution to infer the same for b. Therefore option A identifies the mistake correctly.

From (p^2=3q^2), we only get that (p^2) is divisible by (3). Then by the rule, (p) is written as (3k).

Assume \(\sqrt{3}=a/b\) in lowest terms. Then \(a^2=3b^2\), so \(3\mid a^2\), which implies \(3\mid a\). This subsequently gives \(3\mid b\), a contradiction. Exam tip: remember that for prime \(p\), \(p\mid a^2\Rightarrow p\mid a\).

Since both have common factor (3), the fraction can be reduced. So the lowest form assumption breaks.

Using contradiction, \(3q^2=p^2\) implies that 3 divides \(p\), and then it also divides \(q\), contradicting coprimality. Exam tip: a non-terminating recurring decimal can still be rational.

The decimal expansion of \(\frac{1}{3}\) is infinite but recurring, and it is a ratio of integers, so it is rational. \(\sqrt{2}\), \(\sqrt{3}\), and \(\pi\) are irrational. Exam tip: every recurring decimal represents a rational number.

From \(3q^2=p^2\), \(p^2\) 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: use the prime-divisibility property of squares.

Option A gives the correct relation. Since 3 is prime, if \(p^2\) is divisible by 3, then \(p\) must also be divisible by 3. Therefore, we can write \(p=3k\), where \(k\) is an integer. In the irrationality proof of \(\sqrt{3}\), this ultimately shows that \(q\) is also divisible by 3, contradicting the assumption that \(p\) and \(q\) are coprime. Exam tip: For a prime \(r\), remember that \(r\mid a^2\) implies \(r\mid a\).

The governing idea is proof by contradiction. To test whether √2 or √3 can be rational, we assume it equals a fraction a/b, where a and b are integers, b is non-zero, and gcd(a,b)=1. The lowest-form condition is essential: after squaring and using divisibility, the proof generally shows that both a and b must be even. That is impossible for a coprime pair because a common factor 2 would remain. Therefore the original rational assumption fails. Option A is correct. A zero denominator is forbidden, a decimal approximation cannot establish irrationality, and equal numerator and denominator is not part of the argument.

The correct answer is A. Writing √2 as approximately 1.414 or displaying more decimal digits only gives a numerical approximation. A finite decimal is rational, while an observed non-terminating pattern on a calculator does not by itself prove that no fraction equals the number; calculators also display rounded values. A formal school proof assumes √2 = a/b in lowest terms, squares to obtain a² = 2b², and then shows that both a and b must be even, contradicting their coprime status. Alternatively, a rigorous theorem about decimal expansions may be used, but merely copying digits is insufficient. Options B, C, and D assert conclusions that neither the decimal display nor the irrationality argument logically establishes.

The correct answer is A. In a contradiction proof, assume √2 or √3 equals p/q, where p and q are integers, q is nonzero, and the fraction is already in lowest terms, meaning gcd(p,q) = 1. For √2 the argument eventually shows both numerator and denominator are even; for √3 it shows both are divisible by 3. The contradiction is meaningful only because a lowest-form fraction cannot have a common prime factor. If the fraction is not reduced at the start, the later divisibility result may simply describe a non-reduced representation and no contradiction follows. A denominator cannot be zero, decimals are not a substitute for proof, and numerator and denominator need not be equal.