Proof of irrationality of square root 2 and square root 3

From \(p^2=3q^2\), we get \(3\mid p^2\). Since 3 is prime, the prime-factor property gives \(3\mid p\). Putting \(p=3k\) then shows \(3\mid q\), contradicting coprimality. In exams, state the prime-factor property explicitly.

If \(p\) is even, write \(p=2k\). Then \(4k^2=2q^2\Rightarrow q^2=2k^2\), so \(q\) is also even. Thus both have factor 2, contradicting coprimality. Exam tip: use the fact that an even square has an even root.

Since \(p^2=3q^2\), \(p^2\) is divisible by 3, so \(p=3k\). Substituting gives \(q^2=3k^2\), hence \(q\) is also divisible by 3. This contradicts coprimality. In exams, show both divisibility steps.

From p^2=3q^2, the right conclusion is that 3 divides p. This follows from prime-factor reasoning: the exponent of the prime 3 in a perfect square is even, while the factor 3 on the right makes the exponent in 3q^2 odd unless q also supplies a factor of 3. Thus p must contain a factor 3. However, this does not mean p=3q.

The symbols p and q represent the numerator and denominator in the assumed fraction, and q need not equal p divided by 3. The proof normally writes p=3k for some integer k, then substitutes this into the equation to show that 3 also divides q, producing the contradiction. Therefore option A is correct. The other choices do not describe the valid divisibility argument.

Since \(p^2=3q^2\), \(p^2\) is divisible by 3, so \(p\) is divisible by 3. Put \(p=3k\): then \(q^2=3k^2\), hence \(q\) is also divisible by 3. This contradicts coprimality. Exam tip: use prime divisibility of a square carefully.

From \(p^2=2q^2\), \(p^2\) is even, so \(p\) is even. Put \(p=2k\); then \(q^2=2k^2\), making \(q\) even too. This contradicts coprimality. Exam tip: state that the fraction is in lowest terms.

The governing concept is the prime-factorization property of a perfect square: every prime occurs with an even exponent. In the contradiction proof, suppose √2 = x/y, where x and y are coprime integers and y is non-zero. Squaring gives x² = 2y². Therefore x² has an odd contribution from the factor 2 on the right, so 2 divides x; write x = 2k. Substitution gives 4k² = 2y², hence y² = 2k², so 2 also divides y. This contradicts the assumption that x and y have no common factor. Thus option A states the essential idea; the other choices make false universal claims or give the incorrect value of √2.

The governing principle is that the exponent of every prime in the factorization of a perfect square is even. Assume, for contradiction, that √3 = m/n in lowest terms. Squaring gives m² = 3n², so 3 divides m. Put m = 3r; then 9r² = 3n², which simplifies to n² = 3r². Hence 3 also divides n. The numerator and denominator are therefore both divisible by 3, contradicting the fact that the fraction was in lowest terms. Option B expresses the exact prime-exponent idea behind this contradiction. Option A is too broad, option C confuses a root with the number under the radical, and option D is unrelated to rational representation.

The governing concept is the difference between a preliminary domain condition and the actual contradiction in a proof by contradiction. If √2 is represented as x/y, then y ≠ 0 is required simply because division by zero is undefined. This condition allows the fraction and the subsequent squaring step to make sense, but it does not imply anything about the parity or common factors of x and y. The decisive part comes later: from x² = 2y², one proves that x is even and then that y is even. That conflicts with choosing x/y in lowest terms, where gcd(x,y) = 1. Therefore option A is correct; the other options claim conclusions that do not follow from y ≠ 0.

The condition \(n\ne0\) is necessary because a fraction with denominator zero is not defined. When proving irrationality, we assume \(\sqrt{3}=m/n\) with integers \(m,n\), where \(n\ne0\), and choose the fraction in lowest form. Squaring gives \(m^2=3n^2\), which leads to divisibility by 3.

The divisibility argument first shows that 3 divides \(m\). Writing \(m=3r\) and substituting gives \(n^2=3r^2\), so 3 also divides \(n\). Thus both integers have a common factor 3, contradicting \(\gcd(m,n)=1\). The nonzero-denominator condition does not imply \(m=n\), \(n=0\), or \(\sqrt{3}=3\). Therefore option A is correct.

The proof uses contradiction. To establish that √2 is irrational, assume temporarily that √2 is rational and write it as p/q in lowest terms, with integers p and q and q nonzero. Squaring gives p² = 2q². This implies p is even; writing p = 2k then shows q is also even, contradicting the assumption that p/q was in lowest terms. The contradiction means the initial assumption that √2 is rational must be false. Therefore its negation is true: √2 is irrational. Option B repeats the rejected assumption, while options C and D do not follow from the argument. Hence option A is correct.

From \(3q^2=p^2\), \(p\) is divisible by 3. On putting \(p=3k\), we get \(q^2=3k^2\), so \(q\) is also divisible by 3. This contradicts coprimality. In exams, show divisibility of both integers.

Decimal approximation is not a proof. The coprime fraction argument proves irrationality exactly.

An approximate decimal tells us only that a number is close to a displayed value. It does not describe all of its digits and therefore cannot establish an exact claim about rationality. To prove that \(\sqrt{3}\) is irrational, we need an argument that works for every possible fraction representing it, not just a numerical estimate. Divisibility gives that exact structure.

Assume \(\sqrt{3}=p/q\) in lowest form, where \(q\neq0\). Squaring gives \(p^2=3q^2\), so 3 divides \(p^2\), which implies that 3 divides \(p\). Substituting \(p=3k\) then shows that 3 divides \(q\) as well. Both numbers would share 3, contradicting lowest form. Thus option A is correct.

Both arguments use the same proof architecture: assume that the square root is rational, express it as a fraction in lowest terms, square the equation, and use prime divisibility to force a common factor in the numerator and denominator. For √2, the forced prime is 2; for √3, it is 3. In either case, the new common factor contradicts gcd(numerator, denominator) = 1. Thus option A gives the highest-level similarity while still identifying the essential mechanism. The denominator is not assumed to be zero, decimal termination is not the basis of the proof, and equality of numerator and denominator is never required.

Since \(p^2=3q^2\), \(p^2\) is divisible by 3. As 3 is prime, divisibility of \(p^2\) by 3 implies that \(p\) itself is divisible by 3; write \(p=3k\). Substitution then makes \(q\) divisible by 3 too, contradicting coprimality. Exam tip: use the prime-divides-a-square rule in such proofs.

From \(p^2=2q^2\), \(p^2\) is even. If the square of an integer is even, the integer itself is even; hence \(p=2k\). Substitution then makes \(q\) even too, contradicting coprimality. Exam tip: remember “even square implies even integer.”

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: use prime divisibility from \(p^2\) to \(p\).

Write an integer as 3q, 3q+1, or 3q+2. Their squares leave remainders 0, 1, and 1 respectively, so a square divisible by 3 requires the integer to be divisible by 3. Exam tip: use remainder classes.

If \(3\sqrt{2}\) were rational, dividing it by 3 would make \(\sqrt{2}\) rational. This contradicts the known irrationality of \(\sqrt{2}\). Option C is false because two irrational numbers can have a rational product. Exam tip: rationality is preserved when dividing by a non-zero rational number.