Proof of irrationality of square root 2 and square root 3

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.

Dividing both by (2) is possible only when (2) is a common factor. This contradicts lowest form.

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 such that h = 3r. Therefore writing h = 3r is a definitional substitution: it replaces the divisibility statement with an explicit algebraic form that can be used in the next equation. In the proof, substituting h = 3r into h² = 3k² gives 9r² = 3k², and division by 3 yields k² = 3r². This then proves 3 | k. Option A is correct because the step does not approximate a decimal, make a denominator zero, or finish the proof by itself.

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.

Since \(p^2=3q^2\), \(p^2\) is divisible by 3, so \(p\) is divisible by 3. Put \(p=3k\): \(9k^2=3q^2\), hence \(q^2=3k^2\), so \(q\) is also divisible by 3. This contradicts coprimality. Exam tip: if a prime divides a square, it divides the number.

Assume \(\sqrt{2}+\sqrt{3}=r\), with \(r\) rational. Option A is not a valid conclusion because rational minus irrational can still be irrational. Squaring gives \(r^2=5+2\sqrt{6}\), so \(\sqrt{6}=(r^2-5)/2\) would be rational, a contradiction. Exam tip: square a sum of surds to isolate the remaining radical.

If even a lowest fraction reduces, smaller fractions would keep appearing. This rejects the possibility of a lowest form.

Assume \(r=\sqrt{2}+\sqrt{3}\). Then \(\sqrt{3}=r-\sqrt{2}\); on squaring, \(3=r^2+2-2r\sqrt{2}\). Hence \(\sqrt{2}=(r^2-1)/(2r)\) would be rational, a contradiction. Exam tip: isolate the radical carefully after squaring.

A non-terminating decimal alone does not prove irrationality: \(\frac{1}{3}=0.333\ldots\) goes on forever but is rational. An irrational number has a non-terminating, non-recurring decimal expansion. Exam tip: distinguish recurring from non-recurring decimals.

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.

Every prime factor in \(p^2\) has an even exponent. Thus, if \(3\mid p^2\), the factor 3 must already occur in \(p\), so \(3\mid p\). It need not imply \(9\mid p\). Exam tip: use prime-factor exponents in irrationality proofs.

In lowest terms, \(p\) and \(q\) have no common factor. From \(p^2=2q^2\), \(p\) is even, and then \(q\) is also even, giving a contradiction. Exam tip: link lowest terms with coprime numerator and denominator.

From \(p^2=3q^2\), let \(p=3k\). Substitution gives \(9k^2=3q^2\), so \(q^2=3k^2\) and \(q\) is also divisible by 3. This contradicts coprimality. Exam tip: establish divisibility of both numerator and denominator.