Proof of irrationality of square root 2 and square root 3

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.

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.

Since \((\sqrt{2}+\sqrt{3})(\sqrt{3}-\sqrt{2})=1\), a rational \(x\) would have a rational reciprocal. Their sum would make \(2\sqrt{3}\) rational, a contradiction. Exam tip: first check the product of conjugates.

\(1.732\) is only an approximation of \(\sqrt{3}\), not its exact value, since \(1.732^2=2.999824\), not 3. As 3 is not a perfect square, \(\sqrt{3}\) is irrational. Exam tip: always distinguish \(=\) from \(\approx\).

Since \(p^2=2q^2\), \(p^2\) is even. The square of an integer is even only when the integer itself is even, so \(p\) is even. Substitution then makes \(q\) even too, contradicting coprimality. Exam tip: use parity of squares carefully.

1.732 is a terminating decimal approximation. In fact, \(1.732^2=2.999824\), not 3. Thus \(\frac{1732}{1000}\) is rational, but it is not equal to \(\sqrt{3}\). Exam tip: distinguish an approximate decimal from an exact value.

If \(n=k^2\) for an integer \(k\), then \(\sqrt{n}=k\), which is rational. Being even or composite alone is not enough; for example, \(\sqrt{6}\) is irrational. Exam tip: in prime factorisation, all exponents must be even for a perfect square.

From \(p^2=3q^2\), we get \(3\mid p^2\). Since 3 is prime, if it divides the square of an integer, it divides that integer; hence \(3\mid p\). This later gives \(3\mid q\), contradicting coprimality. Exam tip: explicitly state the prime-divisor property.

If \(p=2k\), then \(4k^2=2q^2\), so \(q^2=2k^2\) and hence \(q\) is even too. Thus both have factor 2, contradicting coprimality. In exams, state this contradiction explicitly.

A mathematical proof requires an exact chain of justified statements, not merely numerical evidence. An approximate decimal value can suggest that √2 or √3 is not an integer, but any finite approximation can be close to a rational number and therefore cannot establish irrationality. The rigorous proofs assume a lowest-term rational fraction and then use equations such as x² = 2y² or h² = 3k² together with prime divisibility to obtain a contradiction. Thus option A is correct: the approximate decimal is useful for intuition, not as the logical basis. Options B, C, and D are essential structural components of the standard proofs.