Proof of irrationality of square root 2 and square root 3

Since \(3\mid p^2\) and 3 is prime, \(3\mid p\). Putting \(p=3k\) then gives \(3\mid q\), contradicting coprimality. Exam tip: use the prime-divides-a-square property.

In lowest terms, \(p\) and \(q\) are coprime. From \(2q^2=p^2\), \(p\) is even and then \(q\) is also even, contradicting coprimality. Exam tip: state this contradiction clearly.

The fundamental exponent rule says that when a number is squared, every prime exponent in its factorisation is doubled. Therefore each prime, including 2, occurs to an even exponent in a perfect square. In the proof, assume √2 = r/s in lowest terms. Squaring gives r² = 2s². The left side is a perfect square and must contain an even exponent of 2, while the right side contains the factor 2 multiplied by the square s², producing an odd exponent of 2 relative to the square structure. This incompatibility leads to the conclusion that the original rational assumption is impossible. Thus option A is correct; the other choices confuse the rule with unrelated or false claims.

The correct idea is that every prime has an even exponent in the factorisation of a perfect square. In particular, the exponent of 3 must be even. Suppose √3 = a/b in lowest terms. Then a² = 3b². Since 3 divides a², it divides a, so let a = 3k. Substitution gives b² = 3k², which similarly forces 3 to divide b. This contradicts the assumption that a and b have no common factor. The exponent viewpoint expresses the same conflict: a square has an even exponent of 3, but multiplication by one additional factor 3 changes the relevant parity. Consequently option C is correct. The other choices make unjustified universal claims or give a false numerical equality.

From \(p^2=3q^2\), we get \(3\mid p^2\), so the prime-factor property gives \(3\mid p\). Put \(p=3k\); then \(q^2=3k^2\), hence \(3\mid q\) too. This contradicts coprimality. Exam tip: explicitly substitute \(p=3k\).

If (m) has no factor (3), then (m^2) also has none. But the equation makes (m^2) divisible by (3).

Non-zero denominator is for defining the fraction. The coprime condition conflicts with the final common factor.

(n\neq0) is necessary for a rational fraction. (\gcd(m,n)=1) breaks when both are divisible by (3).

Since 3 is prime, \(3\mid p^2\) necessarily implies \(3\mid p\). Step: put \(p=3k\); the proof later gives \(3\mid q\), contradicting lowest terms. Exam tip: use the prime-divisor property for a square.

Lowest form makes (m) and (n) coprime. This condition makes both divisible by (3) a contradiction.

From \(p^2=3q^2\), \(3\mid p^2\), so \(3\mid p\). Put \(p=3k\); then \(q^2=3k^2\), hence \(3\mid q\) too. This contradicts lowest terms. Exam tip: state the prime-divisor rule.

Both divisible by (3) is the contradiction found at the end. Initially (\gcd(m,n)=1) is assumed.

Since the prime 3 divides \(p^2\), it must divide \(p\). Let \(p=3k\); then \(9k^2=3q^2\), so \(q^2=3k^2\), making \(q\) divisible by 3 too. This contradicts lowest terms. Exam tip: show a common factor in numerator and denominator.

By the prime-factor property, if the prime 3 divides p², then 3 must divide p. In the proof for √3, this follows from p² = 3q². Exam tip: remember that a prime divisor of a square is also a divisor of its base.

Assume \(q\sqrt{3}\) is rational, with \(q\ne0\) rational. Then \(\sqrt{3}=\frac{q\sqrt{3}}{q}\) would be rational, contradicting the irrationality of \(\sqrt{3}\). Exam tip: division by a non-zero rational preserves rationality.

Assume that \(\sqrt{2}+\sqrt{3}\) is rational. Squaring gives \(5+2\sqrt{6}\), so \(\sqrt{6}\) would have to be rational. But 6 is not a perfect square, hence \(\sqrt{6}\) is irrational. Exam tip: square a sum of surds to test such claims.

For contradiction, write \(\sqrt{3}\) as the lowest-term fraction \(p/q\). From \(p^2=3q^2\), 3 divides \(p\) and then \(q\), contradicting coprimality. Exam tip: always state that the fraction is in lowest terms.

A fraction in lowest form has no common factor greater than 1 between its numerator and denominator. Therefore, if \(\sqrt{3}=u/v\) is assumed to be in lowest form, then \(\gcd(u,v)=1\). The proof begins with \(u^2=3v^2\). Since 3 divides \(u^2\) and is prime, 3 divides \(u\). Substituting this fact back into the equation then proves that 3 divides \(v\) too.

Consequently, both \(u\) and \(v\) are divisible by 3. Their greatest common divisor must therefore be at least 3, so \(\gcd(u,v)\ge3\). This directly contradicts \(\gcd(u,v)=1\), the lowest-form condition. The issue is not that \(v=0\), that \(u=v\), or that the root becomes an integer. Hence option A is decisive.

49 is a perfect square because \(49=7^2\); hence \(\sqrt{49}=7\), a rational number. The numbers 12, 18, and 27 are not perfect squares. Exam tip: in prime factorisation, every exponent must be even for a perfect square.

First (u) is proved divisible by (3). For the final contradiction, (v) must also be divisible by (3).