Proof of irrationality of square root 2 and square root 3

Since \(p^2=2q^2\), \(p^2\) is even. By A, \(p=2k\); substitution gives \(q^2=2k^2\), so \(q\) is also even, contradicting coprimality. Exam tip: use the parity property of a square.

Assume \(2+\sqrt{3}\) is rational. Subtracting 2 then makes \(\sqrt{3}\) rational, contradicting its irrationality. Tip: adding a rational number cannot make an irrational number rational.

The equation is obtained while proving that
sqrt{3}
cannot be rational. From the equality of squares, it is not valid to remove the squares and claim that the two expressions are equal in the same form. In particular, the statement
m=3n
is much stronger than what the equation gives and is generally false.

The correct number-theory conclusion is that 3 divides
m^2
, because
m^2=3n^2
. Since 3 is prime, this implies that 3 divides
m
. Writing
m=3k
and substituting then produces the required contradiction with coprimality. Thus option B identifies the error: it takes an incorrect linear conclusion from a squared equation. The denominator is not being assumed zero, and the contradiction method itself is not the mistake.

From \(a^2=3b^2\), prime divisibility gives \(3\mid a\). Put \(a=3k\): then \(b^2=3k^2\), so \(3\mid b\) too. This contradicts coprimality. Exam tip: explicitly state the prime-divisor rule.

From m² = 3n², the right side is divisible by 3, so m² and therefore m are divisible by 3. Write m = 3k. Substitution gives 9k² = 3n²; after division by 3, n² = 3k². Hence n² is divisible by 3, and because 3 is prime, n is divisible by 3. This complete chain is given only in option A. It is important that the conclusion about n comes after introducing m = 3k and simplifying the substituted equation; it does not follow merely from the original equation without those steps. Once both m and n are divisible by 3, their assumed coprimality is contradicted.

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).

The correct answer is option B. The two conditions have different jobs. In √2=r/s, the condition s≠0 makes the fraction meaningful, because division by zero is undefined. It also allows multiplication by s and ordinary algebraic steps. The condition gcd(r,s)=1 says that r and s have no common factor greater than 1; therefore the fraction is in lowest form. After assuming √2=r/s and squaring, r²=2s². This shows r is even. Writing r=2k then gives s²=2k², so s is even too. Both r and s being even contradicts gcd(r,s)=1. That is why the coprime condition creates the final contradiction, while s≠0 merely protects the definition of the fraction. Option A is wrong because the conditions are not equivalent. Option B is correct. Option C is wrong: gcd(r,s)=1 does not make s zero; it concerns common factors. Option D is wrong: s≠0 does not imply r=s. Memory cue: non-zero denominator starts the proof safely; coprime numerator and denominator finish it.

(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).