Proof of irrationality of square root 2 and square root 3

For coprime numbers, \(\gcd(x,y)=1\). However, if both \(x\) and \(y\) are even, then 2 is a common factor of both, so \(\gcd(x,y)\ge 2\). Thus, \(\gcd(x,y)=1\) and \(\gcd(x,y)\ge2\) cannot hold simultaneously, making option A correct. Options B, C, and D do not follow from the numbers being coprime and even. Exam tip: In a proof by contradiction, if an assumption leads to a result that conflicts with a known fact such as \(\gcd(x,y)=1\), the assumption must be rejected.

The governing definition is that coprime integers have greatest common divisor equal to 1. If both u and v are divisible by 3, there exist integers r and s such that u = 3r and v = 3s. Therefore 3 is a common divisor of u and v, so their greatest common divisor must be at least 3; formally, gcd(u,v) ≥ 3. The original lowest-terms assumption says gcd(u,v) = 1. These two statements cannot both be true, producing the contradiction that proves the rational assumption false. Option A states the exact contradiction. The remaining choices violate the definition or introduce unsupported equations.

In a proof by contradiction, a rational number is written as a fraction such as \(x/y\), where \(x\) and \(y\) are integers. The condition \(y\neq0\) is needed simply because division by zero is not defined. It allows the fraction to represent a number. The condition \(\gcd(x,y)=1\) says that the fraction has already been reduced to lowest terms; it is not merely a requirement for writing the fraction.

For \(\sqrt{2}=x/y\), squaring gives \(x^2=2y^2\). This shows that \(x\) is even, and then \(y\) is also even. Thus both numbers have a common factor 2, contradicting \(\gcd(x,y)=1\). Therefore option A correctly separates the roles: the non-zero denominator keeps the fraction meaningful, while the coprime condition is what the final common factor contradicts.

(v\neq0) is necessary for the fraction. (\gcd(u,v)=1) creates contradiction when both are divisible by (3).

In the standard proof, assume \(\sqrt{2}=x/y\), where \(x\) and \(y\) are coprime integers. From \(2y^2=x^2\), we first conclude that \(x\) is even. Substituting \(x=2k\) then shows that \(y\) is also even. Thus, both \(x\) and \(y\) have the common factor 2, contradicting the assumption that they are coprime. Showing only that \(x\) is even gives no contradiction. Exam tip: In a contradiction proof of irrationality, explicitly reach a violation of the coprime condition.

From \(a^2=2b^2\), \(a^2\), hence \(a\), must be even. Put \(a=2k\); then \(b^2=2k^2\), so \(b\) is also even. A common factor 2 contradicts coprimality. Exam tip: always begin this proof with the fraction in lowest terms.

The lowest-form condition tells that (x,y) are coprime. This creates contradiction when both become even.

From \(p^2=3q^2\), 3 divides \(p^2\). Since 3 is prime, it must divide \(p\). Option B incorrectly demands divisibility by 9. Exam tip: use \(r\mid n^2\Rightarrow r\mid n\) for prime \(r\).

If \(x\) and \(y\) are both even, each is divisible by \(2\). Thus \(\frac{x}{y}=\frac{2m}{2n}=\frac{m}{n}\), so the common factor \(2\) can be cancelled from the numerator and denominator. If \(y=0\), the fraction is undefined; merely having \(x=y\) does not necessarily show that a fraction is reducible. Exam tip: a fraction is reducible when its numerator and denominator have the same non-zero common factor.

Since 3 is prime, if it divides \(p^2=p\times p\), it must divide \(p\). Putting \(p=3k\) then shows that 3 also divides \(q\), contradicting the lowest-form assumption. Exam tip: apply this prime-factor property in irrationality proofs.

The proof first proves factor (2) in the numerator and then in the denominator. This conflicts with coprime condition.

Assume \(\sqrt{2}=p/q\) in lowest terms. Then \(p^2=2q^2\), so \(p^2\) is even and hence \(p\) is even. This later makes \(q\) even too, contradicting coprimality. Exam tip: remember that an even square has an even integer root.

One number having factor (2) gives no contradiction with coprime condition. Both need factor (2).

Only (u) having factor (3) is not a contradiction. The coprime condition breaks when both (u) and (v) have factor (3).

From \(p^2=3q^2\), 3 divides \(p^2\), so 3 divides \(p\). Substitution then shows that 3 also divides \(q\), contradicting lowest terms. Exam tip: use the prime-factor property for a square.

Common factor (3) in both contradicts the coprime condition. Therefore (\sqrt{3}) is irrational.

The proof uses contradiction and begins by assuming √n = u/v, where u and v are integers in lowest terms and v ≠ 0. The lowest-terms condition is essential because it gives gcd(u,v) = 1. The divisibility argument then shows that the same prime divides both u and v, contradicting this condition. Without first reducing the fraction, finding a common factor would not necessarily be a contradiction: unreduced fractions can naturally have common factors. Therefore A is the crucial exam precaution. B is invalid because a denominator cannot be zero, C is only approximation, and D is an unjustified assumption.

Since 3 is prime, the prime-factor property gives \(3\mid p\) whenever \(3\mid p^2\); hence write \(p=3r\). Substitution then shows that \(q\) is also divisible by 3, contradicting coprimality. Exam tip: state the prime-divisor property before substituting.

In contradiction proof, assume \(\sqrt{2}=\frac{p}{q}\) in lowest terms, so \(p\) and \(q\) are coprime. From \(p^2=2q^2\), both become even, contradicting coprimality. Exam tip: always state the lowest-terms condition.

From \(3q^2=p^2\), \(p^2\), hence \(p\), 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 of a square carefully.