Proof of irrationality of square root 2 and square root 3

From \(a^2=3b^2\), \(a^2\) is divisible by 3. Since 3 is prime, \(a\) must be divisible by 3. Then put \(a=3k\) to show that \(b\) is also divisible by 3. Exam tip: if a prime divides a square, it divides the number itself.

Since \(p^2=3q^2\), \(p^2\) is divisible by 3, so \(p\) is divisible by 3. Put \(p=3k\); then \(q\) is also divisible by 3, contradicting lowest terms. Exam tip: use the prime-divisibility property of squares.

Since 3 is prime, \(3\mid p^2\) implies \(3\mid p\). Substituting \(p=3k\) in \(p^2=3q^2\) gives \(3\mid q\), contradicting coprimality. Exam tip: remember this prime-divisibility rule.

A rational number must be expressible as u/v, where u and v are integers, v ≠ 0, and the fraction is in lowest terms, so gcd(u,v) = 1. In the proof, assuming √3 = u/v leads to u² = 3v². The divisibility argument first gives 3 ∣ u and then, after writing u = 3t and substituting, gives 3 ∣ v. Thus u and v have the common factor 3, contradicting their coprime, lowest-form condition. This contradiction rejects the rational assumption. Option B contains the complete sufficient statement; A and C give only partial information, and D does not contradict rationality.

The governing concept is unique prime factorisation. In the square of an integer, every prime exponent is doubled, so every exponent is even. Suppose √2 were rational and write it as a fraction in lowest terms, a/b. Squaring gives a² = 2b². The prime 2 occurs to an odd extra power on the right-hand side, while the left-hand side is a square and must have even prime exponents. Equivalently, divisibility forces 2 ∣ a and then 2 ∣ b, contradicting lowest terms. Therefore option A expresses the key idea. The other options are false or unrelated to the proof.

In a perfect square, all prime exponents are even. This idea strengthens the contradiction for (\sqrt{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 concept is the definition of coprime integers and the meaning of a common divisor. If u and v are coprime, their greatest common divisor is gcd(u,v)=1. If both are divisible by 3, there are integers r and s such that u=3r and v=3s. Hence 3 is a common divisor of u and v, so their greatest common divisor must be at least 3: gcd(u,v)≥3. The statements gcd(u,v)=1 and gcd(u,v)≥3 are incompatible. This contradiction disproves the assumed lowest-terms representation used in the irrationality proof. Option A states the precise contradiction. Zero and negative values do not describe this gcd, and no general rule makes the gcd equal to u+v.

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.

The direct answer is A. These two conditions have different jobs in the proof. To discuss \(\sqrt{3}=u/v\), the denominator must satisfy \(v\neq0\); otherwise the fraction is undefined. The condition \(\gcd(u,v)=1\) says that u and v have no common factor greater than 1, meaning the fraction is in lowest form. In the contradiction proof, squaring gives \(u^2=3v^2\), which eventually shows that both u and v are divisible by 3. That conflicts specifically with \(\gcd(u,v)=1\), not with merely \(v\neq0\). Option A correctly separates the roles. Option B is wrong because \(v\neq0\) only permits division; it does not immediately imply \(u=3t\). Option C is the opposite of the meaning of coprime numbers. Option D is wrong because nonzero denominator and lowest form are separate requirements. Exam cue: nonzero denominator makes the fraction legal; coprime numerator and denominator make the final contradiction possible.

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 direct answer is A. In an irrationality proof, writing \(\sqrt{2}=x/y\) is not enough by itself. We choose integers x and y with \(y\neq0\) and put the fraction in lowest form, so \(\gcd(x,y)=1\). Squaring gives \(x^2=2y^2\). Thus x is even; write \(x=2k\). Substitution gives \(y^2=2k^2\), so y is also even. Now x and y have the common factor 2, contradicting the claim that the fraction was in lowest form. Option A identifies this missing logical link, so it is correct. Option B is wrong because the proof does not show y=0; in fact y must be nonzero. Option C is wrong because the equation does not imply x=y. Option D is the opposite of the intended result: the proof starts by assuming rationality and ends by showing that assumption impossible, so it proves irrationality. The key lesson is that a common factor is a contradiction only after lowest form has been stated. Memory cue: always write “assume a fraction in lowest terms” before beginning this proof.

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 direct answer is A. Assume, for contradiction, that \(\sqrt{2}=x/y\) in lowest form, with \(y\neq0\) and \(\gcd(x,y)=1\). Squaring gives \(x^2=2y^2\), so x is even. Let \(x=2k\). Then \(4k^2=2y^2\), hence \(y^2=2k^2\), so y is even too. Therefore 2 divides both x and y. The number 2 is prime, and it becomes the common prime factor that contradicts the assumption that x and y are coprime. Option A is correct. Option B is wrong because 2 does not make the denominator zero; y was assumed nonzero. Option C is wrong because no step gives x=y. Option D is wrong because the role of 2 is to create a contradiction, not to prove that \(\sqrt{2}\) is rational. The proof actually concludes that the original rational assumption is impossible. Memory cue: in the \(\sqrt{2}\) proof, follow the factor 2 from the square equation into both numerator and denominator.

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.

The direct answer is A. In the usual proof, assume \(\sqrt{2}=x/y\) in lowest form, so x and y have no common factor. From squaring, \(x^2=2y^2\), we learn that x is even. But this alone is not a contradiction: one of two coprime numbers may be even while the other is odd. We must write \(x=2k\), substitute, and obtain \(y^2=2k^2\); this proves y is even as well. Then both x and y contain the common factor 2, which contradicts lowest form. Option A is correct because it identifies the missing second half. Option B is false; x being even is a valid intermediate conclusion. Option C is unnecessary and false in this proof: y is not zero, since it is a denominator. Option D is also false; \(\sqrt{2}\) is not equal to 2. The common mistake is stopping before proving evenness of the denominator. Exam cue: to contradict coprimality, show a common factor in both numbers.

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.

The direct answer is B. Assume \(\sqrt{3}=u/v\) in lowest form, with v nonzero and \(\gcd(u,v)=1\). Squaring gives \(u^2=3v^2\). This shows 3 divides \(u^2\), so 3 divides u; writing \(u=3k\) and substituting gives \(v^2=3k^2\), so 3 also divides v. Both numerator and denominator are therefore divisible by 3, contradicting the lowest-form assumption. Hence the original assumption that \(\sqrt{3}\) is rational is false, and \(\sqrt{3}\) is irrational. Option A is wrong because the equation is used in a contradiction argument and does not prove rationality. Option B is correct because it gives both the conclusion and the exact basis. Option C is wrong: 3 being an integer does not make its square root an integer; for example, \(\sqrt{3}\) lies between 1 and 2. Option D is wrong because a contradiction does not make the square root equal to zero. Memory cue: for \(\sqrt{3}\), track the factor 3 into both u and v.

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.