Proof of irrationality of square root 2 and square root 3

Correct answer: A. 1.732 is an approximation; its rationality does not prove that \(\sqrt{3}\) is rational

Explanation: 1.732 is a terminating rational approximation, not \(\sqrt{3}\) itself. Check: \(1.732^2=2.999824\), not 3. Exam tip: distinguish an approximate value from an exact value.

Correct answer: A. Both numerator and denominator are divisible by 2

Explanation: From \(p^2=2q^2\), \(p\) must be even. Put \(p=2k\); then \(q\) is also even, so both share the factor 2. This contradicts lowest terms. Exam tip: a fraction in lowest terms has coprime numerator and denominator.

Correct answer: B. Both \(m\) and \(n\) are divisible by 3

Explanation: Since \(3\mid m^2\), \(3\mid m\). Let \(m=3k\); then \(n^2=3k^2\), so \(3\mid n\). Both share 3, contradicting coprimality. Tip: use the prime-divisor rule.

Correct answer: A. p and q are coprime

Explanation: Writing \(p/q\) in lowest terms makes \(p\) and \(q\) coprime. From \(p^2=2q^2\), \(p\) is even and then \(q\) is also even, contradicting this condition. Exam tip: always state “lowest terms.”

Correct answer: A. \(\sqrt{3}=\frac{p}{q},\ p,q\in\mathbb{Z},\ \gcd(p,q)=1,\ q\ne0\)

Explanation: If rational, \(\sqrt{3}\) is written as \(p/q\) in lowest terms. From \(3q^2=p^2\), first \(p\) and then \(q\) are divisible by 3, giving a contradiction. Exam tip: always state coprimality.

Correct answer: A. Because only then common factor (3) will appear in (p) and (q)

Explanation: Only (p) being divisible by (3) is not a contradiction. If (q) is also divisible by (3), the coprime condition breaks.

Correct answer: A. Subtracting 5 would make \(\sqrt{3}\) rational, which is a contradiction.

Explanation: If \(5+\sqrt{3}\) were rational, subtracting the rational number 5 would make \(\sqrt{3}\) rational, contradicting its irrationality. Exam tip: rational numbers are closed under subtraction.

Correct answer: B. 1.732 is only an approximation of \(\sqrt{3}\); \(1.732^2\ne3\).

Explanation: Option B is correct. \(1.732^2=2.999824\), not 3, so 1.732 is only an approximation of \(\sqrt{3}\). A terminating decimal describes 1.732, not the exact value of \(\sqrt{3}\). In exams, distinguish a rounded value from an exact value.

Correct answer: A. Coprime numbers have common factor (1) only

Explanation: A fraction in lowest form has numerator and denominator with no common factor greater than 1. In other words, if \(a/b\) is in lowest form, then \(\gcd(a,b)=1\). The number 1 is their only positive common factor. This condition is deliberately used in irrationality proofs so that a common factor found later creates a contradiction.

For \(\sqrt{2}\), the assumption \(\sqrt{2}=a/b\) leads to the conclusion that both \(a\) and \(b\) are even. Therefore 2 divides both numbers, so their greatest common divisor is at least 2, not 1. This is impossible for a fraction in lowest form. Thus option A is correct. The contradiction comes from the coprime condition, not from the denominator being zero or from every square root being an integer.

Correct answer: A. Its decimal expansion is non-terminating and non-recurring.

Explanation: A rational number has a terminating or recurring decimal expansion. \(\sqrt{2}\) is non-terminating and non-recurring, so it is irrational; option D describes a rational decimal. Exam tip: look for “non-recurring”.

Correct answer: A. From lowest (\frac{a}{b}) an even smaller fraction is obtained

Explanation: When both are even, (\frac{a}{b}) can be reduced by (2). This opposes the idea of a lowest fraction.

Correct answer: A. From lowest (\frac{p}{q}) a smaller fraction reducible by (3) is obtained

Explanation: Infinite descent expresses the contradiction as an impossible chain of ever-smaller positive integer examples. Assume that \(\sqrt{3}=p/q\) has been written in lowest terms, with \(q\neq0\) and positive denominator if needed. Squaring gives \(p^2=3q^2\). The divisibility argument shows that 3 divides \(p\), and substituting \(p=3k\) shows that 3 also divides \(q\).

Consequently, the fraction \(p/q\) can be reduced by cancelling a factor 3. The new numerator and denominator are smaller positive integers, yet their ratio is still \(\sqrt{3}\). Repeating the same reasoning would produce an endless sequence of smaller positive denominators, which cannot exist. Hence the assumed lowest fraction is impossible, exactly as stated in option A.

Correct answer: A. In a perfect square the exponent of (2) is even

Explanation: In a perfect square every prime exponent is even. In (2b^2), the exponent of (2) becoming odd explains the contradiction.

Correct answer: A. To ensure that \(p\) and \(q\) have no common factor

Explanation: Assume \(\sqrt{2}=p/q\) with coprime \(p,q\). From \(p^2=2q^2\), \(p\) is even, and then \(q\) is also even, contradicting coprimality. Exam tip: always state that the fraction is in lowest terms.

Correct answer: A. Both numbers being divisible by 3 contradicts their being coprime.

Explanation: Coprime \(m,n\) have no common prime factor. If both are divisible by 3, then \(\gcd(m,n)\ge3\), so \(\frac{m}{n}\) was not in lowest terms. Exam tip: check the claimed common factor.

Correct answer: A. If the diagonal is \(m/n\) with \(m,n\) coprime, then \(m^2=2n^2\) shows that both \(m\) and \(n\) are even, which is impossible.

Explanation: The diagonal is \(\sqrt{2}\). Assuming \(\sqrt{2}=m/n\) in lowest terms gives \(m^2=2n^2\). Put \(m=2k\); then \(n\) is also even, contradicting coprimality. Option C is wrong because an infinite decimal may repeat. Exam tip: state the lowest-terms assumption first.

Correct answer: A. By Pythagoras’ theorem, \(d^2=1^2+1^2=2\), so \(d=\sqrt{2}\), which is irrational.

Explanation: Since \(d^2=2\), \(d=\sqrt{2}\), which is irrational. B is false: a rational square need not have a rational square root. Exam tip: find \(d^2\) first using Pythagoras.

Correct answer: A. Only (p) is divisible by (3)

Explanation: Only (p) being divisible by (3) is not a contradiction to coprime condition. (q) must also be divisible by (3).

Correct answer: A. First write \(p=3k\); substitute to obtain \(q^2=3k^2\) and show that \(q\) is also divisible by 3.

Explanation: From \(3\mid p^2\), the prime-square rule gives \(3\mid p\), so \(p=3k\). Substitution yields \(q^2=3k^2\), hence \(3\mid q\); assuming it directly is invalid. Exam tip: write both divisibility steps.

Correct answer: A. If both \(a\) and \(b\) are even, they cannot be coprime.

Explanation: Since \(a^2=2b^2\), \(a\) is even. Writing \(a=2k\) gives \(b^2=2k^2\), hence \(b\) is even. This contradicts coprimality. Tip: begin with the fraction in lowest terms.

Correct answer: A. (b\neq0) keeps the fraction defined and (\gcd(a,b)=1) creates contradiction

Explanation: When a number is assumed rational, it is represented by a fraction \(a/b\) with \(b\neq0\). This first condition has a basic meaning: the denominator must not be zero, because a fraction with denominator zero is undefined. It says nothing about whether the numerator is even. The separate condition \(\gcd(a,b)=1\) chooses the fraction in lowest terms.

In the proof, \(\sqrt{2}=a/b\) leads to \(a^2=2b^2\). Therefore \(a\) is even; writing \(a=2k\) and substituting back shows that \(b\) is even too. The two numbers then share the factor 2, contradicting \(\gcd(a,b)=1\). Thus option A correctly identifies the denominator condition and the source of the contradiction.

Correct answer: A. (q\neq0) is needed for the fraction and (\gcd(p,q)=1) is the basis of final contradiction

Explanation: (q\neq0) keeps the fraction defined. (\gcd(p,q)=1) gives contradiction when both are divisible by (3).

Correct answer: A. यदि \(3\mid n^2\), तो \(3\mid n\)।

Explanation: From \(p^2=3q^2\), \(3\mid p^2\); since 3 is prime, \(3\mid p\). Put \(p=3k\) to obtain \(3\mid q\), a contradiction. C fails at \(n=3\). Tip: use the prime-divisor rule.

Correct answer: A. \(\sqrt{2}=\frac{p}{q}\), where \(p,q\) are integers, \(q\ne0\), and \(\gcd(p,q)=1\)

Explanation: The fraction must be in lowest terms. From \(2q^2=p^2\), \(p\) is even; substituting back shows that \(q\) is even, contradicting coprimality. Assuming both even already presumes the result. Exam tip: state \(q\ne0\).

Correct answer: A. Put \(p=3k\) to get \(q^2=3k^2\); hence \(3\mid q\) as well.

Explanation: \(3\mid p\) alone does not contradict coprimality. Put \(p=3k\): \(9k^2=3q^2\), so \(q^2=3k^2\) and \(3\mid q\). Thus both have a common factor 3. Exam tip: prove divisibility for both terms.