Proof of irrationality of square root 2 and square root 3

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.

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.

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.

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.”

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.

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

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.

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.

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.

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”.

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

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.

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

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.

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.

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.

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.

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

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.

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.

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.

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

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.

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 \(q\ne0\).

\(3\mid p\) 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.