Proof of irrationality of square root 2 and square root 3

(q\neq0) is necessary for a rational fraction. (\gcd(p,q)=1) breaks when both are divisible by (3).

The standard proof starts with rational assumption and derives contradiction that both are even. This is the correct order.

Since \(p^2=3q^2\), 3 divides \(p\). On writing \(p=3k\), we get \(q^2=3k^2\), so 3 divides \(q\) too. This contradicts coprimality. Tip: prove divisibility of \(p\) before substituting.

If \(r\sqrt{3}=q\) were rational, then since \(r\ne0\), \(\sqrt{3}=q/r\) would also be rational, a contradiction. Thus A is correct. Tip: a non-zero rational multiplier preserves irrationality.

Lowest form makes (p) and (q) coprime. This condition makes both divisible by (3) a contradiction.

From \(a^2=3b^2\), 3 divides \(a^2\), so the prime-divisibility property gives \(3\mid a\). Put \(a=3k\) to obtain \(3\mid b\) too, contradicting coprimality. In exams, state the prime-divisibility step clearly.

Both divisible by (3) is the contradiction found at the end. Initially (\gcd(p,q)=1) is assumed.

Assume \(a\sqrt{3}\) is rational, where \(a\ne0\) is rational. Then \(\frac{a\sqrt{3}}{a}=\sqrt{3}\) would be rational, contradicting its irrationality. Exam tip: always check that division is by a non-zero number.

A proof by contradiction temporarily assumes the opposite of the statement that is to be proved. Here the target is to show that \(\sqrt{3}\) is irrational, so the proof begins by assuming that \(\sqrt{3}\) is rational. It is then written as a fraction \(p/q\) in lowest terms, with \(q\neq0\). The algebraic argument eventually forces both \(p\) and \(q\) to be divisible by 3.

That conclusion conflicts with the choice of a lowest-terms fraction, so the temporary assumption cannot be true. The contradiction does not reject the fact that \(\sqrt{3}>0\), that it is real, or that the denominator is non-zero; those facts remain valid. It rejects only the assumption of rationality. Therefore option A is the correct choice.

Let \(5+\sqrt{3}=r\) be rational. Then \(\sqrt{3}=r-5\) is rational, a contradiction. Adding rational 5 cannot make it rational. Exam tip: state the subtraction step.

Since \(3\mid p^2\) and 3 is prime, \(3\mid p\). Put \(p=3k\); then \(q^2=3k^2\), so \(3\mid q\) as well. This contradicts that \(p\) and \(q\) are coprime. Exam tip: state the prime-factor property explicitly.

The assumption gives \(p^2=3q^2\). Thus 3 divides \(p^2\), so it divides \(p\); substituting back shows that 3 also divides \(q\). This contradicts coprimality. Exam tip: state the prime-divisor property clearly.

An approximate decimal is useful for estimating the size of \(\sqrt{3}\), but it cannot prove whether the number is rational or irrational. Any finite decimal is rational, and a displayed decimal approximation leaves infinitely many possible later digits. Even a long nonterminating-looking calculation cannot establish an exact mathematical property by itself. A proof needs a conclusion that follows with certainty.

The standard proof assumes \(\sqrt{3}=p/q\) in lowest form and obtains an equation such as \(p^2=3q^2\). Divisibility by 3 then forces a corresponding divisibility conclusion for the integers, eventually making both numerator and denominator divisible by 3. That contradicts lowest form. Therefore option A is correct: approximation alone does not provide a complete proof.

Assume coprime \(p,q\) in \(\sqrt{3}=p/q\). Then \(p^2=3q^2\) makes both divisible by 3, a contradiction. Taking both as multiples of 3 initially is invalid. Tip: use lowest terms.

From \(p^2=3q^2\), 3 divides p. Put \(p=3k\); then 3 also divides q. This contradicts p and q being coprime. The denominator need not be prime. Exam tip: state that the fraction is in lowest terms before deriving the contradiction.

Since 3 is prime, \(3\mid p^2\) implies \(3\mid p\). Putting \(p=3k\) then shows that \(q\) is also divisible by 3, contradicting coprimality. Exam tip: use the prime-factor rule for squares.

From \(p^2=2q^2\), \(p^2\), and hence \(p\), is even. Substituting \(p=2k\) shows that \(q\) is also even, contradicting coprimality. Both being odd gives no contradiction. Tip: always begin with lowest terms.

Because \(1.732^2=2.999824\ne3\), 1.732 is an approximation, not the exact value. A decimal point does not make a number irrational. Exam tip: square a claimed exact value.

Since \(2q^2\) is even, \(p^2\) is even. The square of an integer is even only if the integer itself is even; hence \(p\) is even. An even value of \(2q^2\) does not directly imply that \(q\) is even. Exam tip: apply parity rules to the correct factor.

From \(p^2=2q^2\), \(p\) is even. Put \(p=2k\); then \(q^2=2k^2\), so \(q\) is also even. This contradicts only a coprime pair. Exam tip: always begin with a fraction in lowest terms.