Proof of irrationality of square root 2 and square root 3

Assuming \(\sqrt{2}=\frac{p}{q}\) gives \(p^2=2q^2\). Hence \(p\) is even, and then \(q\) is also even. They have a common factor 2, contradicting lowest terms. Exam tip: link the contradiction to coprime numerator and denominator.

Option C is a counterexample: 2 is an integer, but \(\sqrt{2}\) is neither an integer nor rational. If \(\sqrt{2}=p/q\), then \(p^2=2q^2\) makes both p and q even, a contradiction. Exam tip: one counterexample disproves a universal statement.

From \(p^2=3q^2\), \(p^2\) is divisible by 3. Since 3 is prime, divisibility of \(p^2\) by 3 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 such proofs.

A rational number can be written as a ratio of two integers. So (\sqrt{2}=\frac{m}{n}) is a rational assumption.

If a number is written as (\frac{r}{s}), it is being assumed rational. This is the initial assumption of the proof.

Since √12 = 2√3, assuming √12 rational gives √3 = √12 ÷ 2 as rational. This contradicts the known irrationality of √3, so √12 is irrational. Exam tip: A non-zero rational multiple of an irrational number is irrational.

The governing concept is proof by contradiction using a rational number in lowest terms. To assume √3 is rational, write it as r/s where r and s are integers, s is nonzero, and r and s are coprime. Squaring gives r² = 3s². From this relation, 3 divides r, so r = 3k; substitution then shows that 3 also divides s. Thus both r and s have 3 as a common factor, contradicting the original lowest-form condition that their HCF is 1. Option D correctly states this purpose. The aim is not to terminate a decimal, draw a diagram, or make the denominator zero. The lowest-form assumption is essential because without coprimality, common divisibility would not produce a contradiction.

The governing concept is an indirect proof of irrationality. First suppose, contrary to what is to be proved, that √2 is rational. Write √2 = m/n in lowest form, where m and n are integers, n is nonzero, and they are coprime. Squaring gives m² = 2n². This shows m is even; writing m = 2k and substituting then shows n is also even. That conclusion contradicts the lowest-form assumption, because m and n would share the factor 2. Hence the assumption is false and √2 is irrational. Option A gives this correct order. A decimal calculation is not the rigorous proof, and drawing or assuming zero has no role in the argument.

For (\sqrt{3}), after the rational assumption we square. Then both numbers becoming divisible by (3) gives a contradiction.

\(1.7^2=2.89\) is only close to 3, not equal to it. A rational number would need to have square exactly 3. In fact, \(1.7=17/10\), and its square is \(289/100\). Exam tip: never treat an approximation as an exact proof.

1.732 is only an approximation, not the exact value of \(\sqrt{3}\). Check: \((1.732)^2=2.999824\ne3\). In exams, never treat a close decimal approximation as an equality.

From \(p^2=3q^2\), \(p^2\) is divisible by 3, so the prime 3 divides \(p\). Put \(p=3k\); then \(q^2=3k^2\), so 3 also divides \(q\). This contradicts coprimality. Exam tip: remember \(r\mid p^2\Rightarrow r\mid p\) for a prime \(r\).

Both being divisible by (3) shows common factor (3). This contradicts the coprime assumption.

The governing concept is the lowest-form condition in a rational representation. When m/n is in lowest form, m and n are integers, n is nonzero, and they have no common factor greater than 1. If both m and n were even, each would be divisible by 2, so the fraction could be reduced by cancelling 2. That would prove it was not in lowest form. Therefore option D describes the impossible situation. The conditions in options A, B and C are all part of a valid rational representation: the denominator must not be zero, and numerator and denominator are integers. In the √2 proof, both being even is not an initial assumption; it is the contradiction derived from the equation m² = 2n² after assuming √2 rational.

A calculator shows 1.414 only as an approximation, not as the exact value. \(\sqrt{2}=1.414213\ldots\) is non-terminating and non-repeating, so it is irrational. Exam tip: never treat a rounded display as proof of rationality.

This proof uses evenness and the coprime condition instead of decimals. This is what should be written in exams.

The proof of (\sqrt{3}) does not use decimal approximation. It uses divisibility by (3) and contradiction.

The proof starts by assuming that \(\sqrt{2}\) is rational. Write it as \(\frac{p}{q}\) in lowest terms, with q nonzero. Squaring gives \(p^2=2q^2\). This makes \(p^2\) even, so p is even. Substituting \(p=2k\) into the equation gives \(4k^2=2q^2\), hence \(q^2=2k^2\), so q is also even.

The conclusion that both p and q are even contradicts the fact that \(\frac{p}{q}\) was chosen in lowest terms. This contradiction does not show that \(\sqrt{2}\) is non-real, non-positive, or unequal to a positive number. In fact, \(\sqrt{2}\) is a positive real number. What fails is only the starting assumption that it is rational. Therefore option C is correct.

In proof by contradiction, assume that \(\sqrt{3}=p/q\), where \(p\) and \(q\) are coprime integers. On squaring, \(3q^2=p^2\), which shows that \(p\), and then \(q\), are both divisible by 3. This contradicts their being coprime. Hence the assumption that \(\sqrt{3}\) is rational is rejected. \(\sqrt{3}\) is still positive and real; those facts are not rejected by the contradiction. Exam tip: when a contradiction is obtained, reject the initial assumption used in the proof.

If both \(a\) and \(b\) are divisible by 3, they have the common factor 3. This contradicts the assumption that they are coprime, so \(\sqrt{3}\) is irrational. Exam tip: state the common factor clearly in a contradiction proof.