Proof of irrationality of square root 2 and square root 3

The direct answer is option C. To prove \(\sqrt{3}\) irrational, use proof by contradiction. First assume that it is rational. By the definition of a rational number, it can be written as \(\sqrt{3}=p/q\), where \(p\) and \(q\) are integers, \(q\neq0\), and the fraction is in lowest form. Squaring gives \(3=p^2/q^2\), or \(p^2=3q^2\). This eventually forces both \(p\) and \(q\) to be divisible by 3, contradicting their being coprime. Option C states the necessary first rational assumption. Option A is wrong because \(\sqrt{3}\neq3\); its square is 3, not the number itself. Option B is wrong because \(\sqrt{3}\) is positive and not zero. Option D is wrong because \(\sqrt{3}<0\) is false. The first step is only an assumption, and the contradiction comes later. Memory cue: rational numbers are written as an integer fraction with a nonzero denominator.

The governing concept is the lowest or simplest form of a rational number. A rational number is written as p/q with q not equal to zero, and the fraction is in lowest form when p and q have no common factor greater than 1. Equivalently, their highest common factor is 1, so p and q are coprime. Therefore option A is correct. They do not have to be both even; in fact, if both were even, the fraction could be reduced further. They also need not both equal 3, and the denominator cannot be zero because division by zero is undefined. For example, 6/15 is not in lowest form because both terms share 3, whereas 2/5 is in lowest form because 2 and 5 have HCF 1. The signs of the integers do not change this coprime condition.

Being an integer does not guarantee that its square root is rational. \(\sqrt{2}\) cannot be written as a ratio of two integers, so it is irrational. Exam tip: only perfect squares have integer square roots.

\(1.732\) is only an approximation of \(\sqrt{3}\), not its exact value. In fact, \((1.732)^2=2.999824\ne3\). A rational number must have an exact terminating or recurring decimal expansion. Exam tip: never treat an approximation as a proof.

Squaring both sides of \(\sqrt{3}=\frac{r}{s}\) gives \(3=\frac{r^2}{s^2}\). Multiplying both sides by \(s^2\) gives \(r^2=3s^2\), so option C is correct. In option B, the relationship is reversed. Exam tip: after squaring an equation involving a fraction, multiply by the square of the denominator to remove the fraction.

In proof by contradiction, \(\sqrt{3}=\frac{p}{q}\) is taken in lowest terms, so \(p\) and \(q\) must be coprime. Then \(p^2=3q^2\) implies both are divisible by 3, giving a contradiction. Exam tip: lowest terms means coprime.

Given \(r^2=3s^2\). For integers, \(s^2\) is an integer, so \(r^2\) is the product of 3 and the integer \(s^2\). Hence, \(r^2\) is divisible by 3. The equation does not necessarily imply that \(r^2\) is divisible by 2. Exam tip: To prove divisibility by \(k\), express the number as \(k\times\) an integer.

The direct answer is option B, even. Let x be an integer. Every integer is either even or odd. If x is even, x = 2k for some integer k, and x² = 4k² = 2(2k²), so its square is even. Conversely, if x were odd, x = 2k + 1, and x² = 4k² + 4k + 1 = 2(2k² + 2k) + 1, which is odd. Therefore an integer with an even square cannot be odd; it must be even. Option A is wrong because an odd integer always has an odd square. Option B is correct. Option C, prime, is unrelated: the integer could be 2, 4, 6 or another composite even number. Option D, negative, is not implied; 2 and 0 are non-negative examples. This fact is used in the contradiction proof for √2.

If prime factor (3) divides a square, it divides the number too. Therefore (x) is divisible by (3).

From \(3q^2=p^2\), \(p^2\), hence \(p\), is divisible by 3. Putting \(p=3k\) shows that \(q\) is also divisible by 3, contradicting coprimality. Exam tip: a common factor in both numerator and denominator gives the contradiction.

Assume \(\sqrt{2}=\frac{p}{q}\) with coprime integers \(p,q\). From \(p^2=2q^2\), \(p\) is even, and then \(q\) is also even. This contradicts the fraction being in lowest terms; \(p^2\) being even alone is not a contradiction. Exam tip: state the coprime condition clearly.

Given \(m=2k\), substitute it into \(m^2=2n^2\). This gives \((2k)^2=2n^2\), or \(4k^2=2n^2\). Dividing both sides by 2, we get \(n^2=2k^2\). Hence, option B is correct. In \(n^2=k^2\), the factor 2 has been incorrectly omitted. Exam tip: when substituting a term that is squared, square its coefficient as well.

Given \(r=3t\), substitute it into \(r^2=3s^2\). This gives \((3t)^2=3s^2\), or \(9t^2=3s^2\). Dividing both sides by 3 gives \(s^2=3t^2\). Hence, option C is correct. \(t=0\) does not necessarily follow. Exam tip: After substitution, square the expression correctly and then simplify the coefficients.

1.732 is only an approximate decimal value of \(\sqrt{3}\), not its exact value. In fact, \(1.732^2=2.999824\), not 3. Exam tip: distinguish a terminating approximation from an exact decimal expansion.

Since (s^2) is divisible by (3), (s) is also divisible by (3). This leads to the final contradiction.

\(\frac{6}{10}=\frac{3}{5}\), and \(\left(\frac{3}{5}\right)^2=\frac{9}{25}\), not 3. A number equals \(\sqrt{3}\) only if its square is 3; being close is not enough. Exam tip: square a claimed value to verify it.

From \(p^2=2q^2\), \(p^2\) is even, so \(p\) must be even. Substituting \(p=2k\) gives \(q^2=2k^2\), hence \(q\) is also even. This contradicts coprimality. Tip: an even square has an even root.

An infinite decimal alone does not imply irrationality: \(1/3=0.333\ldots\) is rational. For \(\sqrt{3}\), assume \(p/q\) is in lowest terms; the proof shows that both \(p\) and \(q\) are divisible by 3, a contradiction. Exam tip: distinguish non-terminating recurring decimals from non-recurring decimals.

From \(p^2=3q^2\), \(p^2\) is divisible by 3, so the prime-factor rule gives \(p=3k\). Substitution gives \(q^2=3k^2\), making \(q\) divisible by 3 too. This contradicts coprimality. Exam tip: use the prime divisibility rule for squares.

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.