Proof of irrationality of square root 2 and square root 3

In contradiction method we first assume the opposite that (\sqrt{2}) is rational. Later this assumption is shown false.

When assumed rational, (\sqrt{3}) is written as (\frac{p}{q}). Here (p) and (q) are integers.

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.

If an integer has an even square, the integer is also even. This is the key fact in the proof of (\sqrt{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.