Proof of irrationality of square root 2 and square root 3

For side \(s\), \(s^2=2\), so \(s=\sqrt{2}\). If \(\sqrt{2}=a/b\) is in lowest terms, \(a^2=2b^2\) makes both \(a\) and \(b\) even, giving a contradiction. Exam tip: a rational area need not give a rational side.

The correct answer is A. In the prime factorisation of any perfect square, every prime occurs with an even exponent. Thus the left side a² must have an even exponent for the prime 2, as well as for every other prime. On the right side, 2b² contains the factor 2 outside the square b². If the exponent of 2 in b² is 2r, then its exponent in 2b² is 2r + 1, which is odd. The same integer cannot simultaneously have an even and an odd exponent of 2 in its unique prime factorisation. This contradiction is another way to see why the assumed rational representation of √2 is impossible. The other options make false claims about zero, negativity, or decimals.

In lowest terms, \(p\) and \(q\) are coprime. From \(p^2=3q^2\), we get \(3\mid p\), and then \(3\mid q\), creating the contradiction. Exam tip: show that the same prime divides both.

The correct answer is A. Writing √2 as approximately 1.414 or displaying more decimal digits only gives a numerical approximation. A finite decimal is rational, while an observed non-terminating pattern on a calculator does not by itself prove that no fraction equals the number; calculators also display rounded values. A formal school proof assumes √2 = a/b in lowest terms, squares to obtain a² = 2b², and then shows that both a and b must be even, contradicting their coprime status. Alternatively, a rigorous theorem about decimal expansions may be used, but merely copying digits is insufficient. Options B, C, and D assert conclusions that neither the decimal display nor the irrationality argument logically establishes.

Since \(p^2\) is even, \(p\) must be even. Put \(p=2k\): then \(q^2=2k^2\), so \(q\) is also even, contradicting coprimality. Exam tip: begin with the fraction in lowest terms.

Let \(\sqrt{3}=p/q\) be in lowest terms. From \(p^2=3q^2\), \(3\mid p^2\) implies \(3\mid p\); substitution then gives \(3\mid q\), a contradiction. Taking \(p=3\) disproves the other claims. Exam tip: use this rule for primes.

Since \(\sqrt{2}\) is irrational, its decimal expansion is non-terminating and non-repeating. A rational number \(p/q\) always has a terminating or repeating decimal. Exam tip: a non-terminating decimal is irrational only when it is also non-repeating.

From \(p^2=3q^2\), 3 divides \(p\). On writing \(p=3k\), it also follows that 3 divides \(q\), contradicting lowest terms. Exam tip: identify this common-factor contradiction.

2b² is even, hence a² is even. An odd integer has an odd square, so a is even. Substituting a=2k makes b even too, contradicting coprimality. Exam tip: state the odd-square fact.

From \(3q^2=p^2\), \(p^2\) is divisible by 3, so \(p\) is divisible by 3. Substituting \(p=3k\) shows that \(q\) is also divisible by 3, contradicting coprimality. Exam tip: use the prime-divisor property of a square.

Since 3 is prime and \(p^2=p\times p\), Euclid’s lemma gives \(3\mid p\) from \(3\mid p^2\). Applying the same idea to \(q\) contradicts coprimality. Exam tip: this inference specifically requires the divisor to be prime.

From \(p^2=3q^2\), \(3\mid p^2\), so \(3\mid p\). Substituting \(p=3k\) shows that \(q\) is also divisible by 3, contradicting coprimality. Exam tip: state the lowest-terms contradiction clearly.

From \(p^2=2q^2\), \(p^2\) is even, so \(p\) is even. Putting \(p=2k\) gives \(q^2=2k^2\), making \(q\) even too. This contradicts coprimality. Exam tip: if a square is even, its integer root is even.

Both even is not the assumption but the contradiction. At the start (\gcd(a,b)=1) should be assumed.

Both being divisible by (3) is derived later as contradiction. It is not assumed at the beginning.

The correct answer is A. In a contradiction proof, assume √2 or √3 equals p/q, where p and q are integers, q is nonzero, and the fraction is already in lowest terms, meaning gcd(p,q) = 1. For √2 the argument eventually shows both numerator and denominator are even; for √3 it shows both are divisible by 3. The contradiction is meaningful only because a lowest-form fraction cannot have a common prime factor. If the fraction is not reduced at the start, the later divisibility result may simply describe a non-reduced representation and no contradiction follows. A denominator cannot be zero, decimals are not a substitute for proof, and numerator and denominator need not be equal.

Putting \(p=3k\) gives \(9k^2=3q^2\), so \(q^2=3k^2\) and hence \(q\) is also divisible by 3. This contradicts coprimality. In exams, track prime factors in squares carefully.

The key idea is that an irrationality proof begins by temporarily assuming the opposite of what we want to prove. Here, we assume that \(\sqrt{3}\) is rational. A rational number can be written as \(p/q\), where \(p\) and \(q\) are integers, \(q\ne0\), and the fraction is in lowest terms. The contradiction finally shows that this assumption cannot be true.

Thus, the rejected assumption is “\(\sqrt{3}\) is rational,” which is option C. The proof does not reject the fact that \(\sqrt{3}\) is real; it is certainly a real number. Also, \(\sqrt{3}>0\) and \(q\ne0\) are valid facts or conditions used in the argument, not the assumption being disproved. Therefore the supplied answer is correct.

Since both have common factor (2), the fraction can be reduced. This contradicts the coprime lowest form.

The side is \(s=\sqrt{2}\) cm because \(s^2=2\). Thus, the perimeter is \(4s=4\sqrt{2}\) cm, which is irrational. \(2\sqrt{2}\) represents only two sides. Exam tip: the square root of a positive integer is rational only when the integer is a perfect square.