Proof of irrationality of square root 2 and square root 3

Substituting \(a=2r\) gives \(a^2=4r^2\). Using this in \(a^2=2b^2\), we get \(4r^2=2b^2\), or \(b^2=2r^2\). Thus, \(b^2\) is even. If the square of an integer is even, the integer itself must be even; hence \(b\) is even. Being negative is not a reason for \(b\) to be even. Exam tip: this result is used to obtain a contradiction in irrationality proofs.

Given \(p=3k\), substitute it into \(p^2=3q^2\): \((3k)^2=3q^2\), so \(9k^2=3q^2\). Dividing by 3 gives \(q^2=3k^2\); hence \(q^2\) is divisible by 3. Since 3 is prime, if the square of an integer is divisible by 3, then the integer itself is divisible by 3. Therefore, \(q\) is divisible by 3. Exam tip: for a prime \(r\), \(r\mid n^2\) implies \(r\mid n\).

In (\sqrt{2}), (2) is the key factor, and in (\sqrt{3}), (3) is the key factor. Identify the difference by the factor.

At the start the fraction is in lowest form so (a) and (b) are assumed coprime. Both even is the final contradiction.

At the start (\frac{p}{q}) is in lowest form. Both divisible by (3) is derived later as a contradiction.

\(\sqrt{2}\) and \(\sqrt{3}\) are irrational because 2 and 3 are not perfect squares. But \(\sqrt{4}=2\) and \(\sqrt{9}=3\) are rational. Exam tip: the square root of a perfect-square integer is an integer.

As \(3\mid p^2\) and 3 is prime, \(3\mid p\). Put \(p=3k\): \(9k^2=3q^2\), so \(q^2=3k^2\) and \(3\mid q\). Thus both share 3, contradicting coprimality. Exam tip: state the prime-divisor rule clearly.

Assume for contradiction that √3 = p/q, where p and q are integers, q ≠ 0, and the fraction is in lowest terms. Squaring gives p² = 3q². This implies that 3 divides p², and therefore 3 divides p; writing p = 3k then gives q² = 3k², so 3 also divides q. Thus p and q have the common factor 3, contradicting the assumption that p/q was in lowest terms. The contradiction does not make √3 an integer, rational, or zero. Instead, it disproves the rational assumption. Therefore √3 is irrational, and option C is the correct final conclusion.

\(\sqrt{2}\) is irrational. If \(5+\sqrt{2}\) were rational, subtracting the rational number 5 would make \(\sqrt{2}\) rational, a contradiction. Hence the sum is irrational. Exam tip: isolate the irrational term by adding or subtracting a rational number.

In the proof of (\sqrt{3}), the chain uses divisibility by (3). It shows (q) is also divisible by (3).

A rational number is written in lowest form as a ratio of coprime integers. The denominator cannot be zero.

When a number is assumed to be rational, it can be represented as p/q with p and q integers, q ≠ 0, and the fraction reduced to lowest terms. Lowest terms means that p and q are coprime, so they have no common divisor greater than 1. This condition is essential because the proof later derives that both are divisible by 3, producing the contradiction. Option A describes the result obtained later, not the initial assumption. Options C and D are impossible because a denominator cannot be zero and p/q would not be a valid reduced representation if both were zero. Therefore option B correctly states the required starting condition.

Writing the fraction in lowest terms makes \(p\) and \(q\) coprime. From \(p^2=3q^2\), 3 divides \(p\), and then \(q\), creating a contradiction. Exam tip: state “lowest terms” explicitly.

The correct answer is option A: because 3 is a prime factor. Start with the fact that if a prime number divides a square, it must divide the number whose square was taken. Here, the proof gives that 3 divides p². The prime-factor rule therefore says that 3 divides p, so p can be written as p=3k for some integer k. This is not because p is zero, equal to q, or always even. Option A is correct because 3 is prime. Option B is wrong because p=0 is neither given nor suitable in the rational representation. Option C is wrong because p=q is not obtained and is unnecessary. Option D is wrong because p need not be even; divisibility by 3 has nothing to do with being even. Remember: for a prime r, if r divides n², then r divides n.

The square root of a rational number is not always rational. If \(\sqrt{3}=p/q\) in lowest form, then \(p^2=3q^2\), so 3 divides \(p\) and then \(q\), a contradiction. Exam tip: the square root of a non-perfect-square integer is irrational.

If \(\frac{3+\sqrt{2}}{5}\) were rational, multiplying by 5 and subtracting 3 would make \(\sqrt{2}\) rational, a contradiction. B, C and D equal 2, 1 and 0. Exam tip: use closure of rational numbers to test such expressions.

Assume \(\sqrt{3}=p/q\) in lowest terms. From \(p^2=3q^2\), \(p\) is divisible by 3, and then \(q\) is also divisible by 3. This contradicts coprimality. Exam tip: state “lowest terms” first.

Let \(p=3k\). Substituting in \(p^2=3q^2\) gives \(9k^2=3q^2\), so \(q^2=3k^2\). Hence \(q\) is also divisible by 3, contradicting that \(p\) and \(q\) are coprime. Exam tip: if a prime divides a square, it divides the number itself.

Since 3 is prime, \(3\mid a^2\) gives \(3\mid a\). Set \(a=3k\); then \(b^2=3k^2\), so \(3\mid b\), contradicting lowest terms. Exam tip: state the prime-divisor rule clearly.

Assuming \(\sqrt{3}=a/b\) gives \(a^2=3b^2\). Thus \(a^2\), hence \(a\), is divisible by 3; putting \(a=3k\) shows that \(b\) is also divisible by 3. This contradicts coprimality. Exam tip: if a square is divisible by 3, its root number is divisible by 3.

The proof uses contradiction. It begins by assuming that √2 is rational and can be written as a/b in lowest terms. The equation a² = 2b² then forces a to be even. Substituting a = 2r gives b² = 2r², so b is also even. This means that a and b share the factor 2, contradicting the claim that a/b was in lowest terms. Consequently, the assumption that √2 is rational is false, and the correct conclusion is that √2 is irrational. Positivity and reality are not disproved: √2 is a positive real number. Thus only option A identifies the statement rejected by the proof.

The direct answer is Option B: the assumption that √3 is rational. In a contradiction proof, we temporarily assume the opposite of what we want to prove. Suppose √3 = p/q, where p and q are coprime integers and q is not zero. Squaring gives p² = 3q². Therefore p² is divisible by 3, so p is divisible by 3; write p = 3r. Substitution gives q² = 3r², so q is also divisible by 3. This contradicts the statement that p and q have no common factor. Thus the temporary assumption that √3 is rational must be rejected, and √3 is irrational. Option A and Option D merely state positivity, which is true but is not the rejected assumption. Option C says it is real; that is also true and is not contradicted. Memory cue: reject the starting assumption that creates the impossible common factor.

From \(a^2=3b^2\), 3 divides \(a^2\), so since 3 is prime, it divides \(a\) first. Put \(a=3k\); then 3 also divides \(b\), contradicting coprimality. Exam tip: track the numerator first.

Assume √3 = p/q in lowest terms, with q ≠ 0. Squaring gives p² = 3q². Since 3 divides p², the prime-divisibility property implies that 3 divides p. Let p = 3k; substitution gives q² = 3k², so 3 divides q as well. This creates a common factor 3 in p and q, contradicting the lowest-terms assumption. Therefore the initial assumption that √3 is rational is false, and √3 is irrational. Option C merely repeats an intermediate equation and does not establish rationality. Options A and D confuse the contradiction with an unrelated conclusion. Hence option B gives both the correct conclusion and its reason.

Coprime numbers have 1 as their only common factor, so their HCF is 1. If both a and b are divisible by 2, then 2 is a common factor and they cannot be coprime. In contrast, one number may be even and the other odd, such as 2 and 3, which are coprime. Exam tip: For coprime numbers, check that no common factor other than 1 divides both numbers.