Proof of irrationality of square root 2 and square root 3

\(1.732^2=2.999824\), not 3, so it is only an approximation. If \(\sqrt{3}=p/q\), then \(p^2=3q^2\) makes both \(p\) and \(q\) divisible by 3, contradicting lowest terms. Exam tip: never treat a rounded decimal as an exact value.

The equation p = 3k is the standard algebraic way to express that 3 divides p. In the proof, p² = 3q² first shows that 3 divides p². Since 3 is prime, a prime-divisibility result gives 3 | p, so there is an integer k such that p = 3k. Substituting this form into p² = 3q² gives 9k² = 3q² and hence q² = 3k², which then shows that 3 also divides q. This eventually contradicts the assumption that p and q are coprime. Thus option D is correct. The equation does not say p is zero, equal to q, or even; those claims are not implied by p = 3k.

Since \(3q^2=p^2\), \(p^2\) is divisible by 3. As 3 is prime, \(p\) must be divisible by 3. Substituting \(p=3k\) later makes \(q\) divisible by 3 too, contradicting coprimality. Exam tip: use prime divisibility of squares.

In proof by contradiction, assume \(\sqrt{2}=\frac{p}{q}\) in lowest terms. If both are even, write \(p=2m, q=2n\); cancelling 2 gives a smaller equivalent fraction, a contradiction. Exam tip: always state that \(p/q\) is in lowest terms.

A proof by contradiction begins by temporarily assuming the statement opposite to the desired conclusion. To prove that √2 is irrational, we assume that √2 is rational and write it as p/q in lowest form. The algebra and divisibility arguments then force both p and q to be divisible by 2, contradicting their coprimality. The contradiction therefore rejects the temporary assumption that √2 is rational. It does not reject the facts that √2 is real, positive, or greater than zero; those facts remain true. Hence option C is correct. Options A, B, and D describe valid properties of √2 and are not the assumption targeted by the contradiction.

From \(p^2=2q^2\), \(p^2\), and hence \(p\), is even. Put \(p=2k\); then \(q^2=2k^2\), so \(q\) is also even. This contradicts coprimality. Exam tip: establish evenness for both integers.

In the rational assumption, (\sqrt{2}) is written as (\frac{m}{n}) in lowest form. Therefore (m) and (n) are coprime integers.

When (\sqrt{3}) is assumed rational, (\frac{p}{q}) is taken in lowest form. Therefore (p) and (q) are coprime integers.

Assume that \(\sqrt{2}=\frac{m}{n}\), where m and n have no common factor. Squaring gives \(m^2=2n^2\). Since the right side is even, \(m^2\) is even, and therefore m is even. Write \(m=2k\). Substitution gives \(4k^2=2n^2\), so \(n^2=2k^2\), which means n is also even.

Thus the proof’s important middle objective is to establish, in sequence, that m is even and then n is even. If both are even, they share the factor 2, contradicting the assumption that \(\frac{m}{n}\) was in lowest terms. The proof does not aim to show m equals n, either variable is zero, or \(\sqrt{2}=2\). Therefore option C correctly describes the central intermediate step.

Both being divisible by (3) gives contradiction with coprime condition. This is the middle goal of the proof.

From \(a^2=2b^2\), the right side is even, so \(a^2\) is even. The valid standard conclusion is that a is even; write \(a=2k\). Substituting gives \(4k^2=2b^2\), and after dividing by 2 we get \(b^2=2k^2\). This shows that b is even as well. These parity conclusions are enough to produce the contradiction when a and b were initially assumed to have no common factor.

It is not valid to conclude directly that \(a=2b\). Taking square roots would give a relation involving \(\sqrt{2}\), not the equation \(a=2b\); moreover, the variables need not have that particular relationship. The correct step is to infer that a is divisible by 2, then use substitution to infer the same for b. Therefore option A identifies the mistake correctly.

From (p^2=3q^2), we only get that (p^2) is divisible by (3). Then by the rule, (p) is written as (3k).

Assume \(\sqrt{3}=a/b\) in lowest terms. Then \(a^2=3b^2\), so \(3\mid a^2\), which implies \(3\mid a\). This subsequently gives \(3\mid b\), a contradiction. Exam tip: remember that for prime \(p\), \(p\mid a^2\Rightarrow p\mid a\).

Since both have common factor (3), the fraction can be reduced. So the lowest form assumption breaks.

Using contradiction, \(3q^2=p^2\) implies that 3 divides \(p\), and then it also divides \(q\), contradicting coprimality. Exam tip: a non-terminating recurring decimal can still be rational.

The decimal expansion of \(\frac{1}{3}\) is infinite but recurring, and it is a ratio of integers, so it is rational. \(\sqrt{2}\), \(\sqrt{3}\), and \(\pi\) are irrational. Exam tip: every recurring decimal represents a rational number.

From \(3q^2=p^2\), \(p^2\) is divisible by 3, so \(p\) is divisible by 3. Put \(p=3k\); then \(q^2=3k^2\), making \(q\) divisible by 3 too. This contradicts coprimality. Exam tip: use the prime-divisibility property of squares.

Option A gives the correct relation. Since 3 is prime, if \(p^2\) is divisible by 3, then \(p\) must also be divisible by 3. Therefore, we can write \(p=3k\), where \(k\) is an integer. In the irrationality proof of \(\sqrt{3}\), this ultimately shows that \(q\) is also divisible by 3, contradicting the assumption that \(p\) and \(q\) are coprime. Exam tip: For a prime \(r\), remember that \(r\mid a^2\) implies \(r\mid a\).

The governing idea is proof by contradiction. To test whether √2 or √3 can be rational, we assume it equals a fraction a/b, where a and b are integers, b is non-zero, and gcd(a,b)=1. The lowest-form condition is essential: after squaring and using divisibility, the proof generally shows that both a and b must be even. That is impossible for a coprime pair because a common factor 2 would remain. Therefore the original rational assumption fails. Option A is correct. A zero denominator is forbidden, a decimal approximation cannot establish irrationality, and equal numerator and denominator is not part of the argument.

A calculator shows only a rounded approximation. Since \(1^2<3<2^2\), \(\sqrt{3}\) lies between 1 and 2, but its decimal expansion is non-terminating and non-repeating, so it is irrational. Exam tip: never treat a finite display as the exact decimal expansion.