Proof of irrationality of square root 2 and square root 3

Assume \(5+\sqrt{3}\) is rational. Subtracting 5 makes \(\sqrt{3}\) rational, a contradiction. Thus 5 being rational is insufficient. Exam tip: rationals are closed under subtraction.

From \(p^2=3q^2\), we get \(3\mid p^2\). Since 3 is prime, if it divides a square, it must divide its base \(p\); hence \(3\mid p\). Only after writing \(p=3k\) can we show \(3\mid q\), contradicting coprimality. Exam tip: use the prime-divisor property carefully.

The final contradiction completes only when both are divisible by (2). This breaks the coprime assumption.

The governing concept is proof by contradiction together with the lowest-form condition for a rational number. Assume √3 = p/q, where p and q are integers, q ≠ 0, and gcd(p,q) = 1. Squaring gives p² = 3q². From this equation, 3 divides p, so p = 3k; substitution then shows that 3 also divides q. Thus p and q have a common factor 3, contradicting gcd(p,q) = 1. Therefore the rational assumption must be rejected. Option A is incomplete because divisibility of p alone does not contradict lowest form; options C and D do not produce the required contradiction.

The direct answer is A. A square has an important prime-factor property: every prime factor occurs with an even exponent. For example, if cn=m^2c, then in the prime factorisation of n, each exponent is doubled. In the proof of csqrt{2}c, assume csqrt{2}=p/qc in lowest terms. Squaring gives cp^2=2q^2c. The right side contains one extra factor 2 in its prime-factor pattern, while the left side is a square and must have an even exponent of 2. Equivalently, the equation forces both p and q to be even, contradicting lowest terms. Option A states this hidden principle correctly. Option B is false because a fraction’s denominator must be non-zero, not zero. Option C is false because some square roots, such as csqrt{4}=2c, are integers, but not every square root is one. Option D is false because rational numbers may be odd, even or non-integers; “even” applies to integers. Memory cue: squares have even prime exponents.

In a perfect square, every prime exponent is even. This principle strengthens the proof of (\sqrt{3}).

From \(p^2=3q^2\), \(p^2\) is divisible by 3. Since 3 is prime, divisibility of \(p^2\) by 3 implies divisibility of \(p\) by 3. Put \(p=3k\); then \(q\) is also divisible by 3, contradicting lowest terms. Exam tip: use the prime-factor property for squares.

Since \(\sqrt{8}=2\sqrt{2}\), \(\frac{\sqrt{2}}{\sqrt{8}}=\frac12\), which is rational. But \(\sqrt{2}\sqrt{3}=\sqrt{6}\) is irrational and \(\sqrt{3}-\sqrt{3}=0\). Tip: simplify radicals first.

From \(p^2=3q^2\), \(p^2\), and hence \(p\), is divisible by 3. Substituting \(p=3k\) shows that \(q\) is also divisible by 3, contradicting lowest terms. Exam tip: look for a common factor contradiction.

The direct answer is A. In proving csqrt{3}c irrational, we first suppose csqrt{3}=p/qc, where p and q are integers, q is not zero, and p/q is in lowest terms. The condition cq\neq0c is needed simply because division by zero is undefined; it allows p/q to be a valid fraction. After squaring, cp^2=3q^2c, and divisibility reasoning shows that 3 divides p. Writing cp=3rc and substituting then shows that 3 also divides q. Thus p and q have a common factor 3, contradicting the coprime condition. Option A is correct because it distinguishes the definition condition from the contradiction-producing lowest-terms condition. Option B is wrong: q being non-zero alone does not prove cp=3kc. Option C reverses the result; coprimality does not give q=0. Option D is wrong because the two conditions serve different purposes. Memory cue: non-zero denominator makes the fraction legal; coprime numerator and denominator make the contradiction possible.

From \(p^2=3q^2\), \(p^2\) is divisible by 3, so \(p\) is divisible by 3. Substituting \(p=3k\) shows that \(q\) is also divisible by 3, contradicting lowest terms. Exam tip: use the prime-divisibility rule.

Only (p) being divisible by (3) is not enough. The contradiction completes only when (q) is also divisible by (3).

A rational number can be represented as a/b, but for this contradiction proof the fraction must be chosen in lowest terms, meaning gcd(a,b) = 1 and b ≠ 0. Starting with √2 = a/b and squaring gives a² = 2b². The parity argument then shows that a is even and b is also even. That conclusion is a contradiction only because the original representation was assumed to have no common factor. Without explicitly stating gcd(a,b) = 1, the proof has not established the condition that the final conclusion violates. Therefore option A is the best mathematical comment. The omission does not prove b = 0 or rationality, and it does not make the proof complete.

If \(r\) were rational, \(r^2\) would also be rational. From \(r^2=5+2\sqrt6\), \(\sqrt6=(r^2-5)/2\) would be rational, impossible because 6 is not a perfect square. Exam tip: retain the cross term \(2\sqrt6\).

If both are divisible by (2), numerator and denominator can be reduced by (2). This opposes lowest form.

A non-terminating decimal is not sufficient: \(\frac{1}{3}=0.333...\) is non-terminating but rational. An irrational decimal is non-terminating and non-recurring. Exam tip: check recurrence too.

In the proof of \(\sqrt{2}\), the number 2 is important because it is prime. Assuming \(\sqrt{2}=a/b\) in lowest form gives \(a^2=2b^2\). Since the square \(a^2\) is divisible by the prime 2, \(a\) itself must be divisible by 2. Writing \(a=2k\) and substituting back shows that \(b^2\), and hence \(b\), is also divisible by 2.

Therefore 2 becomes a common factor of both numerator and denominator. This contradicts the statement that \(a/b\) was in lowest form, or that \(\gcd(a,b)=1\). It does not make the denominator zero, prove \(a=b\), or prove rationality. Hence option A correctly describes the role of 2.

To prove that \(\sqrt{3}\) is irrational, assume the opposite and write \(\sqrt{3}=\frac{p}{q}\), where \(pq\) are coprime integers and \(q\neq0\). Squaring gives \(p^2=3q^2\). This shows that 3 divides \(p^2\), and because 3 is prime, it must divide \(p\). Write \(p=3k\).

Substitution gives \(9k^2=3q^2\), so \(q^2=3k^2\). Hence 3 also divides \(q\). Thus 3 is a common factor of both \(p\) and \(q\), contradicting their being coprime. Therefore option A correctly describes the role of 3: it produces the common factor and the final contradiction.

In the contradiction proof, we assume \(\sqrt{2}=a/b\), where \(a\) and \(b\) are coprime. The equation first shows that \(a\) is even; substituting this result back then shows that \(b\) is also even. Saying only that \(a\) is even does not contradict the coprime condition. The contradiction arises because both numbers would have 2 as a common factor. Exam tip: always state the final contradiction explicitly in a proof.

The proof begins by assuming √3 = p/q in lowest terms, so gcd(p,q) = 1. After squaring, p² = 3q², which implies that 3 divides p. However, that fact alone is not a contradiction: a numerator may be divisible by 3 while the denominator is not. The argument must continue by writing p = 3k and substituting into the equation. This leads to q² being divisible by 3, and hence q is divisible by 3 as well. Then p and q share the factor 3, contradicting gcd(p,q) = 1. Thus option A identifies the missing step; the other options either deny a valid result or introduce irrelevant conditions.

Since \(p^2=3q^2\), \(3\mid p^2\), and as 3 is prime, \(3\mid p\). Writing \(p=3k\) gives \(3\mid q\) too. Thus \(p\) and \(q\) are not coprime. Exam tip: start with a fraction in lowest terms.

The direct answer is B. The standard proof assumes, for contradiction, that \(\sqrt{3}=p/q\), where p and q are integers, q is nonzero, and \(\gcd(p,q)=1\). Squaring gives \(p^2=3q^2\). Hence 3 divides \(p^2\), so 3 divides p; write \(p=3k\). Substitution gives \(q^2=3k^2\), so 3 also divides q. Therefore both numerator and denominator are divisible by 3, contradicting the lowest-form condition. Option A reverses the conclusion: the equation is part of the assumption and eventually produces a contradiction, so it does not prove rationality. Option B correctly gives both the conclusion, irrationality, and its basis, the common factor 3. Option C is false because an integer being under a square root does not make its square root an integer. Option D is false because a contradiction does not mean \(\sqrt{3}=0\). Exam cue: for \(\sqrt{3}\), the common factor is 3, not 2.

Since 3 is prime, Euclid’s lemma gives \(3\mid p\) from \(3\mid p^2\). A conclusion about \(q\) follows only later after substituting \(p=3k\). In exams, state this prime-divisibility step clearly.

Since 3 is prime and \(3\mid a^2\), Euclid’s lemma gives \(3\mid a\). Writing \(a=3k\) then gives \(3\mid b\), contradicting coprimality. Exam tip: apply this converse directly only when the divisor is prime.

The contradiction comes from the way the rational number is initially represented. In a standard irrationality proof, assume √3 = p/q, where p and q are integers, q ≠ 0, and the fraction is in lowest terms. The last condition is expressed as gcd(p,q) = 1. If the algebra later proves that both p and q are divisible by 3, then 3 is a common divisor and gcd(p,q) is at least 3, not 1. This directly contradicts the initial lowest-form condition. Positivity of √3 is true but irrelevant, p = q was never assumed, and q = 0 is forbidden rather than an initial condition. Hence option C is correct.