Proof of irrationality of square root 2 and square root 3

From \(2q^2=p^2\), \(p\) is even. Putting \(p=2k\) then shows \(q\) is also even, contradicting lowest terms. Exam tip: this contradiction is the key proof step.

Since 3 is prime, \(3\mid p^2\) implies \(3\mid p\). Put \(p=3k\); then \(q^2=3k^2\), so \(3\mid q\) too. This contradicts coprime terms. Exam tip: apply prime divisibility carefully.

Assume \(\sqrt{2}=p/q\) with coprime integers \(p,q\). Then \(p^2=2q^2\), so \(p^2\), and hence \(p\), is even. Putting \(p=2k\) makes \(q\) even too, contradicting coprimality. Exam tip: always begin the contradiction proof with the fraction in lowest terms.

The direct answer is Option A: √2 = a/b, where a and b are coprime integers and b ≠ 0. This is the standard form used when assuming a rational number. A rational number can be expressed as a ratio of two integers, and the fraction is taken in lowest terms so that the numerator and denominator have no common factor. The denominator cannot be zero because division by zero is undefined. Option A is correct. Option B is invalid because a/0 has no defined value. Option C, a + b, is a sum of integers, not the required general ratio form; it also does not set up the proof correctly. Option D, 2a, is simply twice an integer and does not represent the necessary rational fraction with a nonzero denominator. After using Option A, squaring leads to a² = 2b² and then to the contradiction that both a and b are even. Memory cue: rational means “integer over integer,” lowest terms, denominator not zero.

The correct answer is option B: sqrt{3}=p/q, where p and q are coprime integers and q≠0. To prove irrationality by contradiction, assume that sqrt{3} is rational. Every rational number can be expressed as a fraction of integers in lowest terms, so write sqrt{3}=p/q with q nonzero and with no common factor between p and q. Option B states exactly these conditions. Option A is impossible because division by zero is undefined. Option C is not the standard rational form and does not provide a denominator or the needed coprime condition. Option D is merely an expression involving p and does not represent the required general rational number. The lowest-terms condition is important because the proof later shows that both p and q are divisible by 3, creating a contradiction. Memory cue: rational assumption means ‘integer over nonzero integer, reduced fully’.

From \(p^2=3q^2\), 3 divides \(p^2\), so prime-factor reasoning gives \(3\mid p\). Put \(p=3k\) to obtain \(3\mid q\) too. This contradicts lowest terms. In exams, show both divisibility steps.

From \(p^2=2q^2\), \(p^2\) is even, so \(p\) is even. Put \(p=2k\); then \(q^2=2k^2\), making \(q\) even too. Thus both share factor 2, contradicting coprimality. Exam tip: an even square implies the number itself is even.

The standard proof of irrationality assumes, for contradiction, that the square root can be written as a fraction in lowest form. For \(\sqrt{2}\), this assumption leads to both numerator and denominator being divisible by 2. For \(\sqrt{3}\), it leads to both being divisible by 3. In each case, this contradicts the requirement that the fraction is already in lowest form.

Therefore neither \(\sqrt{2}\) nor \(\sqrt{3}\) can be rational. Both are irrational numbers, so option B is correct. They are also not integers, because 2 and 3 are not perfect squares. The supplied explanation correctly refers to the contradiction involving coprime numerator and denominator and reaches the correct common conclusion.

The governing concept is the parity property used in the irrationality proof of √2. From b² = 2r², the square b² is divisible by 2, so b² is even. An integer with an even square must itself be even; therefore b can be written as b = 2s for some integer s. Since the earlier step already gave a = 2r, both a and b are even. This contradicts the original assumption that a/b was in lowest terms, because a common factor 2 can be cancelled. Thus option A is the correct next conclusion. The equation does not show that b is odd, zero, or equal to a; none of those statements follows from the divisibility information.

(q^2) is divisible by (3) so (q) is also divisible by (3). This contradicts the coprime assumption.

If \(1+\sqrt{3}\) were rational, subtracting 1 would make \(\sqrt{3}\) rational, contradicting its irrationality. Option D is false since sums are not always irrational. Exam tip: isolate the square root in contradiction proofs.

Having common factor (3) breaks the condition of lowest fraction. Therefore (\sqrt{3}) cannot be rational.

Since \(p^2\) is divisible by 3, \(p=3k\). Substitution gives \(9k^2=3q^2\), so \(q^2=3k^2\); hence \(q\) is also divisible by 3. This contradicts coprimality. Exam tip: establish divisibility of both integers.

In lowest rational form, numerator and denominator are coprime. Later both becoming divisible by (3) gives the contradiction.

In contradiction method, the opposite of the statement is assumed first. So (\sqrt{2}) is assumed rational to get a contradiction.

The correct answer is option D. Begin with p²=3q². This equation says that p² is divisible by 3. Since 3 is prime, a prime divisor of a square must divide the original integer; therefore 3 divides p. Hence there is an integer k such that p=3k. Option D states both the exact fact and the correct conclusion. Option A is wrong because p=q was never given. Option B is wrong because q=0 is forbidden in the rational form p/q; also, it does not explain divisibility of p. Option C is wrong because being even concerns divisibility by 2, while the relevant prime here is 3. The next proof step substitutes p=3k into p²=3q², giving 9k²=3q² and then q²=3k². Thus q is also divisible by 3, contradicting that p and q were chosen coprime. Memory cue: square divisible by a prime means the base is divisible by that prime.

From \(p^2=3q^2\), 3 divides \(p^2\), so 3 divides \(p\). Put \(p=3k\); then 3 also divides \(q\). This contradicts \(p/q\) being in lowest terms. Exam tip: lowest terms means \(p\) and \(q\) are coprime.

(q^2) is divisible by (3), so (q) is also divisible by (3). This gives common factor (3) in both (p) and (q).

From \(p^2=3q^2\), \(p^2\) 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: if a prime divides \(p^2\), it divides \(p\).

If both \(p\) and \(q\) are divisible by \(3\), then they have \(3\) as a common factor. Hence, they cannot be coprime. In the irrationality proof of \(\sqrt{3}\), \(p\) and \(q\) are assumed to be coprime; showing that both are divisible by \(3\) gives the required contradiction. Therefore, option C is the wrong conclusion. Exam tip: coprime numbers always have HCF \(1\).

50 is not a perfect square, and \(50=25\times2\), so \(\sqrt{50}=5\sqrt{2}\). Since \(\sqrt{2}\) is irrational, \(\sqrt{50}\) is also irrational. In contrast, 16, 36, and 64 are perfect squares. Exam tip: check for perfect squares first.

From (q^2=3k^2), (q) is also divisible by (3). Both (p) and (q) divisible by (3) contradicts the coprime condition.

To prove √2 irrational, assume √2 = m/n, where m and n are coprime integers and n is non-zero. Squaring gives m² = 2n². The right side is even, so m² is even; consequently m is even. Write m = 2r. Substitution gives 4r² = 2n², hence n² = 2r², so n is also even. Thus m and n share a factor 2, contradicting the assumption that the fraction was in lowest terms. Option A presents this essential chain. Option B belongs to the analogous √3 argument, while C and D omit or contradict the required reasoning.

For the contradiction proof of √3, assume √3 = p/q in lowest terms, with q not equal to zero. Squaring gives p² = 3q². Therefore p² is divisible by the prime 3, and the prime-divisibility property implies that p is divisible by 3. Write p = 3k. Substituting and simplifying gives 3k² = q², so q², and hence q, is also divisible by 3. Both p and q then have a common factor 3, contradicting lowest terms. Option C states this correct chain. Option A uses the factor 2 and belongs to √2; the remaining options do not establish the contradiction.

From q² = 3k², 3 divides q². Since 3 is prime, it must divide q. Also, p = 3k shows that 3 divides p, contradicting that p and q are coprime. Exam tip: if a prime divides a square, it divides the original number.