Proof of irrationality of square root 2 and square root 3

The direct answer is option A. Under the rational assumption, \(\sqrt{3}=p/q\) gives \(p^2=3q^2\). Suppose, as the option says, that 3 does not divide \(p\). Then 3 also cannot divide \(p^2\): a prime factor appears in a square only when it already appears in its base. However, the equation \(p^2=3q^2\) has a factor 3 on the right, so the right side is divisible by 3. Equality would then say that \(p^2\) is divisible by 3, creating a contradiction. Therefore the assumption that 3 does not divide \(p\) is impossible; 3 must divide \(p\), and the proof continues to show that it also divides \(q\). Option A is correct. Option B is wrong because the equation does not directly prove \(q=0\). Option C is wrong because it does not prove \(p=q\). Option D is wrong because it does not make \(\sqrt{3}=1\). The key idea is divisibility, not guessing values. Memory cue: check whether a prime factor on one side must appear in the square on the other side.

The proof begins by assuming, for contradiction, that √2 can be written as a rational number p/q in lowest form, where p and q are integers, q ≠ 0, and gcd(p,q) = 1. Rearranging and comparing prime factors shows that both p and q must be divisible by 2. That contradicts the choice that the fraction was already in lowest form. In a proof by contradiction, the contradiction rejects the original assumption, not the valid algebraic steps or the fact that √2 is real and positive. Hence the correct conclusion is that √2 is irrational. Option A states the opposite, while C and D do not express the logical conclusion.

The rational assumption makes both (p) and (q) divisible by (3). This contradiction proves (\sqrt{3}) irrational.

In both proofs, the opposite is assumed and contradiction with coprime condition is derived. Hence contradiction method is correct.

\(1.732\) is a terminating decimal and hence rational, but it is only an approximation to \(\sqrt{3}\). Check: \(1.732^2=2.999824\), not \(3\). In exams, carefully distinguish an exact equality from an approximation.

Although \((\sqrt{2})^2=2\) is rational, the converse statement is invalid. Assuming \(\sqrt{2}=p/q\) in lowest terms gives \(p^2=2q^2\), making both \(p\) and \(q\) even—a contradiction. Exam tip: check whether a converse is justified.

The direct answer is Option A: n also becomes even. Start with m² = 2n². Because the right side is divisible by 2, m² is even, and therefore m is even. Write m = 2r. Substitute this into the equation: (2r)² = 2n², so 4r² = 2n². Dividing both sides by 2 gives n² = 2r². Thus n² is divisible by 2, so n is also even. The proof has now shown that both m and n are even. But the fraction m/n was assumed to be in lowest terms, meaning m and n should have no common factor. Their common factor 2 is the contradiction. Option A is correct. Option B, n = 0, does not follow and would make the original rational form invalid. Option C, m = n, is not derived. Option D, √2 = 0, is false because √2 is positive. Memory cue: in this proof, “one even” leads to “the other even,” contradicting coprimality.

From \(a^2=2b^2\), \(a\) is even; write \(a=2k\). Substitution gives \(b^2=2k^2\), so \(b\) is also even. Thus both share factor 2, contradicting coprimality. Exam tip: state the lowest-terms contradiction clearly.

Assume \(\sqrt{3}=p/q\) with coprime integers \(p,q\). From \(p^2=3q^2\), first \(p\), and then \(q\), is divisible by 3. This contradicts coprimality. Exam tip: identify the common factor obtained in both numerator and denominator.

From \(3q^2=p^2\), 3 divides \(p^2\), so it divides \(p\). Putting \(p=3k\) then shows that 3 also divides \(q\). This contradicts coprimality. Exam tip: always assume the fraction is in lowest terms first.

In both proofs, the rational assumption clashes with the coprime condition. Therefore both (\sqrt{2}) and (\sqrt{3}) are irrational.

From \(3q^2=p^2\), \(p^2\) is divisible by 3, so \(p\) is divisible by 3. Substitution then shows that \(q\) is also divisible by 3, contradicting lowest terms. Exam tip: check for a common factor in both numerator and denominator.

Put \(p=3k\). Then \(3q^2=p^2=9k^2\), so \(q^2=3k^2\) and \(q\) is also divisible by 3. Thus \(p\) and \(q\) have a common factor, contradicting lowest terms. Exam tip: state the common-factor contradiction clearly.

In option C, the number of zeros between 1s keeps increasing, so no fixed repeating block occurs. An irrational number has a non-terminating, non-repeating decimal expansion. A and D terminate, while B repeats. Exam tip: terminating or recurring decimals are rational.

Assume \(\sqrt{2}=p/q\) in lowest terms. From \(p^2=2q^2\), p is even; writing \(p=2k\) then shows q is even too. This contradicts coprimality. Exam tip: begin with lowest terms.

For contradiction, assume \(\sqrt{3}\) is rational in lowest form \(p/q\). Squaring gives \(p^2=3q^2\), which ultimately makes both \(p\) and \(q\) divisible by 3. Exam tip: the coprime condition creates the contradiction.

From p² = 3q², p is divisible by 3. Substituting p = 3k gives q² = 3k², so q is also divisible by 3. This contradicts p and q being coprime. Exam tip: always state that p/q is in lowest terms.

Since \(\sqrt{12}=2\sqrt{3}\), we get \(\sqrt{3}=\frac{\sqrt{12}}{2}\). If \(\sqrt{12}\) were rational, division by 2 would make \(\sqrt{3}\) rational, a contradiction. Exam tip: state the contradiction explicitly.

The correct answer is option D: q is divisible by 3. From q²=3k², the right side is divisible by 3, so q² is divisible by 3. Because 3 is prime, divisibility of the square implies divisibility of its base; therefore 3 divides q. We may write q=3m for some integer m. Option A is wrong because the equation concerns divisibility by 3, not by 2, so q need not be even. Option B is wrong because no sign information is given; q may be positive or negative, and in a fraction q is simply nonzero. Option C is wrong because q=0 is not allowed in p/q and was never deduced. Option D is correct and is the second half of the contradiction: earlier p=3k showed that p is divisible by 3, and now q is also divisible by 3. Thus p and q share a common factor 3, contradicting their being coprime. Memory cue: repeat the same prime-divisor rule for q².

In a contradiction proof, assume \(\sqrt{2}=p/q\) in lowest terms, with coprime integers \(p,q\). From \(p^2=2q^2\), both \(p\) and then \(q\) are even, contradicting coprimality. Exam tip: always state that the fraction is in lowest terms.

In lowest form, there should be no common factor except (1). If both are even, (2) is common.

To prove √3 irrational, assume that √3 = p/q, where p and q are integers, q ≠ 0, and the fraction is in lowest form, so gcd(p,q) = 1. Squaring gives p² = 3q². Because 3 is prime and divides p², it must divide p; write p = 3k. Substitution then gives 9k² = 3q², so q² = 3k², which implies that 3 divides q as well. Thus both p and q have the common factor 3, contradicting the lowest-form condition. Therefore option D identifies the contradiction. Options A, B, and C are assumptions used in the proof, not contradictions.

The direct answer is option A. In a proof by contradiction, we begin by temporarily assuming the opposite of what we want to prove: suppose \(\sqrt{2}\) is rational. Then write it in lowest terms as \(a/b\), where \(a\) and \(b\) have no common factor. Squaring gives \(2=a^2/b^2\), so \(a^2=2b^2\). Thus \(a^2\) is even, so \(a\) is even; write \(a=2k\). Substitution shows \(b^2=2k^2\), so \(b\) is also even. Both then have a common factor 2, contradicting lowest terms. Therefore \(\sqrt{2}\) is irrational. Option A lists the correct order: assume rational, square, and obtain the both-even contradiction. Option B is insufficient because a decimal expansion alone does not prove irrationality. Option C is a construction activity, not this proof. Option D assumes an unrelated equality and supplies no contradiction. Exam cue: always include the lowest-terms assumption and the final contradiction.

For (\sqrt{3}), we first assume rationality and square. Then divisibility by (3) conflicts with coprime condition.

Any integer is of the form 3q, 3q+1, or 3q+2. The last two forms have squares leaving remainder 1 on division by 3, so a square divisible by 3 requires the integer to be 3q. Exam tip: use remainders.