Proof of irrationality of square root 2 and square root 3

First (p) is proved divisible by (3) from (p^2). Then after putting (p=3k), the conclusion for (q) follows.

Assume \(5+\sqrt{3}\) is rational. Subtracting the rational number 5 would then make \(\sqrt{3}\) rational, contradicting its irrationality. Exam tip: rational minus rational is always rational.

The relevant number-theory fact is Euclid’s lemma for a prime: if a prime divides the square of an integer, it divides the integer itself. In this proof, √3 = p/q leads to p² = 3q², so 3 divides p². Since 3 is prime, 3 must divide p, and we may write p = 3k for some integer k. This step is essential because substituting p = 3k into the equation subsequently shows that 3 also divides q, producing the contradiction with gcd(p,q)=1. Therefore option D is correct. Option A uses the wrong prime, while B and C do not follow from divisibility and are not valid number-theory conclusions.

From \(p^2=3q^2\), we get \(3\mid p^2\). Since 3 is prime, divisibility of a square by 3 implies \(3\mid p\). Substituting \(p=3k\) then makes \(q\) divisible by 3 too, contradicting coprimality. Exam tip: use the prime-divides-square rule in such proofs.

Every rational number can be represented as a quotient of integers, and any common factor can be cancelled. Consequently, for a proof by contradiction, we choose √3 = p/q in lowest form, meaning p and q are coprime and q ≠ 0. This choice is crucial: from p² = 3q² the proof shows that 3 divides p and then that 3 divides q. If p and q were already coprime, their both being divisible by 3 would be impossible, giving the required contradiction. Therefore option B is correct. Option A is an unsupported claim, C violates the definition of a fraction, and D is not required for a rational representation.

The correct answer is option D: both p and q have the common factor 3. Coprime integers are integers whose only positive common factor is 1. In the irrationality proof, p/q is deliberately chosen in lowest terms, so p and q must not share any factor other than 1. The equations first show that 3 divides p and then that 3 divides q. Therefore 3 is a common factor, which is impossible under the coprime assumption. Option A, q≠0, is not impossible; it is actually required so that p/q is defined. Option B, both are integers, is also required in a rational representation. Option C is simply the meaning of p in the chosen fraction and causes no contradiction. Option D is impossible because it violates the lowest-terms condition. This contradiction shows that the original assumption that sqrt{3} is rational must be false. Memory cue: ‘coprime’ means no common factor survives except 1.

In \(p^2=3q^2\), the right side is divisible by 3, so \(p^2\) is divisible by 3. Since 3 is prime, this implies that \(p\) is divisible by 3. Substitution then shows \(q\) is also divisible by 3, contradicting lowest terms. Exam tip: use prime divisibility of a square carefully.

If (p) has no factor (3), then (p^2) also has none. But the equation shows (p^2) divisible by (3).

To prove that \(\sqrt{2}\) is irrational, assume \(\sqrt{2}=p/q\), where \(p\) and \(q\) are coprime. Squaring gives \(p^2=2q^2\), so \(p\) is even. This then shows that \(q\) is also even, contradicting the fact that \(p\) and \(q\) are coprime. Hence, no decimal expansion is needed. Exam tip: In a contradiction proof, state the coprime condition clearly.

From \(p^2=3q^2\), \(p^2\) is divisible by 3. Since 3 is prime, divisibility of \(p^2\) by 3 implies that \(p\) is divisible by 3. Only after writing \(p=3k\) can we show that \(q\) is also divisible by 3, contradicting coprimality. Exam tip: use the prime-divisor property for squares.

The key idea is that a number is irrational when it cannot be written as a fraction of two integers in lowest form. The standard contradiction proofs assume that \(\sqrt{2}\) or \(\sqrt{3}\) is rational and then show that the numerator and denominator must share a prime factor. This contradicts the assumption that the fraction was already in lowest terms.

For \(\sqrt{2}\), the equation becomes \(p^2=2q^2\), forcing both \(p\) and \(q\) to be even. For \(\sqrt{3}\), it becomes \(p^2=3q^2\), forcing both to be divisible by 3. Thus neither square root is rational. Therefore option D, both are irrational, follows.

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

The fraction \(\frac{p}{q}\) is taken in lowest terms, so p and q are coprime. The proof gives \(3\mid p^2\Rightarrow3\mid p\), and then \(3\mid q\), a contradiction. Exam tip: always state “lowest terms.”

Squaring gives \(3\mid p^2\). Since 3 is prime, it must divide \(p\). Putting \(p=3k\) then shows that 3 divides \(q\) too, contradicting lowest terms. Exam tip: state the prime-divisor rule.

Assume \(\sqrt{2}=p/q\) with coprime integers. Squaring gives \(2q^2=p^2\); this eventually makes both integers even, contradicting coprimality. Exam tip: always state that the fraction is in lowest terms.

From \(p^2=3q^2\), \(p\) is divisible by 3, so let \(p=3k\). Substitution gives \(9k^2=3q^2\), hence \(q^2=3k^2\), so \(q\) is also divisible by 3. This contradicts coprimality. Exam tip: complete the argument by proving a common factor of both \(p\) and \(q\).

The correct answer is option B. Assume for contradiction that sqrt{3} is rational and write it as p/q in lowest terms, with p and q integers, q≠0, and no common factor. Squaring gives p²=3q². Since 3 divides p² and 3 is prime, 3 divides p. Write p=3k. Substitution gives 9k²=3q², so q²=3k². Applying the same rule again shows that 3 divides q. Thus both p and q are divisible by 3, contradicting their being coprime. Therefore the rational assumption is false and sqrt{3} is irrational. Option A is wrong because p=q is not assumed and would not prove irrationality. Option C is wrong because q=0 is forbidden, not a valid proof reason. Option D is wrong because the size of a decimal does not determine rationality; rational numbers can have large decimals, and irrationality requires proof. Option B gives both the reason and conclusion. Exam cue: contradiction with lowest terms proves irrationality.

\(1.732^2=2.999824\), 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.