Proof of irrationality of square root 2 and square root 3

In contradiction method, rationality is assumed first. Then squaring gives the contradiction that both are even.

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.

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.

Coprime numbers have no common factor except (1). Having common factor (3) is a contradiction.

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 \(q\).

Common factor (3) in both contradicts the coprime condition. Therefore (\sqrt{3}) is irrational.