Proof of irrationality of square root 2 and square root 3

From \(p^2=3q^2\), 3 divides \(p^2\), so 3 divides \(p\); substituting back shows that 3 also divides \(q\). This contradicts coprimality. Exam tip: use the prime-factor property of a square carefully.

By Pythagoras’ theorem, \(d^2=1^2+1^2=2\), so \(d=\sqrt{2}\). It cannot be written as a ratio of integers. If both terms are even, the fraction is not in lowest form. Exam tip: write \(d^2\) first for a square’s diagonal.

Substituting \(a=2k\) into \(a^2=2b^2\) gives \((2k)^2=2b^2\), or \(4k^2=2b^2\). Dividing both sides by 2 gives \(b^2=2k^2\). The relation \(b^2=3k^2\) introduces an unjustified factor of 3, so it is incorrect. Exam tip: after substitution, expand the square first and then cancel common factors carefully.

Substitute \(p=3r\) into \(p^2=3q^2\). This gives \((3r)^2=3q^2\), or \(9r^2=3q^2\). Dividing both sides by 3 gives \(q^2=3r^2\). Hence, option B is correct. There is no basis for \(q^2=2r^2\). Exam tip: while substituting, remember that \((3r)^2=9r^2\).

Since the right-hand side of (b^2=2k^2) is a multiple of 2, (b^2) is even. The square of an integer is even only when the integer itself is even; hence (b) is even. If (b) were odd, its square would also be odd, so option A is incorrect. Exam tip: in a contradiction proof of irrationality, this step helps show that both integers have the common factor 2.

Given q² = 3r², q² is divisible by 3. Since 3 is prime, if 3 divides the square of an integer, it must also divide the integer itself. Therefore, q is divisible by 3. Divisibility of q² by 3 does not imply that q is even or zero. Exam tip: For a prime p, use p | a² ⇒ p | a.

From \(p^2=3q^2\), \(p^2\) is divisible by 3; since 3 is prime, \(p\) is divisible by 3. Put \(p=3k\) to obtain that \(q\) is also divisible by 3, contradicting coprimality. Exam tip: show this substitution clearly.

From \(2q^2=p^2\), \(p^2\), and hence \(p\), is even. Put \(p=2k\); then \(q^2=2k^2\), so \(q\) is also even. This contradicts coprimality. Exam tip: identify the common factor 2 to state the contradiction.

The governing concept is proof by contradiction, also called reductio ad absurdum. To prove a statement such as the irrationality of a square root, we temporarily assume the opposite—for example, that the number is rational and can be written as a fraction in lowest terms. Algebraic reasoning then produces a result that conflicts with the original lowest-form or divisibility condition. Since a valid assumption cannot logically lead to an impossibility, the assumed opposite statement must be false. Therefore, option A is correct. Being small, containing a decimal, or lacking a diagram has no role in deciding whether the assumption is false.

Being an integer is not enough; only perfect squares have integer square roots. If \(\sqrt{3}=p/q\), then \(p^2=3q^2\), so both \(p\) and \(q\) become divisible by 3, a contradiction. Exam tip: check perfect squares and prime factors.

If prime (3) divides the square, it divides the number too. This is the main step in the proof.

If \(\sqrt{3}=\frac{26}{15}\), squaring both sides would give \(26^2=3\times15^2\). But \(676\ne675\), so the claim is false. Exam tip: verify a claimed square root fraction by squaring it.

An irrational number cannot be written in the form \(p/q\), where \(p\) and \(q\) are integers and \(q\ne0\). Hence, proving that \(\sqrt{3}\) is irrational directly means that \(\sqrt{3}\) is not rational. It is neither an integer nor zero nor negative; in fact, \(\sqrt{3}>0\). Exam tip: remember that “irrational” means “not rational.”

From \(p^2=3q^2\), \(p^2\) is divisible by 3, so \(p\) is divisible by 3. Put \(p=3k\); then \(q^2=3k^2\), so \(q\) is also divisible by 3. This contradicts coprimality. Exam tip: look for a common factor in both terms.

First (\sqrt{3}) is assumed rational and written as (\frac{p}{q}). Then squaring gives (p^2=3q^2).

Assume \(\sqrt{3}=a/b\) for integers \(a,b\). Then \(a^2=3b^2\), so \(3\mid a^2\), which implies \(3\mid a\). Substitution then gives \(3\mid b\), creating the contradiction. Exam tip: remember that for a prime \(p\), \(p\mid a^2\Rightarrow p\mid a\).

Coprime numbers have no common factor except (1). Both becoming divisible by (3) clashes with this.

In the proof of (\sqrt{2}), it is first assumed rational. Then squaring gives the contradiction of both even.

The governing concept is the contradiction proof that √3 is irrational. Begin by assuming √3 is rational, so √3 = a/b where a and b are coprime integers and b is non-zero. Squaring gives 3b² = a². This shows that 3 divides a, so write a = 3k; substitution then shows that 3 also divides b. That contradicts the assumption that a and b have no common factor. Therefore the sequence in option B is the correct short order. A diagram, decimal approximation, zero assumption or subtraction does not establish irrationality rigorously.

From \(3q^2=p^2\), \(p\) must be divisible by 3. Putting \(p=3k\) then shows that \(q\) is also divisible by 3, contradicting coprimality. Exam tip: always assume the fraction is in lowest terms.