Proof of irrationality of square root 2 and square root 3

First (a) is found divisible by (3), and then (b) is also found divisible by (3). This creates the contradiction.

Both proofs start with rational assumption and a coprime fraction. Finally a common factor gives contradiction.

Since 3 is prime, \(3\mid p^2\) implies \(3\mid p\). On writing \(p=3k\), the proof further shows that \(q\) is also divisible by 3, contradicting that \(p\) and \(q\) are coprime. Exam tip: use this prime-factor property carefully.

The direct answer is B: assume that \\(\\sqrt{3}\\) is rational. To prove irrationality by contradiction, we temporarily assume the opposite of what we want to prove. Thus write \\(\\sqrt{3}=p/q\\), where p and q are integers, q is not zero, and the fraction is in lowest terms. Squaring gives \\(3q^2=p^2\\). This eventually leads to the conclusion that both p and q are divisible by 3, contradicting the assumption that p/q was in lowest terms. Therefore \\(\\sqrt{3}\\) is irrational. Option A, saying it is a perfect square, is not the starting assumption and is not the required opposite statement. Option B is correct because rationality is assumed temporarily. Option C, saying it is zero, is false since its square is 3. Option D, saying it is a natural number, is also false and is not the standard contradiction assumption. Memory cue: in an irrationality proof by contradiction, first assume rational, then derive an impossibility.

From \(3n^2=m^2\), \(m^2\), and hence \(m\), is divisible by 3. Putting \(m=3k\) shows that \(n\) is also divisible by 3, contradicting coprimality. Exam tip: identify the common factor causing the contradiction.

In proof by contradiction, assume \(\sqrt{2}=p/q\) in lowest terms. Squaring gives \(p^2=2q^2\), so \(p\), and then \(q\), must both be even. This contradicts coprimality. Exam tip: always state that the fraction is in lowest terms.

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: use prime divisibility of squares.

\(1.414\) is a terminating decimal and hence rational, but it is not exactly equal to \(\sqrt{2}\). Check: \(1.414^2=1.999396\ne2\). Thus it is only an approximation. In exams, distinguish “approximately equal” from “equal.”

In \(a^2=2b^2\), \(a^2\) is 2 multiplied by the integer \(b^2\). Hence \(a^2\) is divisible by 2, so it is even. The relation does not necessarily show that \(a^2\) is prime or zero. Exam tip: A number expressible as \(2\times\text{an integer}\) is even.

The equality is correct because \(\sqrt{12}=2\sqrt{3}\) and \(\frac{6}{2\sqrt{3}}=\sqrt{3}\). However, a rational number must be expressed as \(a/b\) with integers \(a,b\); the denominator here is irrational. Check the denominator in such claims.

If the square of an integer is even then the integer is also even. This is the key fact in the proof of (\sqrt{2}).

If prime factor (3) divides the square then it also divides the number. This is used in the proof of (\sqrt{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.”