Proof of irrationality of square root 2 and square root 3

If \(a\) is odd, then its square \(a^2\) is also odd. On the other hand, \(2b^2\) is even for every integer value of \(b\), since it has 2 as a factor. Thus, the equation would make the same number both odd and even, which is the contradiction. Option B is not valid because the equation does not prove that \(b^2\) is odd. Exam tip: the square of an odd integer is odd, and the square of an even integer is even.

For a prime \(p\), \(p\mid n^2\Rightarrow p\mid n\). From \(a^2=2b^2\), 2 divides \(a\), and then \(b\). But \(p^2\mid n\) need not hold. Exam tip: state this lemma first.

\(\sqrt{12}=2\sqrt3\). A non-zero rational times an irrational is irrational, so A works. B wrongly uses \(\sqrt{a+b}=\sqrt a+\sqrt b\). Exam tip: extract perfect-square factors first.

For side \(s\), \(s^2=2\), so \(s=\sqrt{2}\). If \(\sqrt{2}=a/b\) is in lowest terms, \(a^2=2b^2\) makes both \(a\) and \(b\) even, giving a contradiction. Exam tip: a rational area need not give a rational side.

In lowest terms, \(p\) and \(q\) are coprime. From \(p^2=3q^2\), we get \(3\mid p\), and then \(3\mid q\), creating the contradiction. Exam tip: show that the same prime divides both.

Since \(p^2\) is even, \(p\) must be even. Put \(p=2k\): then \(q^2=2k^2\), so \(q\) is also even, contradicting coprimality. Exam tip: begin with the fraction in lowest terms.

Let \(\sqrt{3}=p/q\) be in lowest terms. From \(p^2=3q^2\), \(3\mid p^2\) implies \(3\mid p\); substitution then gives \(3\mid q\), a contradiction. Taking \(p=3\) disproves the other claims. Exam tip: use this rule for primes.

Since \(\sqrt{2}\) is irrational, its decimal expansion is non-terminating and non-repeating. A rational number \(p/q\) always has a terminating or repeating decimal. Exam tip: a non-terminating decimal is irrational only when it is also non-repeating.

From \(p^2=3q^2\), 3 divides \(p\). On writing \(p=3k\), it also follows that 3 divides \(q\), contradicting lowest terms. Exam tip: identify this common-factor contradiction.

2b² is even, hence a² is even. An odd integer has an odd square, so a is even. Substituting a=2k makes b even too, contradicting coprimality. Exam tip: state the odd-square fact.

From \(3q^2=p^2\), \(p^2\) is divisible by 3, so \(p\) is divisible by 3. Substituting \(p=3k\) shows that \(q\) is also divisible by 3, contradicting coprimality. Exam tip: use the prime-divisor property of a square.

Since 3 is prime and \(p^2=p\times p\), Euclid’s lemma gives \(3\mid p\) from \(3\mid p^2\). Applying the same idea to \(q\) contradicts coprimality. Exam tip: this inference specifically requires the divisor to be prime.

From \(p^2=3q^2\), \(3\mid p^2\), so \(3\mid p\). Substituting \(p=3k\) shows that \(q\) is also divisible by 3, contradicting coprimality. Exam tip: state the lowest-terms contradiction clearly.

From \(p^2=2q^2\), \(p^2\) is even, so \(p\) is even. Putting \(p=2k\) gives \(q^2=2k^2\), making \(q\) even too. This contradicts coprimality. Exam tip: if a square is even, its integer root is even.

Both even is not the assumption but the contradiction. At the start (\gcd(a,b)=1) should be assumed.

Both being divisible by (3) is derived later as contradiction. It is not assumed at the beginning.

Putting \(p=3k\) gives \(9k^2=3q^2\), so \(q^2=3k^2\) and hence \(q\) is also divisible by 3. This contradicts coprimality. In exams, track prime factors in squares carefully.

The key idea is that an irrationality proof begins by temporarily assuming the opposite of what we want to prove. Here, we assume that \(\sqrt{3}\) is rational. A rational number can be written as \(p/q\), where \(p\) and \(q\) are integers, \(q\ne0\), and the fraction is in lowest terms. The contradiction finally shows that this assumption cannot be true.

Thus, the rejected assumption is “\(\sqrt{3}\) is rational,” which is option C. The proof does not reject the fact that \(\sqrt{3}\) is real; it is certainly a real number. Also, \(\sqrt{3}>0\) and \(q\ne0\) are valid facts or conditions used in the argument, not the assumption being disproved. Therefore the supplied answer is correct.

The direct answer is A. In a proof by contradiction, we first suppose that \\(\\sqrt{2}=a/b\\), where a and b have no common factor and b is not zero. Squaring gives \\(a^2=2b^2\\). This shows that a is even, so write \\(a=2r\\). Substitution gives \\(4r^2=2b^2\\), hence \\(b^2=2r^2\\), so b is also even and can be written \\(b=2s\\). Therefore both a and b have the common factor 2. But that contradicts the original statement that the fraction was already in lowest terms. Equivalently, \\(a/b=(2r)/(2s)=r/s\\), so it could be reduced. Option A expresses this precise conclusion. B is wrong because the argument proves that \\(\\sqrt{2}\\) is not an integer. C is wrong because b is a denominator and must not be zero; the proof does not show b=0. D is wrong because nothing proves a and b are equal. Exam cue: a common factor obtained after assuming lowest form is the contradiction.

The side is \(s=\sqrt{2}\) cm because \(s^2=2\). Thus, the perimeter is \(4s=4\sqrt{2}\) cm, which is irrational. \(2\sqrt{2}\) represents only two sides. Exam tip: the square root of a positive integer is rational only when the integer is a perfect square.

\(\sqrt{3}\) is irrational, so its decimal expansion is non-terminating and non-recurring. Terminating or recurring decimals are always rational. Exam tip: use this decimal property to identify irrational numbers.

If \(p^2=3q^2\), then 3 divides \(p\); substituting \(p=3k\) shows that 3 also divides \(q\). This contradicts coprimality. Exam tip: state that the fraction is in lowest terms.

Assume \(1+\sqrt{2}\) is rational. Rational numbers are closed under subtraction, so \((1+\sqrt{2})-1=\sqrt{2}\) would be rational. This contradicts the known irrationality of \(\sqrt{2}\). Exam tip: adding or subtracting a rational number cannot make an irrational number rational.

From a² = 3b², 3 divides a². Since 3 is prime, it must divide a. Then 3 also divides b, contradicting that a/b is in lowest terms. Exam tip: use the prime-divisibility rule for squares.

The correct answer is option B: q must also be proved divisible by 3. To prove that √3 is irrational, assume √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. Therefore 3 divides p², and hence 3 divides p. Write p=3k. Substitution gives 9k²=3q², so q²=3k². Thus 3 also divides q. Now both p and q have the common factor 3, contradicting gcd(p,q)=1. Merely showing that p is divisible by 3 does not yet contradict coprimality, because p alone may have a factor that q does not have. Option A is wrong because p=0 is not the required contradiction and would not follow appropriately. Option B is correct because both numerator and denominator must share 3. Option C is wrong because p=q is neither required nor obtained. Option D is wrong because q=0 is forbidden from the beginning, not the final contradiction. Memory cue: in irrationality proofs, “both numerator and denominator share a factor” breaks lowest form.