Proof of irrationality of square root 2 and square root 3

The governing concept is extracting perfect-square factors from radicals and then combining like surds. Factor 242 as 121 × 2, so √242 = √121 × √2 = 11√2. Factor 128 as 64 × 2, so √128 = √64 × √2 = 8√2. Both simplified terms contain the same radical √2, so they are like surds and their coefficients can be subtracted: 11√2 − 8√2 = (11 − 8)√2 = 3√2. Therefore option C is correct. Option A would come from adding the coefficients instead of subtracting them. Option B reflects an incorrect factorisation or subtraction. Option D incorrectly applies √a − √b = √(a−b), an identity that is not generally valid. Each radical must first be simplified separately.

The governing concept is rationalisation of a denominator containing a square root. Because x = 3 − √5, its reciprocal is 1/(3 − √5). Multiply numerator and denominator by the conjugate 3 + √5. The denominator becomes (3 − √5)(3 + √5) = 3² − (√5)² = 9 − 5 = 4, while the numerator becomes 3 + √5. Hence 1/x = (3 + √5)/4, making option A correct. Option B omits the denominator 4. Option C retains the original sign and does not use the conjugate correctly. Option D reverses the required division and multiplies by 4 instead. Substitution also confirms that (3 − √5)(3 + √5)/4 = 1.

The governing concept is extracting the greatest perfect-square factor from under a radical. Factor 200 as 100 × 2, where 100 is a perfect square. Therefore √200 = √(100 × 2) = √100 × √2 = 10√2, so option A is correct. Option C, 5√8, is numerically equivalent because √8 = 2√2 and hence 5√8 = 10√2, but it is not fully simplified: 8 still contains the perfect-square factor 4. Option B would square to 800, not 200, and option D is much too large. A radical is in simplest form when no perfect-square factor greater than 1 remains inside the square-root sign. This criterion selects option A.

The governing concept is the quotient rule for square roots: for positive b, √a ÷ √b = √(a/b). Applying it directly gives √20 ÷ √2 = √(20/2) = √10. Thus option B is correct. The result can also be checked by writing √20 = 2√5; combining the radicals through the quotient rule still produces √10. Option A, 5, would arise from an incorrect treatment of the radicands and is not equal to √10. Option C divides 20 by 2 but forgets to retain the square root, while option D incorrectly subtracts 2 from 20 inside the radical. Since 10 is not a perfect square, √10 cannot be reduced further and remains an irrational number.

The governing geometric concept is that the area of a square equals the square of its side: A = s². With side s = 2√7 units, calculate A = (2√7)² = 2²(√7)² = 4 × 7 = 28 square units. Therefore option C is correct. The irrational factor in the side disappears when the side is squared, so the area is a rational integer. Option A is only half of the correct product and may result from multiplying 2 by 7 without squaring the coefficient. Option B has no valid derivation from the square-area formula. Option D treats the expression as though the coefficient 2 were absent or mishandles the multiplication. The units must be square units because area is two-dimensional.

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.

The standard proof uses contradiction. Assume that √2 is rational and write √2=p/q, where p and q are coprime integers and q≠0. Squaring gives p²=2q². Thus p² is even, which means p is even; let p=2k. Substitution gives 4k²=2q², so q²=2k², and q is also even. This contradicts the assumption that p and q have no common factor. Therefore the assumption is false and √2 is irrational. Option C states this conclusion. Options A and B conflict with the contradiction proof, and √2 is clearly not zero because its square is 2.

The governing concept is the lowest-terms condition used in the contradiction proof that √2 is irrational. At the beginning, √2 is assumed to equal a/b, where a and b have no common factor. If the derivation shows that both a and b are even, each is divisible by 2; hence the fraction has a common factor and cannot actually be in lowest form. This produces the required contradiction. Therefore, option A is the conclusion that should not be taken. Options B, C and D correctly describe the common factor, the failure of coprimality and the resulting contradiction.

The governing concept is proof by contradiction using a rational number in lowest terms. To assume √3 is rational, write it as r/s where r and s are integers, s is nonzero, and r and s are coprime. Squaring gives r² = 3s². From this relation, 3 divides r, so r = 3k; substitution then shows that 3 also divides s. Thus both r and s have 3 as a common factor, contradicting the original lowest-form condition that their HCF is 1. Option D correctly states this purpose. The aim is not to terminate a decimal, draw a diagram, or make the denominator zero. The lowest-form assumption is essential because without coprimality, common divisibility would not produce a contradiction.

The governing concept is an indirect proof of irrationality. First suppose, contrary to what is to be proved, that √2 is rational. Write √2 = m/n in lowest form, where m and n are integers, n is nonzero, and they are coprime. Squaring gives m² = 2n². This shows m is even; writing m = 2k and substituting then shows n is also even. That conclusion contradicts the lowest-form assumption, because m and n would share the factor 2. Hence the assumption is false and √2 is irrational. Option A gives this correct order. A decimal calculation is not the rigorous proof, and drawing or assuming zero has no role in the argument.

The governing concept is the lowest-form condition in a rational representation. When m/n is in lowest form, m and n are integers, n is nonzero, and they have no common factor greater than 1. If both m and n were even, each would be divisible by 2, so the fraction could be reduced by cancelling 2. That would prove it was not in lowest form. Therefore option D describes the impossible situation. The conditions in options A, B and C are all part of a valid rational representation: the denominator must not be zero, and numerator and denominator are integers. In the √2 proof, both being even is not an initial assumption; it is the contradiction derived from the equation m² = 2n² after assuming √2 rational.

The governing concept is proof by contradiction with a fraction in lowest form. To begin the proof, it is legitimate to assume that √2 = m/n is rational, with m and n coprime integers and n nonzero. Squaring this equation is also a valid algebraic step, and deriving a contradiction is the intended conclusion. However, assuming from the beginning that m and n are both even is a mistake. Their being both even must be obtained from the equation, not inserted as an initial premise. If it were assumed at the start, the argument would be circular and would not demonstrate anything. Once both-evenness is derived, it conflicts with the lowest-form condition, because 2 would be a common factor. Hence option A is the correct choice.

The governing concept is parity in the contradiction proof of the irrationality of √2. From a² = 2b², the right-hand side is twice an integer, so it is divisible by 2. Consequently, a² is even. The parity property of integers then allows the next step: if the square of an integer is even, the integer itself is even, so one may write a = 2r for some integer r. The question asks only for the immediate conclusion, which is that a² is even; therefore option A is correct. The equation does not establish that b² is odd, because b may be either odd or even at this stage. It also gives no basis for a = b, and b cannot be zero because a/b must be a valid fraction.

The direct answer is option B: p is divisible by 3. From √3 = p/q, squaring gives p² = 3q². Therefore p² is divisible by 3. Since 3 is prime, if 3 divides the square p², then 3 must divide p itself. In prime-factor language, any prime factor appearing in a square appears with an even exponent; because 3 appears in p², it must already be a factor of p. We may write p = 3k for some integer k. Option A, p even, does not follow: the relevant prime is 3, not 2. Option B is correct. Option C, p = 0, is not allowed in the usual fraction proof because p and q are taken as nonzero coprime integers; √3 is not zero anyway. Option D, p negative, is not forced; p may be chosen positive, and divisibility does not determine its sign. Memory cue: if a prime divides n², it divides n.

\(1.7^2=2.89\) gives only an approximation to \(\sqrt{3}\), not equality. To prove rationality, one must establish \(\sqrt{3}=\frac{p}{q}\); closeness is insufficient. Exam tip: always distinguish “approximately equal” from “equal.”

Since \(\sqrt{3}\) is irrational, assume \(2\sqrt{3}\) is rational. Dividing by the non-zero rational number 2 gives \((2\sqrt{3})/2=\sqrt{3}\), a contradiction. Hence A is correct. Exam tip: divide by a non-zero rational factor to test such claims.

At the beginning of the proof, √2 is assumed to be a/b in lowest terms. This means that a and b have no common factor other than 1; in other words, they are coprime. The algebraic steps eventually show that a is even and b is also even. Thus 2 divides both numbers, so 2 is a common factor. That directly contradicts the original lowest-terms assumption. It does not mean that a and b become equal, zero, or negative. The contradiction invalidates the assumption that √2 can be represented as a rational fraction, leading to the conclusion that √2 is irrational. Therefore option A gives both the reason and the required logical link.

Assuming either \(\sqrt{2}\) or \(\sqrt{3}\) equals \(p/q\) in lowest terms leads to a contradiction. For example, \(p^2=2q^2\) makes both \(p\) and \(q\) even. Exam tip: the square root of a non-square prime is irrational.

The direct answer is Option C: assume rationality, square, and then obtain a contradiction. The proof uses contradiction. First suppose that √2 is rational, so it can be written as a/b, where a and b are coprime integers and b is not zero. Squaring gives 2 = a²/b², so a² = 2b². Thus a² is even, so a is even; write a = 2r. Substitution then shows b² = 2r², so b is also even. This contradicts that a and b were coprime. Option C is correct. Option A reverses the logical order because a = 2r is obtained after proving a is even. Option B is not a complete proof; a decimal representation does not by itself establish the required contradiction. Option D simply asserts the result without proof. Exam cue: contradiction proofs usually follow assume, transform, contradiction, conclude.

From \\(p^2=3q^2\\), 3 divides p²; since 3 is prime, it divides p. Substitution then shows that 3 also divides q, contradicting coprimality. Exam tip: identify the common factor of p and q that creates the contradiction.

The statement “if \(x^2\) is even, then \(x\) is even” is a useful divisibility fact. If a square is even, it is divisible by 2. A square of an odd integer is always odd, so the number whose square is even cannot be odd; it must be even. This fact helps convert information about a square into information about its original integer.

In the proof of the irrationality of \(\sqrt{2}\), assume \(\sqrt{2}=a/b\) in lowest form. Squaring gives \(a^2=2b^2\), so \(a^2\) is even and hence \(a\) is even. Substituting an even value of \(a\) then shows that \(b\) is also even, contradicting lowest form. Therefore option A is correct.

In the proof of (\sqrt{3}), divisibility of (p) by (3) is derived from (p^2). This leads to contradiction.

Any rational number can be written as a fraction a/b with b ≠ 0, and common factors can be cancelled. Choosing lowest terms ensures that a and b are coprime before the contradiction argument begins. From a² = 2b², the proof shows that a is even; writing a = 2r then leads to b² = 2r², so b is even as well. Both numbers therefore have the common factor 2, which is impossible for a coprime pair. Without the lowest-terms condition, finding a common factor would not itself contradict anything, because the original fraction might already have had a common factor. Hence option C states the essential purpose.

Although 3 is rational, its square root need not be rational. If \(\sqrt{3}=p/q\), then \(p^2=3q^2\) makes both \(p\) and \(q\) divisible by 3. Tip: test the square root separately.

Assume \(4+\sqrt{3}\) is rational. Since 4 is rational, subtracting it would make \(\sqrt{3}\) rational, contradicting its irrationality. Hence the sum is irrational. Exam tip: adding or subtracting a rational number does not change irrationality.