Proof of irrationality of square root 2 and square root 3

The standard proof assumes, for contradiction, that \(\sqrt{2}\) is rational and writes it as \(\frac{p}{q}\), where p and q are integers with no common factor. Squaring gives \(p^2=2q^2\). The right side is even, so \(p^2\) is even. A key elementary result is that if the square of an integer is even, then the integer itself is even. Therefore p is even.

Writing \(p=2k\) and substituting back shows that \(q^2\), and hence q, is also even. This means p and q have a common factor 2, contradicting the assumption that the fraction was in lowest terms. Thus the statement in option A is the central parity fact used in the proof. The other statements are false or unrelated to this argument.

In the proof of (\sqrt{3}), divisibility of (a) by (3) is concluded from (a^2). This is the key fact.

Let \(\sqrt{3}=p/q\) in lowest terms. From \(p^2=3q^2\), \(3\mid p^2\) implies \(3\mid p\), since 3 is prime. Substituting \(p=3k\) also gives \(3\mid q\), a contradiction. Exam tip: always state that \(p,q\) are coprime.

Assume that \(5\sqrt{3}\) is rational. Since 5 is a non-zero rational number, dividing by 5 would make \(\sqrt{3}\) rational, a contradiction. The fact that 5 is prime is irrelevant. Exam tip: use division by a non-zero rational number.

Rational numbers remain rational under subtraction. If \(5+\sqrt{3}\) were rational, then \(\sqrt{3}=(5+\sqrt{3})-5\) would be rational, a contradiction. Exam tip: use closure properties to test such expressions.

Being non-terminating alone is not enough: \(0.333\ldots=\frac{1}{3}\) is rational. The decimal expansion of \(\sqrt{2}\) is non-repeating, or a lowest-form fraction assumption gives a contradiction. Exam tip: distinguish non-terminating from non-terminating non-repeating decimals.

A contradiction proof begins by assuming the negation of the statement to be proved. The reasoning then proceeds logically until it produces an impossible result or a conclusion that conflicts with a known fact or with an earlier condition. The purpose of the final step is to identify that conflict and reject the initial assumption. In the irrationality proof, assuming √2 or √3 is rational eventually forces the numerator and denominator to share a factor, even though they were chosen in lowest form. Therefore option A is correct. The final step does not create a new number, require a diagram, or replace reasoning with memorisation; it completes the logical disproof of the opposite assumption.

A calculator gives a rounded approximation, so 1.732 is not the exact value of \(\sqrt{3}\). For proof, assume \(\sqrt{3}=a/b\) in lowest terms and derive a contradiction. Exam tip: never use a displayed decimal as proof of rationality.

The correct statement is that √3 is irrational. If √3 were rational, write it as a/b in lowest form, where a and b are coprime integers and b is non-zero. Squaring gives a² = 3b². Hence 3 divides a², so 3 divides a; write a = 3k. Substitution then gives b² = 3k², so 3 divides b as well. This contradicts the assumption that a and b are coprime. Therefore √3 cannot be expressed as a ratio of integers and is irrational, making option B correct. It is not an integer or a natural number because its square is 3, which is not the square of an integer; it is also clearly not zero because 0² is 0.

If 3 divides both \(a\) and \(b\), then \(\frac{a}{b}\) is not in lowest terms. This contradicts the original assumption, so \(\sqrt{3}\) is irrational. Exam tip: look for a common factor in numerator and denominator.

Assume √3 = a/b, where a and b are coprime integers and b ≠ 0. Squaring and multiplying by b² gives a² = 3b². The right-hand side is a multiple of 3, so a² is divisible by 3. Since 3 is prime, if 3 divides a², then 3 must divide a. Let a = 3k; substitution gives 9k² = 3b², hence b² = 3k², so 3 also divides b. This produces the common factor that contradicts the lowest-form assumption. Therefore option A identifies the important immediate consequence. The equation does not imply that a is zero, that b is negative, or that a and b are equal.

A fraction p/q is in lowest form when the numerator p and denominator q have no common factor greater than 1. Equivalently, their greatest common divisor is 1, so p and q are coprime; q must also be non-zero. For example, 6/8 is not in lowest form because both numbers are divisible by 2, while 3/4 is in lowest form. This condition is crucial in irrationality proofs: if later reasoning forces both p and q to be divisible by 2 or by 3, a contradiction is obtained. Thus option A is correct. The numbers need not both be even, divisible by 3, or equal. Those properties would actually show that the fraction can be reduced further in some cases.

The number \(2\) is rational. If \(2+\sqrt{3}\) were rational, then \((2+\sqrt{3})-2=\sqrt{3}\) would be rational, contradicting the irrationality of \(\sqrt{3}\). Hence the original number is irrational. Exam tip: rational minus rational is rational.

\(1.414\) is only an approximation, not the exact value of \(\sqrt{2}\). Check: \((1.414)^2=1.999396\), not \(2\). In exams, always distinguish an approximate decimal from an exact value.

A rational number can be written as a ratio of two integers. So assuming (\sqrt{2}=\frac{p}{q}) means assuming it rational.

Writing √3 as a/b means assuming that √3 is rational, because a rational number is defined as a number that can be expressed as the quotient of two integers, with a non-zero denominator. In the proof, a and b are selected as coprime integers so that a/b is in lowest form. Squaring the assumed equality gives a² = 3b², and divisibility by 3 eventually forces both a and b to have a factor 3. That contradicts their coprime condition and proves that the assumption was impossible. Therefore option B is correct. Being written as a fraction does not mean the number is an integer, zero, or negative; those are different properties.

The correct conclusion is that 3 divides a. On dividing a by 3, the possible remainders are 0, 1, and 2; their squares leave remainders 0, 1, and 1. Thus \(a^2\) is divisible by 3 only when a is divisible by 3. In exams, use this prime-divisibility fact carefully.

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, m² is even, so m is even. Put m=2k: 4k²=2n², hence n²=2k² and n is also even. This contradicts m/n being in lowest terms. Exam tip: use the fact that an even square has an even root.

Assuming \(\sqrt{3}=p/q\) gives \(p^2=3q^2\), so 3 divides \(p\). Put \(p=3k\); then \(q\) is also divisible by 3. Thus they have a common factor 3, contradicting coprimality. Exam tip: after this contradiction, state that the rationality assumption is false.

\(\sqrt{2}\) is irrational because it cannot be written as \(p/q\), where \(p,q\) are integers and \(q\ne0\). Its decimal expansion is non-terminating and non-repeating. Exam tip: a square root is rational only when the number is a perfect square.

To prove that \(\sqrt{2}\) is irrational, we use proof by contradiction. We begin by assuming the opposite of what we want to prove: suppose \(\sqrt{2}\) is rational and can be written in lowest terms as \(p/q\), with nonzero integers p and q having no common factor. Squaring then gives a relation that forces both p and q to be even, contradicting their being in lowest terms.

Thus the logical order is to assume rationality first, square the expression, and then derive a contradiction from the parity of the integers. More specifically, \(2=p^2/q^2\) gives \(p^2=2q^2\), so p is even; writing \(p=2k\) then shows q is also even. This contradicts the assumption that the fraction was reduced. Therefore option A states the correct order. The other choices do not describe this standard proof.

In the proof of (\sqrt{3}), divisibility by (3) is used after squaring. Finally the coprime assumption is contradicted.

The general form of a rational number is (\frac{p}{q}) where (p) and (q) are integers. This starts the proof.

In a fraction written as \(a/b\), the denominator tells how many equal parts are being considered. Division by zero is not defined, so a denominator equal to zero would not represent a valid fraction. Therefore, when a rational number is written in the form \(a/b\), it is essential to state that \(b\neq0\). This condition is about the meaning of the fraction, not about whether \(b\) is positive, negative, even, or equal to 3.

In the proof, we suppose that \(\sqrt{3}\) is rational and write it as \(a/b\). The symbol \(a/b\) is meaningful only when \(b\neq0\). Thus option A is correct because the denominator of a fraction cannot be zero. This condition is separate from the later condition that \(a\) and \(b\) have no common factor.

\(\sqrt{3}+5\) is irrational. If it were rational, subtracting the rational number 5 would make \(\sqrt{3}\) rational, which is impossible. The other expressions equal 3, 1, and 5. Exam tip: a rational number plus an irrational number is always irrational.