Proof of irrationality of square root 2 and square root 3

Lowest terms means that \(p\) and \(q\) are coprime. The proof shows that both are even, contradicting this assumption. Exam tip: read “lowest terms” as “common factor is 1.”

From \(p^2=3q^2\), \(p^2\) is divisible by 3, so \(p\) is divisible by 3. Put \(p=3k\); then \(q\) is also divisible by 3, contradicting coprimality. Exam tip: use the prime-divisibility property of squares.

If \(2\sqrt{2}=r\) were rational, then \(\sqrt{2}=r/2\) would also be rational, a contradiction. A non-zero rational multiple of an irrational remains irrational. Exam tip: divide by the rational factor to test the claim.

An irrational number cannot be written as a ratio of two integers. These proofs establish exactly that.

Assuming either (\sqrt{2}) or (\sqrt{3}) rational gives a contradiction. Therefore both are irrational.

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.

Both being divisible by (3) is derived later in the proof. At the start they are only assumed coprime.

A rational number is written as a ratio of two integers. Therefore (m) and (n) are assumed integers.

A rational number is written as a ratio of two integers. Therefore (r) and (s) are assumed integers.

The equation a² = 2b² says that a² is exactly twice the integer b². Therefore a² has 2 as a factor and must be even. By the parity property of integers, if the square of an integer is even, the integer itself is even; hence the proof will next write a = 2r for some integer r. The immediate conclusion asked here is only that a² is even, so option A is correct. Option B does not follow: b² may be odd or even at this stage. The equality a = b is unsupported, and b cannot be zero because a/b is a valid rational representation and the denominator must be nonzero.

If (p^2) is divisible by (3), then (p) is also divisible by (3). Remember this prime factor rule.

\(1.7^2=2.89\) 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.

In contradiction method we first assume rationality. Then squaring gives the contradiction that both are even.

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.