Proof of irrationality of square root 2 and square root 3

From

, 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.

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.

In contradiction method (\sqrt{3}) is first assumed rational. Later this assumption is shown false.

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}).