Proof of irrationality of square root 2 and square root 3

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: look for a common factor in both terms.

First (\sqrt{3}) is assumed rational and written as (\frac{p}{q}). Then squaring gives (p^2=3q^2).

Assume \(\sqrt{3}=a/b\) for integers \(a,b\). Then \(a^2=3b^2\), so \(3\mid a^2\), which implies \(3\mid a\). Substitution then gives \(3\mid b\), creating the contradiction. Exam tip: remember that for a prime \(p\), \(p\mid a^2\Rightarrow p\mid a\).

Coprime numbers have no common factor except (1). Both becoming divisible by (3) clashes with this.

The direct answer is option A. To prove that \(\sqrt{2}\) is irrational, begin by assuming the opposite: suppose \(\sqrt{2}\) is rational. Write it in lowest form as \(\sqrt{2}=p/q\), where \(p,q\) are integers, \(q\neq0\), and they have no common factor. Squaring gives \(2=p^2/q^2\), so \(p^2=2q^2\). Thus \(p^2\) is even, so \(p\) is even; write \(p=2k\). Substitution gives \(q^2=2k^2\), so \(q\) is also even. This contradicts the lowest-form condition. Therefore \(\sqrt{2}\) is irrational. Option A gives this correct short order. Option B is wrong because drawing a picture does not prove irrationality. Option C is wrong because there is no reason to assume the square root is zero. Option D is wrong because a decimal approximation or guess cannot be a proof. Memory cue: assume rational, square, show both even, contradict lowest form.

From \(3q^2=p^2\), \(p\) must be divisible by 3. Putting \(p=3k\) then shows that \(q\) is also divisible by 3, contradicting coprimality. Exam tip: always assume the fraction is in lowest terms.

A rational number has either a terminating or a non-terminating recurring decimal expansion. For example, \(\frac{1}{3}=0.333\ldots\) repeats, whereas \(\sqrt{2}=1.414\ldots\) is non-repeating and irrational. Exam tip: look carefully for the word “recurring.”

Write \(p=2k\) because \(p\) is even. Then \(4k^2=2q^2\), so \(q^2=2k^2\) and \(q\) is also even. This contradicts coprimality. Exam tip: an even square has an even base.

For integers, evenness of the square is linked with evenness of the number. If the square is not even, the number is not even.

Prime factor (3) appears in the square only when it appears in the number. This supports the proof of (\sqrt{3}).

In proof by contradiction, assume \(\sqrt{3}=p/q\) in lowest terms. From \(p^2=3q^2\), first \(p\), then \(q\), is divisible by 3, contradicting lowest terms. Exam tip: common divisibility by the same prime signals the contradiction.

1.732 is not the exact value of \(\sqrt{3}\); it is only an approximation. Check: \(1.732^2=2.999824\), not 3. The decimal expansion of \(\sqrt{3}\) is non-terminating and non-repeating. In exams, do not treat an approximate value as an exact value.

From \(3q^2=p^2\), \(p^2\) is divisible by 3, so \(p\) is divisible by 3. Put \(p=3k\); then \(q^2=3k^2\), making \(q\) divisible by 3 too. This contradicts coprimality. Exam tip: use prime divisibility of squares carefully.

Assume \(\sqrt{2}=p/q\), where \(p\) and \(q\) are coprime. Then \(p^2=2q^2\), so \(p^2\) is even and hence \(p\) is even. This also makes \(q\) even, contradicting coprimality. Exam tip: remember that an even square implies an even integer.

Since 3 divides \(p\), write \(p=3k\). Substituting in \(p^2=3q^2\) gives \(9k^2=3q^2\), so \(q^2=3k^2\); hence 3 also divides \(q\). This contradicts lowest form. Exam tip: track prime factors after squaring.

In the standard contradiction proof, assume that \(\sqrt{3}=p/q\), where \(p\) and \(q\) are integers and the fraction is in lowest form. Lowest form means that \(p\) and \(q\) have no common factor. Squaring gives \(p^2=3q^2\). This shows that 3 divides \(p^2\), and therefore 3 divides \(p\). Writing \(p=3k\) then shows that 3 also divides \(q\).

That conclusion contradicts the original statement that \(p/q\) was in lowest form. The contradiction proves that \(\sqrt{3}\) cannot be rational. Thus option B is correct. Without the coprime assumption, finding a common factor would not create a contradiction, so taking lowest form is essential. The supplied explanation is accurate.

1.73 is only \(\sqrt{3}\) rounded to two decimal places, not its exact value. Since \(\sqrt{3}\approx1.732\ldots\) is non-terminating and non-repeating, it is irrational. Exam tip: never infer rationality from a rounded decimal.

Since \(1^2<2<2^2\), 2 is not a perfect square, so \(\sqrt{2}\) is irrational. In contrast, \(\sqrt{4}=2\) is rational. Exam tip: first check whether the radicand is a perfect square.

To prove that a square root such as \(\sqrt{2}\) or \(\sqrt{3}\) is irrational, the proof begins by assuming the opposite: that the number is rational. Every rational number can be written as a fraction \(\frac{m}{n}\), where m and n are integers, n is nonzero, and the fraction is in lowest terms. The lowest-terms condition means m and n have no common factor.

This form is essential because the later argument shows that both m and n must be divisible by the same number, usually 2 for \(\sqrt{2}\) or 3 for \(\sqrt{3}\). That contradicts their being coprime. A decimal or percentage form does not provide this useful coprime condition. Therefore option A, the simplest fraction form, is correct.

\(\sqrt{3}\) is indeed irrational, but merely saying that 3 is not a perfect square is not a complete formal proof. Assume \(\sqrt{3}=p/q\); then \(p^2=3q^2\), so both \(p\) and \(q\) are divisible by 3, a contradiction. Exam tip: state that \(p,q\) are coprime.

A number expressible as \(\frac{p}{q}\), with integers \(p,q\) and \(q\neq0\), is rational. Coprime \(p,q\) indicate lowest terms; not every rational number is whole or natural. Exam tip: the denominator can never be zero.

An irrational number cannot be written as a ratio of two integers. Both proofs establish this fact.

In \(a^2=2b^2\), the right-hand side is a multiple of \(2\), so \(a^2\) is even. The next step is to conclude that \(a\) is also even. We cannot conclude at this stage that \(b\) is odd. Exam tip: In such proofs, first compare the parity of both sides of the equation.

If \(p=2k\), then \(4k^2=2q^2\), so \(q^2=2k^2\) and \(q\) is even too. Thus both numerator and denominator have factor 2, contradicting coprimality. Exam tip: use the fact that an even square has an even root.

In contradiction method we first assume the opposite that (\sqrt{2}) is rational. Later this assumption is shown false.