Proof of irrationality of square root 2 and square root 3

An integer leaves remainder 0, 1, or 2 on division by 3. The squares of remainders 1 and 2 both leave remainder 1, so a square divisible by 3 requires the integer itself to be divisible by 3. Exam tip: list possible square remainders modulo 3.

From ext{\(p^2=3q^2\)} , ext{\(p^2\)} and hence ext{\(p\)} are divisible by 3. Put ext{\(p=3k\)} to get ext{\(q^2=3k^2\)} , so ext{\(q\)} is also divisible by 3. This contradicts coprimality. Exam tip: use the prime-divides-a-square property.

Since \(m^2=2n^2\), \(m^2\) is even, so \(m=2k\). Substitution gives \(n^2=2k^2\), making \(n\) even too. This contradicts coprimality. Exam tip: establish that both numerator and denominator share 2.

Since \(12=4\times3\), \(\sqrt{12}=\sqrt4\sqrt3=2\sqrt3\). As \(\sqrt3\) is irrational, multiplying it by non-zero rational 2 still gives an irrational number. \(3\sqrt2\) squares to 18, not 12. Exam tip: extract perfect-square factors first.

Since 3 is prime,

b^2=3a^2 shows that 3 divides

b^2. By the prime divisor property, 3 must divide

b. This leads to a contradiction with the lowest-form assumption. In exams, state the prime-property step explicitly.

From \(p^2=3q^2\), we get \(3\mid p^2\). Since 3 is prime, \(3\mid p\); put \(p=3k\) to obtain \(q^2=3k^2\), so \(3\mid q\) too. This contradicts coprimality. Exam tip: state the prime-divides-square rule first.

If both have common factor (3), they cannot remain coprime. This makes the rational assumption false.

The equation m² = 2n² shows that m² is even, and the parity theorem then shows that m is even. We may write m = 2k, but the equation does not imply the much stronger statement m = 2n. That direct equality is an unjustified shortcut, so option D is wrong. The other steps are valid parts of the contradiction proof.

The governing concept is careful algebra combined with prime divisibility. From a² = 3b², the right side is a multiple of 3, so a² is divisible by 3; option A is valid. Since 3 is prime, 3 ∣ a² implies 3 ∣ a, making B valid as a stated theorem. Writing a = 3r and substituting gives 9r² = 3b², hence b² = 3r², so C is also valid. However, a² = 3b² does not imply a = 3b. Squaring a = 3b would give a² = 9b², not 3b². Therefore D is the wrong shortcut and the correct answer.

From \(a^2=3b^2\), 3 divides \(a^2\), so 3 divides \(a\). Put \(a=3k\); then \(b^2=3k^2\), so 3 also divides \(b\). This contradicts coprimality. Exam tip: use the prime-divides-a-square rule carefully.

If (a) has no factor (3), then (a^2) also has none. But (a^2=3b^2) makes (a^2) divisible by (3).

The proof begins by assuming the opposite of what must be shown: suppose \(\sqrt{2}\) is rational and write it as \(m/n\) in lowest form, with \(n\neq0\). Squaring gives \(m^2=2n^2\). This equation shows that \(m^2\), and therefore \(m\), is even. Writing \(m=2k\) and substituting back shows that \(n^2\), and therefore \(n\), is also even.

Thus both \(m\) and \(n\) have 2 as a common factor. That contradicts the assumption that \(m/n\) was in lowest form, meaning the two integers were coprime. This contradiction proves that the original assumption was false. Hence option A gives the correct logical chain; the other options omit the essential algebra and contradiction.

The proof of \(\sqrt{3}\) follows a chain of divisibility by \(3\). Finally the coprime assumption breaks.

The lowest-form condition states coprimality. Without it, both even will not form a decisive contradiction.

If \(7+\sqrt{2}\) were rational, subtracting the rational number 7 would make \(\sqrt{2}\) rational. This contradicts the proven irrationality of \(\sqrt{2}\). Exam tip: rational ± irrational is always irrational.

From \(m=2k\), \(m\) is even. Also, \(n^2=2k^2\) shows that \(n^2\) is even; if the square of an integer is even, the integer itself must be even. Hence \(n\) is also even, so both \(m\) and \(n\) are even. This contradicts the assumption that the fraction \(m/n\) was in lowest terms. Exam tip: In irrationality proofs, remember that an even square implies an even integer.

The fraction is taken in lowest terms, so \(p\) and \(q\) must be coprime. The proof then shows that both are divisible by 3, contradicting this condition. Exam tip: explicitly state that the fraction is in lowest terms before starting the proof.

The claim is true. On division by 3, an integer leaves remainder 0, 1, or 2. Squaring remainders 1 and 2 gives remainders 1 and 4, i.e. 1 modulo 3, never 0. Hence a square divisible by 3 has a base divisible by 3. Exam tip: use remainders to test divisibility claims quickly.

The basis is the prime-divisor property: if a prime p divides x², then p divides x. Here p = 3, so 3 ∣ a² implies 3 ∣ a. This can be seen from prime factorisation. If 3 did not divide a, then no factor 3 would occur in the factorisation of a, and consequently no factor 3 could occur in a², contradicting 3 ∣ a². In the √3 proof, this result permits writing a = 3r; substitution then leads to b² = 3r² and eventually proves 3 ∣ b. Thus the primality of 3 is essential. Options B, C, and D are not assumptions of the proof and do not justify the implication.

Assume \(r=\sqrt{2}+\sqrt{3}\) is rational. Then \(r^2=5+2\sqrt6\), which would make \(\sqrt6\) rational, an impossibility. Thus the conclusion is correct. However, two irrational numbers need not always have an irrational sum; avoid this overgeneralisation.