Proof of irrationality of square root 2 and square root 3

An infinite decimal alone does not prove irrationality: \(0.\overline{3}=1/3\) is infinite but rational. For \(\sqrt{2}\), establish non-repetition or use contradiction. Exam tip: every recurring decimal is rational.

In a contradiction proof, first assume \(\sqrt{3}=\frac{m}{n}\) in lowest terms. Later, both \(m\) and \(n\) become divisible by 3, contradicting coprimality. Exam tip: always state that the fraction is in lowest terms.

A rational representation in lowest terms is written as \(a/b\), with \(b\neq0\) and \(\gcd(a,b)=1\). In the proof of \(\sqrt{3}\), squaring the assumed equality gives \(a^2=3b^2\). Since 3 divides the square \(a^2\), it follows that 3 divides \(a\). Substituting this fact back into the equation then shows that 3 also divides \(b\).

Thus both numerator and denominator have the common factor 3. This directly contradicts the original choice of the fraction in lowest terms, namely \(\gcd(a,b)=1\). It does not mean that \(b=0\), that \(a=b\), or that \(\sqrt{3}=3\). Therefore option B identifies the exact assumption that is broken.

If \(7-\sqrt{3}\) were rational, then \(7-(7-\sqrt{3})=\sqrt{3}\) would also be rational, a contradiction. Also, \((\sqrt{3})^2=3\) is rational. In exams, use closure properties with rational numbers carefully.

An infinite decimal alone is insufficient because \(1/3=0.333\ldots\) is rational. For \(\sqrt{2}\), establish non-repetition or derive a contradiction from a lowest-term fraction. Exam tip: distinguish infinite decimals from non-repeating decimals.

A proof by contradiction must reach a statement that directly conflicts with an assumption. In the usual proof, write \(\sqrt{2}=m/n\) in lowest form. Squaring gives \(m^2=2n^2\). This first shows that \(m^2\), and therefore \(m\), is even. However, the fact that only \(m\) is even is not yet a contradiction, because there is no problem with one numerator being even.

Substituting \(m=2k\) back into the equation shows that \(n\) is also even. Now both numbers have a common factor 2, contradicting \(\gcd(m,n)=1\). Thus option A is the incomplete statement. Option B or C expresses the actual contradiction, while D is the conclusion drawn after it.

Only (a) being divisible by (3) is not the final contradiction. (b) must also be proved divisible by (3).

From \(a^2=3b^2\), \(3\mid a^2\), so prime-factor reasoning gives \(3\mid a\). Put \(a=3k\) to obtain \(b^2=3k^2\), hence \(3\mid b\), contradicting coprimality. Exam tip: always show the substitution step.

Under the assumption, both \(s\) and \(1/s=\sqrt{3}-\sqrt{2}\) are rational. Option A adds them to obtain \(2\sqrt{3}\), which would make \(\sqrt{3}\) rational—a contradiction. In B, subtraction gives \(2\sqrt{2}\), not \(2\sqrt{3}\). Exam tip: first check the product of conjugates.

1.732 is a terminating rational approximation, not \(\sqrt{3}\) itself. Check: \(1.732^2=2.999824\), not 3. Exam tip: distinguish an approximate value from an exact value.

From \(p^2=2q^2\), \(p\) must be even. Put \(p=2k\); then \(q\) is also even, so both share the factor 2. This contradicts lowest terms. Exam tip: a fraction in lowest terms has coprime numerator and denominator.

Since \(3\mid m^2\), \(3\mid m\). Let \(m=3k\); then \(n^2=3k^2\), so \(3\mid n\). Both share 3, contradicting coprimality. Tip: use the prime-divisor rule.

Writing \(p/q\) in lowest terms makes \(p\) and \(q\) coprime. From \(p^2=2q^2\), \(p\) is even and then \(q\) is also even, contradicting this condition. Exam tip: always state “lowest terms.”

If rational, \(\sqrt{3}\) is written as \(p/q\) in lowest terms. From \(3q^2=p^2\), first \(p\) and then \(q\) are divisible by 3, giving a contradiction. Exam tip: always state coprimality.

Only (p) being divisible by (3) is not a contradiction. If (q) is also divisible by (3), the coprime condition breaks.

If \(5+\sqrt{3}\) were rational, subtracting the rational number 5 would make \(\sqrt{3}\) rational, contradicting its irrationality. Exam tip: rational numbers are closed under subtraction.

Option B is correct. \(1.732^2=2.999824\), not 3, so 1.732 is only an approximation of \(\sqrt{3}\). A terminating decimal describes 1.732, not the exact value of \(\sqrt{3}\). In exams, distinguish a rounded value from an exact value.

A fraction in lowest form has numerator and denominator with no common factor greater than 1. In other words, if \(a/b\) is in lowest form, then \(\gcd(a,b)=1\). The number 1 is their only positive common factor. This condition is deliberately used in irrationality proofs so that a common factor found later creates a contradiction.

For \(\sqrt{2}\), the assumption \(\sqrt{2}=a/b\) leads to the conclusion that both \(a\) and \(b\) are even. Therefore 2 divides both numbers, so their greatest common divisor is at least 2, not 1. This is impossible for a fraction in lowest form. Thus option A is correct. The contradiction comes from the coprime condition, not from the denominator being zero or from every square root being an integer.

A rational number has a terminating or recurring decimal expansion. \(\sqrt{2}\) is non-terminating and non-recurring, so it is irrational; option D describes a rational decimal. Exam tip: look for “non-recurring”.

When both are even, (\frac{a}{b}) can be reduced by (2). This opposes the idea of a lowest fraction.