Proof of irrationality of square root 2 and square root 3

For (\sqrt{3}), after the rational assumption we square. Then both numbers becoming divisible by (3) gives a contradiction.

\(1.7^2=2.89\) is only close to 3, not equal to it. A rational number would need to have square exactly 3. In fact, \(1.7=17/10\), and its square is \(289/100\). Exam tip: never treat an approximation as an exact proof.

1.732 is only an approximation, not the exact value of \(\sqrt{3}\). Check: \((1.732)^2=2.999824\ne3\). In exams, never treat a close decimal approximation as an equality.

From \(p^2=3q^2\), \(p^2\) is divisible by 3, so the prime 3 divides \(p\). Put \(p=3k\); then \(q^2=3k^2\), so 3 also divides \(q\). This contradicts coprimality. Exam tip: remember \(r\mid p^2\Rightarrow r\mid p\) for a prime \(r\).

The direct answer is option B: the common factor is 3. Assume \(\sqrt{3}=p/q\) in lowest form. Squaring gives \(p^2=3q^2\), so 3 divides \(p^2\). Since 3 is prime, if it divides \(p^2\), it divides \(p\); write \(p=3k\). Substitution gives \(9k^2=3q^2\), hence \(q^2=3k^2\), so 3 also divides \(q\). Therefore both numbers have the common factor 3, contradicting the assumption that \(p,q\) are coprime. Option B is correct. Option A is wrong because 2 is not the factor forced by this equation. Option C is wrong because the equation does not force both numbers to be divisible by 4. Option D is wrong because divisibility by 3 does not automatically mean divisibility by 6; no factor 2 has been established. The contradiction specifically uses the common factor 3. Memory cue: in the \(\sqrt{3}\) proof, follow the factor 3.

A calculator shows 1.414 only as an approximation, not as the exact value. \(\sqrt{2}=1.414213\ldots\) is non-terminating and non-repeating, so it is irrational. Exam tip: never treat a rounded display as proof of rationality.

The direct answer is option B. The proof of √2 is not a numerical approximation exercise; it is a proof about whether a number can be written as a ratio of integers. Assume, for contradiction, that √2 = p/q, where p and q are integers with no common factor and q is nonzero. Squaring gives p² = 2q². Hence p² is even, so p is even; write p = 2k. Substitution gives 4k² = 2q², so q² = 2k², and q is also even. Thus p and q have a common factor 2, contradicting their lowest-term condition. Option B is correct because divisibility, parity and contradiction do the work. Option A is false: √2 is not zero. Option C is false because √2 is not an integer. Option D is false because the proof does use the fraction p/q. Decimal approximations may illustrate the value, but they are not needed for this logical proof.

The proof of (\sqrt{3}) does not use decimal approximation. It uses divisibility by (3) and contradiction.

The proof starts by assuming that \(\sqrt{2}\) is rational. Write it as \(\frac{p}{q}\) in lowest terms, with q nonzero. Squaring gives \(p^2=2q^2\). This makes \(p^2\) even, so p is even. Substituting \(p=2k\) into the equation gives \(4k^2=2q^2\), hence \(q^2=2k^2\), so q is also even.

The conclusion that both p and q are even contradicts the fact that \(\frac{p}{q}\) was chosen in lowest terms. This contradiction does not show that \(\sqrt{2}\) is non-real, non-positive, or unequal to a positive number. In fact, \(\sqrt{2}\) is a positive real number. What fails is only the starting assumption that it is rational. Therefore option C is correct.

In proof by contradiction, assume that \(\sqrt{3}=p/q\), where \(p\) and \(q\) are coprime integers. On squaring, \(3q^2=p^2\), which shows that \(p\), and then \(q\), are both divisible by 3. This contradicts their being coprime. Hence the assumption that \(\sqrt{3}\) is rational is rejected. \(\sqrt{3}\) is still positive and real; those facts are not rejected by the contradiction. Exam tip: when a contradiction is obtained, reject the initial assumption used in the proof.

If both \(a\) and \(b\) are divisible by 3, they have the common factor 3. This contradicts the assumption that they are coprime, so \(\sqrt{3}\) is irrational. Exam tip: state the common factor clearly in a contradiction proof.

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.

The direct answer is option C. The aim is to prove that √2 and √3 are irrational numbers. A rational number is one that can be written as p/q, where p and q are integers and q is not zero. In each proof, we assume that the square root has such a fraction form, choose p and q in lowest terms, square the equation, and use divisibility to show that both p and q must share a factor. That contradicts the lowest-terms assumption. Therefore the assumed fraction form is impossible. Option C states exactly this purpose. Option A is not the aim: decimal conversion may give an approximation but does not prove irrationality. Option B is false because neither square root is an integer. Option D is false because both numbers are positive, not negative. Exam cue: “cannot be expressed as p/q” is the definition and proof target for irrationality.

The direct answer is option D: both are irrational. To understand this, a rational number can be written as p/q using integers p and q with q ≠ 0. The standard contradiction proof assumes √2 or √3 has this form in lowest terms. For √2, squaring gives p² = 2q², forcing both p and q to be even. For √3, squaring gives p² = 3q², forcing both p and q to be divisible by 3. In either case this contradicts lowest terms, so the assumed rational form is impossible. Therefore both numbers are irrational. Option A is wrong because √2 and √3 are not integers. Option B is wrong because rationality is precisely what the proofs disprove. Option C is wrong because their squares are 2 and 3, not 0. Option D is correct. Remember: square roots of 2 and 3 are classic examples of irrational numbers.

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.

A proof by contradiction begins by assuming the opposite of the statement to be proved. Here the assumption is that √2 is rational. If valid reasoning from that assumption produces an impossibility, the assumption must be false. Therefore √2 is not rational; it is irrational, so option A is correct. The contradiction does not imply that √2 is zero, an integer, or a natural number. In fact, every integer and every natural number is rational, so options B and D conflict with the conclusion. Option C is also numerically false because the square of zero is 0, not 2. The logical structure is assumption, contradiction, rejection of assumption, and conclusion.