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Suppose that the number V6 is rational. Then 6 = for some p, q e ZZ with q + 0. Suppose that the fraction 2 is in its simplest form, i.e. all common factors between p and q have been cancelled in the fraction. This implies that 6q? = p?, and therefore p? is a multiple of 3. This in turn means that p that 6q2 = 9p², and thus we arrive at an equation 2q2 = 3n2. Since this implies that q2 is a multiple of 3. we may set q = 3r for somere Z. Therefore both integers p and q are multiples of 3. Contradiction. This proves that 6 is an irrational number. 3n for some n EZ. .e., p? = 9n2, which implies True O False

Suppose that the number V6 is rational. Then 6 = for some p, q e ZZ with q + 0. Suppose that the fraction 2 is in its simplest form, i.e. all common factors between p and q have been cancelled in the fraction. This implies that 6q? = p?, and therefore p? is a multiple of 3. This in turn means that p that 6q2 = 9p², and thus we arrive at an equation 2q2 = 3n2. Since this implies that q2 is a multiple of 3. we may set q = 3r for somere Z. Therefore both integers p and q are multiples of 3. Contradiction. This proves that 6 is an irrational number. 3n for some n EZ. .e., p? = 9n2, which implies True O False

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Suppose that the number 6 is rational. Then 6 = 2 for some p, q e Z with q 0. Suppose that the fraction 2 is in its simplest form, i.e. all common factors between p and q have been cancelled in the fraction. This implies that 6q? = p?, and therefore p2 is a multiple of 3. This in turn means that p = 3n for some n e Z. I.e., p? = 9n2, which implies that 6q? = 9p?, and thus we arrive at an equation 2q² = 3n2. Since this implies that q? is a multiple of 3. we may set q = 3r for some re Z. Therefore both integers p and q are multiples of 3. Contradiction. This proves that 6 is an irrational number. O True O False
Transcribed Image Text:Suppose that the number 6 is rational. Then 6 = 2 for some p, q e Z with q 0. Suppose that the fraction 2 is in its simplest form, i.e. all common factors between p and q have been cancelled in the fraction. This implies that 6q? = p?, and therefore p2 is a multiple of 3. This in turn means that p = 3n for some n e Z. I.e., p? = 9n2, which implies that 6q? = 9p?, and thus we arrive at an equation 2q² = 3n2. Since this implies that q? is a multiple of 3. we may set q = 3r for some re Z. Therefore both integers p and q are multiples of 3. Contradiction. This proves that 6 is an irrational number. O True O False
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