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ange these steps in the correct order to prove that there are no solutions in integers x and y to the equation 2x2+5y2 = 14. Rank the options below. In the latter case to 2x² = 9. so, the only possible values of y to try are 0 and 11. If yl 22, then 2x2 +52 ≥ 2x2+20 220, so, there are no solutions to the original equation. In the former case, we would be looking for solutions to 2x² = 14. Clearly, there are no integer solutions to these equations,

ange these steps in the correct order to prove that there are no solutions in integers x and y to the equation 2x2+5y2 = 14. Rank the options below. In the latter case to 2x² = 9. so, the only possible values of y to try are 0 and 11. If yl 22, then 2x2 +52 ≥ 2x2+20 220, so, there are no solutions to the original equation. In the former case, we would be looking for solutions to 2x² = 14. Clearly, there are no integer solutions to these equations,

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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12: Secs 1.7-1.8 Ⓡ Arrange these steps in the correct order to prove that there are no solutions in integers x and y to the equation 2x2 + 5₁2² = 14. Rank the options below. In the latter case to 2x² = 9. so, the only possible values of y to try are 0 and ±1. If (yl ≥ 2, then 2x² +5² ≥ 2x² + 20 ≥20, so, there are no solutions to the original equation. In the former case, we would be looking for solutions to 2x² = 14. Clearly, there are no integer solutions to these equations, Saved < Prev 15 of 16 *** Next h Help Save &
Transcribed Image Text:12: Secs 1.7-1.8 Ⓡ Arrange these steps in the correct order to prove that there are no solutions in integers x and y to the equation 2x2 + 5₁2² = 14. Rank the options below. In the latter case to 2x² = 9. so, the only possible values of y to try are 0 and ±1. If (yl ≥ 2, then 2x² +5² ≥ 2x² + 20 ≥20, so, there are no solutions to the original equation. In the former case, we would be looking for solutions to 2x² = 14. Clearly, there are no integer solutions to these equations, Saved < Prev 15 of 16 *** Next h Help Save &
2: Secs 1.7-1.8 Put the steps in correct order to prove that 2 is irrational. Barch Rank the options below. Substituting into the equation p3=2q³,853 = 2q3, which simplifies to 45³ =q³. Hence, 2 =p³ / q³, or, equivalently, p3= 2q3. This is obtained by cubing on both sides. Suppose that 21/3 or 2 2 is the rational number p/ q, where p and q are positive integers with no common factors. Hence, the assumption that 2 √2 is rational is wrong. It is irrational. Since q³ is even, q must be even. Hence, p and q are both even, that is, that 2 is a common divisor of p and q. Thus p3 is even. Since the product of odd numbers is odd, p is even, and hence p = 2s. This contradicts the choice of p/ q, such that p and q have no common factors. < Prev E 16 of 16 X Next ******* Me A Help Save & Exit
Transcribed Image Text:2: Secs 1.7-1.8 Put the steps in correct order to prove that 2 is irrational. Barch Rank the options below. Substituting into the equation p3=2q³,853 = 2q3, which simplifies to 45³ =q³. Hence, 2 =p³ / q³, or, equivalently, p3= 2q3. This is obtained by cubing on both sides. Suppose that 21/3 or 2 2 is the rational number p/ q, where p and q are positive integers with no common factors. Hence, the assumption that 2 √2 is rational is wrong. It is irrational. Since q³ is even, q must be even. Hence, p and q are both even, that is, that 2 is a common divisor of p and q. Thus p3 is even. Since the product of odd numbers is odd, p is even, and hence p = 2s. This contradicts the choice of p/ q, such that p and q have no common factors. < Prev E 16 of 16 X Next ******* Me A Help Save & Exit
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