Use the Completeness Axiom to prove that there is a real numbersuch that x 2 =7.

Question

Use the Completeness Axiom to prove that there is a real numbersuch that x2 =7. Completeness Axiom: Each nonempty set of real numbersthat is bounded above has a supremum. a)If y is a rational number such that y2 >2, then y is an upper bound of S. b)Every rational number that is an upper bound of S is greater than1. c)The number q is rational. Use the Completeness Axiom to prove that there is a real numbersuch that x2 =7. Completeness Axiom: Each nonempty set of real numbersthat is bounded above has a supremum. a)If y is a rational number such that y2 >2, then y is an upper bound of S. b)Every rational number that is an upper bound of S is greater than1. c)The number q is rational. Completeness Axiom: Each nonempty set of real numbersthat is bounded above has a supremum. a)If y is a rational number such that y2 >2, then y is an upper bound of S. b)Every rational number that is an upper bound of S is greater than1. c)The number q is rational.

Explanation / Answer

Let S = { x in R |  x2 7 }. Clearly the set S is bounded above by3 since 7 < 32 . Then by completenessaxiom S has supremum    Let sup S = a.    Now we show that a2  = 7.Suppose a2 < 7.   Then7-a2 >0. Then we can find a large n such that7-a2 > 2a/n + 1/n2.     Then 7 > a2 + 2a/n +1/n2 = (a + 1/n )2.     That is a+1/n belong to S,  a contradiction to the fact that a is thesupremum of S.    Hence a2 = 7.    Let sup S = a.    Now we show that a2  = 7.Suppose a2 < 7.   Then7-a2 >0. Then we can find a large n such that7-a2 > 2a/n + 1/n2.     Then 7 > a2 + 2a/n +1/n2 = (a + 1/n )2.     That is a+1/n belong to S,  a contradiction to the fact that a is thesupremum of S.    Hence a2 = 7.

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