Find and Prove the inf and sup:

{x is a rational number: x^2 < 2 }

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- Sep 17th 2009, 09:32 AMJdg6057infimums and supremums
Find and Prove the inf and sup:

{x is a rational number: x^2 < 2 } - Sep 20th 2009, 03:25 AMdavidlyness
If you were taking x in the real numbers, then your supremum would be $\displaystyle \sqrt 2 $ and your infimum would be $\displaystyle - \sqrt 2 $.

However, these are both irrational numbers. There does not exist a supremum (least upper bound) in the rational numbers. Say it did exist, and call it a. Then $\displaystyle {a^2} > 2$. Define $\displaystyle b = \frac{{2a + 2}}{{a + 2}}$. Then b is also an upper bound of your set, and $\displaystyle b < a$, which contradicts our assumption that a was our LEAST upper bound.

A similar argument works for the infimum.