Prove that if we have a positive real number x satisfying its square i

1.	Prove that if we have a positive real number x satisfying its square is less than 2 , then one can always have a number greater than x such that it's square is also less than 2.
Answer» ANSWER: $0 < x < 1. And \: now \: squaring \: it \: we \: get. 0 < x^2 < 1.$

Prove that if we have a positive real number x satisfying its square is less than 2 , then one can always have a number greater than x such that it's square is also less than 2.

Answer»

ANSWER:

$0 < x < 1. And \: now \: squaring \: it \: we \: get. 0 < x^2 < 1.$

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Prove that if we have a positive real number x satisfying its square is less than 2 , then one can always have a number greater than x such that it's square is also less than 2.​

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Prove that if we have a positive real number x satisfying its square is less than 2 , then one can always have a number greater than x such that it's square is also less than 2.