Question

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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Answer 1

Answer:

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

Answer 2

Answer:

(a)40sqcm...........

Step-by-step explanation:

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