Given that \(a<b\) and \( a \times b=p\)

so we can put \(a<\sqrt{p}<b\)

also \( \frac{2 \times a \times b}{a + b} <\sqrt{p} < \frac{a+b}{2} \)

this is a great methode to frames any squart number increasingly precise

Given that \(a<b\) and \( a \times b=p\)

so we can put \(a<\sqrt{p}<b\)

also \( \frac{2 \times a \times b}{a + b} <\sqrt{p} < \frac{a+b}{2} \)

this is a great methode to frames any squart number increasingly precise

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