Question

if p is a prime number and p divides b² where b is a +ve integer, then ​

Answer 1

$\pink{ Theorem : }}$

Let 'p' be a prime number. If 'p' divides b², where b is a positive integer,then p divides b.

$\green{ Proof : }}$

$Let \: p_{1}, p_{2}, p_{3}, \ldots , p_{n} \:be \\the \: factors \: of \: b .$

$\implies b = p_{1} p_{2} p_{3}\ldots p_{n}\\where \: p_{1}, p_{2}, p_{3}, \ldots , p_{n}\: are \\prime \:not \: necessarily \: all \: distinct .$

$\therefore ( p_{1} p_{2} p_{3}\ldots p_{n})(p_{1} p_{2} p_{3}\ldots p_{n})\\= p_{1}^{2}\cdot p_{2}^{2}\cdot p_{3}^{2}\ldots \cdot p_{n}^{2}\\p \: divides \: b^{2} \: ---(1)$

$\implies p \:is \: a \: prime \:factor \: of \: b^{2}$

$( from \: fundamental \: theorem \: of \\ arithmetic)$

$The \:prime \:factors \:of \:b^{2} \:are \:only \\p_{1}, p_{2}, p_{3}, \ldots , p_{n}\:-----(2)$

$( Uniqueness \: of \:factors \:of \\ fundamental\: theorem \:of \: arithmetic )$

$From \: (1);\:and \: (2) , we \:have \:'p' \:is \:a \\prime \:factor \:of \: p_{1}, p_{2}, p_{3}, \ldots , p_{n}$

$\green { p \: divides \: b }$

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