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Question Number 8164 by 666225 last updated on 02/Oct/16

if p is prima show that (√p) irasional

$${if}\:\boldsymbol{{p}}\:{is}\:{prima}\:{show}\:{that}\:\sqrt{\boldsymbol{{p}}}\:{irasional} \\ $$

Commented by 123456 last updated on 02/Oct/16

1∣p,p∣p,p>1  p∣a^2 ⇔p∣a

$$\mathrm{1}\mid{p},{p}\mid{p},{p}>\mathrm{1} \\ $$$${p}\mid{a}^{\mathrm{2}} \Leftrightarrow{p}\mid{a} \\ $$

Commented by 123456 last updated on 02/Oct/16

(√p)=x=(a/b),(a,b)=1  p=x^2 =(a^2 /b^2 )  a^2 =pb^2 ⇒p∣a^2 ⇒p∣a  a=rp  (rp)^2 =pb^2   pb^2 =r^2 p^2     (p≠0)  b^2 =r^2 p⇒p∣b^2 ⇒p∣b  b=sp  (a,b)=(rp,sp)=(r,s)p≠1

$$\sqrt{{p}}={x}=\frac{{a}}{{b}},\left({a},{b}\right)=\mathrm{1} \\ $$$${p}={x}^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} } \\ $$$${a}^{\mathrm{2}} ={pb}^{\mathrm{2}} \Rightarrow{p}\mid{a}^{\mathrm{2}} \Rightarrow{p}\mid{a} \\ $$$${a}={rp} \\ $$$$\left({rp}\right)^{\mathrm{2}} ={pb}^{\mathrm{2}} \\ $$$${pb}^{\mathrm{2}} ={r}^{\mathrm{2}} {p}^{\mathrm{2}} \:\:\:\:\left({p}\neq\mathrm{0}\right) \\ $$$${b}^{\mathrm{2}} ={r}^{\mathrm{2}} {p}\Rightarrow{p}\mid{b}^{\mathrm{2}} \Rightarrow{p}\mid{b} \\ $$$${b}={sp} \\ $$$$\left({a},{b}\right)=\left({rp},{sp}\right)=\left({r},{s}\right){p}\neq\mathrm{1} \\ $$

Answered by prakash jain last updated on 02/Oct/16

answer in comments.

$$\mathrm{answer}\:\mathrm{in}\:\mathrm{comments}. \\ $$

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