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Question-166911




Question Number 166911 by rexford last updated on 02/Mar/22
Answered by MathsFan last updated on 02/Mar/22
suppose (√2) is rational  (√2)=(p/q) ⇒ 2=(p^2 /q^2 ) ⇒  2q^2 =p^2 .....(1)   2 divides p   2 divides p^2   let  r=(p/2) ⇒  p=2r.......(2)  2q^2 =4r^2   ⇒   q^2 =2r^2   2 divides q  2 divides q^2   p and p have 2 as common factor  but contradict the fact that p and q  are coprime  hence  (√2) is an irrational number.
$$\boldsymbol{{suppose}}\:\sqrt{\mathrm{2}}\:\boldsymbol{{is}}\:\boldsymbol{{rational}} \\ $$$$\sqrt{\mathrm{2}}=\frac{\boldsymbol{{p}}}{\boldsymbol{{q}}}\:\Rightarrow\:\mathrm{2}=\frac{\boldsymbol{{p}}^{\mathrm{2}} }{\boldsymbol{{q}}^{\mathrm{2}} }\:\Rightarrow\:\:\mathrm{2}\boldsymbol{{q}}^{\mathrm{2}} =\boldsymbol{{p}}^{\mathrm{2}} …..\left(\mathrm{1}\right) \\ $$$$\:\mathrm{2}\:\boldsymbol{{divides}}\:\boldsymbol{{p}} \\ $$$$\:\mathrm{2}\:\boldsymbol{{divides}}\:\boldsymbol{{p}}^{\mathrm{2}} \\ $$$$\boldsymbol{{let}}\:\:\boldsymbol{{r}}=\frac{\boldsymbol{{p}}}{\mathrm{2}}\:\Rightarrow\:\:\boldsymbol{{p}}=\mathrm{2}\boldsymbol{{r}}…….\left(\mathrm{2}\right) \\ $$$$\mathrm{2}\boldsymbol{{q}}^{\mathrm{2}} =\mathrm{4}\boldsymbol{{r}}^{\mathrm{2}} \:\:\Rightarrow\:\:\:\boldsymbol{{q}}^{\mathrm{2}} =\mathrm{2}\boldsymbol{{r}}^{\mathrm{2}} \\ $$$$\mathrm{2}\:\boldsymbol{{divides}}\:\boldsymbol{{q}} \\ $$$$\mathrm{2}\:\boldsymbol{{divides}}\:\boldsymbol{{q}}^{\mathrm{2}} \\ $$$$\boldsymbol{{p}}\:\boldsymbol{{and}}\:\boldsymbol{{p}}\:\boldsymbol{{have}}\:\mathrm{2}\:\boldsymbol{{as}}\:\boldsymbol{{common}}\:\boldsymbol{{factor}} \\ $$$$\boldsymbol{{but}}\:\boldsymbol{{contradict}}\:\boldsymbol{{the}}\:\boldsymbol{{fact}}\:\boldsymbol{{that}}\:\boldsymbol{{p}}\:\boldsymbol{{and}}\:\boldsymbol{{q}} \\ $$$$\boldsymbol{{are}}\:\boldsymbol{{coprime}} \\ $$$$\boldsymbol{{hence}}\:\:\sqrt{\mathrm{2}}\:\boldsymbol{{is}}\:\boldsymbol{{an}}\:\boldsymbol{{irrational}}\:\boldsymbol{{number}}. \\ $$

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