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