Class 10th CBSE maths prove that √2 is irrational.

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× Prove that √2 is irrational.

Solution :-

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${Let}\:\sqrt{\mathrm{2}}\:{be}\:{an}\:{irrational}\:{number}, \\ $$ $${then}\:\sqrt{\mathrm{2}}\:=\:\frac{{p}}{{q}}\:\left(\:{where}\:{q}\:\neq\:\mathrm{0},\:{p}\:{and}\:{q}\:{are}\:{co}-{primes}\:\right) \\ $$ $$\:{squaring}\:{both}\:{sides}, \\ $$ $$\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}\:\:} =\:\left(\:\frac{{p}}{{q}}\right)^{\mathrm{2}} \\ $$ $$\:\:\mathrm{2}\:=\:\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{2}} } \\ $$ $$\:\:\frac{{p}^{\mathrm{2}} }{\mathrm{2}}\:=\:{q}^{\mathrm{2}\:} \:-\:{equation}\:\left[{a}\right] \\ $$ $$\Rightarrow\:\mathrm{2}\:{divides}\:{p}^{\mathrm{2}} \:{then}\:\mathrm{2}\:{will}\:{also}\:{divide}\:{p}.\left({theorem}\:\mathrm{1}.\mathrm{2}\right)\:-\:{equation}\:\left[{b}\right] \\ $$ $${Let}\:\frac{{p}}{\mathrm{2}}\:=\:{x}\:\left({where}\:{x}\:{is}\:{an}\:{integer}\right) \\ $$ $$\Rightarrow\:{p}\:=\:\mathrm{2}{x} \\ $$ $${put}\:{p}\:=\:\mathrm{2}{x}\:{in}\:{equation}\:\left[{a}\right], \\ $$ $$\:\:\frac{\left(\mathrm{2}{x}\right)^{\mathrm{2}} }{\mathrm{2}}\:=\:{q}^{\mathrm{2}} \\ $$ $$\:\:\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{2}}\:=\:{q}^{\mathrm{2}} \\ $$ $$\:\:\mathrm{2}{x}^{\mathrm{2}\:} =\:{q}^{\mathrm{2}} \\ $$ $$\:\:\frac{{q}^{\mathrm{2}} }{\mathrm{2}}\:=\:{x}^{\mathrm{2}} -\:{equation}\:\left[{c}\right] \\ $$ $$\Rightarrow\mathrm{2}\:{divides}\:{q}^{\mathrm{2}\:} {then}\:{it}\:{will}\:{also}\:{divide}\:{q}\:\left({theorem}\:\mathrm{1}.\mathrm{2}\right)\:-\:{equation}\:\left[{d}\right] \\ $$ $$\: \\ $$ $${From}\:\left[{b}\right]\:{and}\:\left[{d}\right]\:{we}\:{get}\:{p}\:{and}\:{q}\:{both}\:{have}\:\mathrm{2}\:{as}\:{their}\:{common}\:{factor},\:{but}\:{we}\:{assumed}\:{p}\:{and}\:{q}\:{as}\:{co}-{primes}. \\ $$ $${This}\:{contradiction}\:{occurs}\:{due}\:{to}\:{our}\:{false}\:{assumption}. \\ $$ $$\therefore\:{conclusion}\:−\:\sqrt{\mathrm{2}}\:{is}\:{irrat}\boldsymbol{{i}}{onal}. \\$

Class 10th CBSE maths prove that √2 is irrational.

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