Question Number 45712 by malwaan last updated on 15/Oct/18
$$\mathrm{prove}\:\mathrm{that}\:: \\ $$$$\sqrt{\mathrm{2}}\:\mathrm{is}\:\mathrm{irrationl}\:\mathrm{number} \\ $$
Commented by maxmathsup by imad last updated on 15/Oct/18
$${let}\:{suppose}\:\sqrt{\mathrm{2}}=\frac{{p}}{{q}}\:{with}\:{p}\:{and}\:{q}\:{integrsn}\:{natural}\:{and}\:\Delta\left({p},{q}\right)=\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{2}=\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{2}} }\:\Rightarrow{p}^{\mathrm{2}} =\mathrm{2}{q}^{\mathrm{2}} \:\:\Rightarrow\mathrm{2}/{p}^{\mathrm{2}} \:\Rightarrow\mathrm{2}/{p}\:\Rightarrow\:\exists{m}\:\in{N}\:/{p}=\mathrm{2}{m}\:\Rightarrow\mathrm{4}{m}^{\mathrm{2}} \:=\mathrm{2}{q}^{\mathrm{2}} \:\Rightarrow{q}^{\mathrm{2}} =\mathrm{2}{m}^{\mathrm{2}} \:\Rightarrow \\ $$$$\mathrm{2}/{q}^{\mathrm{2}} \:\Rightarrow\mathrm{2}/{q}\:\:\Rightarrow\mathrm{2}\in{D}_{{p}} \cap{D}_{{q}} \:\:{but}\:{this}\:{is}\:{impossible}\:{because}\:\Delta\left({p},{q}\right)=\mathrm{1}\: \\ $$$${finally}\:\:\sqrt{\mathrm{2}}\:\notin{Q}\:. \\ $$
Commented by malwaan last updated on 15/Oct/18
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by maxmathsup by imad last updated on 16/Oct/18
$${you}\:{are}\:{welcome}\:{sir}. \\ $$