Question Number 54909 by Hassen_Timol last updated on 14/Feb/19
$${Do}\:{you}\:{know}\:{a}\:{geometrical}\:{proof} \\ $$$${of}\:{the}\:{irrationality}\:{of}\:{number}\:\sqrt{\mathrm{2}}\:? \\ $$
Commented by kaivan.ahmadi last updated on 14/Feb/19
https://jeremykun.com/2011/08/14/the-square-root-of-2-is-irrational-geometric-proof/
Commented by kaivan.ahmadi last updated on 14/Feb/19
$$\mathrm{you}\:\mathrm{can}\:\mathrm{see}\:\mathrm{in}\:\mathrm{above}\:\mathrm{page} \\ $$
Commented by Hassen_Timol last updated on 14/Feb/19
$${Thank}\:{you}\:{sir}\:! \\ $$
Commented by maxmathsup by imad last updated on 14/Feb/19
$${if}\:\sqrt{\mathrm{2}}{is}\:{rational}\:{we}\:{get}\:\sqrt{\mathrm{2}}=\frac{{p}}{{q}}\:\:\:{with}\:{p}\:{and}\:{q}\:{from}\:{Q}^{+\bigstar} \:\:{we}\:{can}\:{take}\: \\ $$$$\Delta\left({p},{q}\right)=\mathrm{1}\:\:\Rightarrow\mathrm{2}{q}^{\mathrm{2}} ={p}^{\mathrm{2}} \:\:\Rightarrow\:{p}^{\mathrm{2}} {is}\:{even}\:\Rightarrow{p}\:{even}\:\Rightarrow\exists\:{m}\:\in{N}\:/{p}=\mathrm{2}{m}\:\Rightarrow \\ $$$$\mathrm{2}{q}^{\mathrm{2}} =\mathrm{4}{m}^{\mathrm{2}} \:\Rightarrow{q}^{\mathrm{2}} =\mathrm{2}{m}^{\mathrm{2}} \:\Rightarrow{q}^{\mathrm{2}} \:{is}\:{even}\:\Rightarrow\:{q}\:{is}\:{even}\:\Rightarrow\exists{m}^{'} \:/{q}=\mathrm{2}{m}^{'} \:\Rightarrow \\ $$$$\mathrm{2}\:{divide}\:{p}\:{and}\:{divide}\:{q}\:\:{but}\:{this}\:{is}\:{impossible}\:{because}\:\Delta\left({p},{q}\right)=\mathrm{1}\:. \\ $$
Commented by maxmathsup by imad last updated on 14/Feb/19
$${this}\:{proof}\:{is}\:{algebric}. \\ $$