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Question-159015




Question Number 159015 by Armindo last updated on 11/Nov/21
Answered by mr W last updated on 12/Nov/21
say (√2)=(p/q) with gcd(p,q)=1  2=(p^2 /q^2 )  p^2 =2q^2  ⇒p=even=2k  4k^2 =2q^2   q^2 =2k^2  ⇒q=even=2h  ⇒both p and q are even, gcd(p,q)=2  ⇒contradiction with gcd(p,q)=1
$${say}\:\sqrt{\mathrm{2}}=\frac{{p}}{{q}}\:{with}\:{gcd}\left({p},{q}\right)=\mathrm{1} \\ $$$$\mathrm{2}=\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{2}} } \\ $$$${p}^{\mathrm{2}} =\mathrm{2}{q}^{\mathrm{2}} \:\Rightarrow{p}={even}=\mathrm{2}{k} \\ $$$$\mathrm{4}{k}^{\mathrm{2}} =\mathrm{2}{q}^{\mathrm{2}} \\ $$$${q}^{\mathrm{2}} =\mathrm{2}{k}^{\mathrm{2}} \:\Rightarrow{q}={even}=\mathrm{2}{h} \\ $$$$\Rightarrow{both}\:{p}\:{and}\:{q}\:{are}\:{even},\:{gcd}\left({p},{q}\right)=\mathrm{2} \\ $$$$\Rightarrow{contradiction}\:{with}\:{gcd}\left({p},{q}\right)=\mathrm{1} \\ $$

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