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Question Number 12822 by fawadalamawan@gmail.com last updated on 02/May/17

prove by contradiction 9+13(√(3 )) is irrational

$${prove}\:{by}\:{contradiction}\:\mathrm{9}+\mathrm{13}\sqrt{\mathrm{3}\:} \\ $$$${is}\:{irrational} \\ $$

Answered by mrW1 last updated on 03/May/17

let us assume 9+13(√3) is rational,.i.e. there exist integer numbers a and b, b≠0, 9+13(√3)=(a/b) ⇒(√3)=((a−9b)/(13))=((integer)/(integer))=rational that is to say: if 9+13(√3) is assumed to be rational, then (√3) is also rational, but this is not true, therefore 9+13(√3) is irrational.

$${let}\:{us}\:{assume}\:\mathrm{9}+\mathrm{13}\sqrt{\mathrm{3}}\:{is}\:{rational},.{i}.{e}. \\ $$$${there}\:{exist}\:{integer}\:{numbers}\:{a}\:{and}\:{b},\:{b}\neq\mathrm{0}, \\ $$$$\mathrm{9}+\mathrm{13}\sqrt{\mathrm{3}}=\frac{{a}}{{b}} \\ $$$$\Rightarrow\sqrt{\mathrm{3}}=\frac{{a}−\mathrm{9}{b}}{\mathrm{13}}=\frac{{integer}}{{integer}}={rational} \\ $$$${that}\:{is}\:{to}\:{say}:\:{if}\:\mathrm{9}+\mathrm{13}\sqrt{\mathrm{3}}\:{is}\:{assumed} \\ $$$${to}\:{be}\:{rational},\:{then}\:\sqrt{\mathrm{3}}\:{is}\:{also}\:{rational}, \\ $$$${but}\:{this}\:{is}\:{not}\:{true},\:{therefore} \\ $$$$\mathrm{9}+\mathrm{13}\sqrt{\mathrm{3}}\:{is}\:{irrational}. \\ $$

Commented by Joel577 last updated on 04/May/17

how to prove that (√3) is irrational?

$${how}\:{to}\:{prove}\:{that}\:\sqrt{\mathrm{3}}\:{is}\:{irrational}? \\ $$

Commented by mrW1 last updated on 04/May/17

assume (√3) were rational, i.e. (√3)=(a/b), and (a/b) can not be further simplified, i.e. a and be have no common factor. 3=(a^2 /b^2 ) a^2 =3b^2 if b is even, i.e. b=2n a^2 =3×4n^2 =even ⇒a=even=2m ⇒a and b have common factor 2, (a/b) can be further simplified to (m/n), ⇒b is not even! if b is odd, i.e. b=2n+1 b^2 is also odd, 3b^2 is also odd, i.e. a^2 is odd, a must be also add, i.e. a=2m+1 (2m+1)^2 =3(2n+1)^2 4m^2 +4m+1=12n^2 +12n+3 2(m^2 +m)=6(n^2 +n)+1 even=odd ⇒ b is not odd. ⇒there exist no integer a and b to fulfill (√3)=(a/b), ⇒(√3) is irrational

$${assume}\:\sqrt{\mathrm{3}}\:{were}\:{rational},\:{i}.{e}. \\ $$$$\sqrt{\mathrm{3}}=\frac{{a}}{{b}},\:{and}\:\frac{{a}}{{b}}\:{can}\:{not}\:{be}\:{further} \\ $$$${simplified},\:{i}.{e}.\:{a}\:{and}\:{be}\:{have}\:{no} \\ $$$${common}\:{factor}. \\ $$$$ \\ $$$$\mathrm{3}=\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} } \\ $$$${a}^{\mathrm{2}} =\mathrm{3}{b}^{\mathrm{2}} \\ $$$$ \\ $$$${if}\:{b}\:{is}\:{even},\:{i}.{e}.\:{b}=\mathrm{2}{n} \\ $$$${a}^{\mathrm{2}} =\mathrm{3}×\mathrm{4}{n}^{\mathrm{2}} ={even} \\ $$$$\Rightarrow{a}={even}=\mathrm{2}{m} \\ $$$$\Rightarrow{a}\:{and}\:{b}\:{have}\:{common}\:{factor}\:\mathrm{2}, \\ $$$$\frac{{a}}{{b}}\:{can}\:{be}\:{further}\:{simplified}\:{to}\:\frac{{m}}{{n}}, \\ $$$$\Rightarrow{b}\:{is}\:{not}\:{even}! \\ $$$$ \\ $$$${if}\:{b}\:{is}\:{odd},\:{i}.{e}.\:{b}=\mathrm{2}{n}+\mathrm{1} \\ $$$${b}^{\mathrm{2}} \:{is}\:{also}\:{odd},\:\mathrm{3}{b}^{\mathrm{2}} \:{is}\:{also}\:{odd},\:{i}.{e}.\: \\ $$$${a}^{\mathrm{2}} \:{is}\:{odd},\:{a}\:{must}\:{be}\:{also}\:{add},\:{i}.{e}.\:{a}=\mathrm{2}{m}+\mathrm{1} \\ $$$$\left(\mathrm{2}{m}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{3}\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}{m}^{\mathrm{2}} +\mathrm{4}{m}+\mathrm{1}=\mathrm{12}{n}^{\mathrm{2}} +\mathrm{12}{n}+\mathrm{3} \\ $$$$\mathrm{2}\left({m}^{\mathrm{2}} +{m}\right)=\mathrm{6}\left({n}^{\mathrm{2}} +{n}\right)+\mathrm{1} \\ $$$${even}={odd} \\ $$$$\Rightarrow\:{b}\:{is}\:{not}\:{odd}. \\ $$$$ \\ $$$$\Rightarrow{there}\:{exist}\:{no}\:{integer}\:{a}\:{and}\:{b}\:{to}\:{fulfill}\:\sqrt{\mathrm{3}}=\frac{{a}}{{b}}, \\ $$$$\Rightarrow\sqrt{\mathrm{3}}\:{is}\:{irrational} \\ $$

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