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Question Number 191525 by mehdee42 last updated on 25/Apr/23

$$Q:Showthatthenumbers\sqrt{3},2\mathrm{\&}\sqrt{8}cannotbetermsofanarithmeticsequence.$$

Answered by mr W last updated on 26/Apr/23

$$\sqrt{3}<2<\sqrt{8}$$ $$assumetheyaretermsofanA.P.,$$ $$thenwehave$$ $$2=\sqrt{3}+md$$ $$\sqrt{8}=2+nd$$ $$withm,n\in N,d\in R.$$ $$d=\frac{2-\sqrt{3}}{m}=\frac{\sqrt{8}-2}{n}$$ $$\Rightarrow 2(m+n)=m\sqrt{8}+n\sqrt{3}$$ $$\Rightarrow 4{(m+n)}^2=8m^2+3n^2+4mn\sqrt{6}$$ $$\Rightarrow \sqrt{6}=\frac{4{(m+n)}^2-8m^2-3n^2}{4mn}=\frac{integer}{integer}$$ $$i.e.\sqrt{6}isrational,but\sqrt{6}isinfactnot$$ $$rational.$$ $$contradiction!$$ $$\Rightarrow \sqrt{3},2,\sqrt{8}cannotbetermsofanA.P.!$$

Commented bymehdee42 last updated on 27/Apr/23

$$beravosirW$$ $$theresultofcontradictioncanalsobeobtainedfrometherelation\frac{2-\sqrt{3}}{m}=\frac{\sqrt{8}-2}{n}\Rightarrow \frac{2-\sqrt{3}}{\sqrt{8}-2}=\frac{m}{n},$$ $$becausewehaverationalnumberononesideanfdumbnimberonthrotherside$$

Commented bymehdee42 last updated on 27/Apr/23

$$anyway,yourproofforthiscontradictionisappriciated$$

Commented bymr W last updated on 27/Apr/23

$$butwecannotgenerallysaythat$$ $$\frac{irrationalnumber}{irrationalnumber}isalsoirrational.$$

Commented bymehdee42 last updated on 28/Apr/23

$$exactly$$

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