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

Q : Show that the numbers (√(3 )) , 2 & (√8) cannot be terms of an arithmetic sequence.

$${Q}\::\:{Show}\:{that}\:{the}\:{numbers}\:\sqrt{\mathrm{3}\:}\:,\:\mathrm{2}\:\&\:\sqrt{\mathrm{8}}\:{cannot}\:{be}\:{terms}\:{of}\:{an}\:{arithmetic}\:{sequence}. \\ $$

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

(√3)<2<(√8) assume they are terms of an A.P., then we have 2=(√3)+md (√8)=2+nd with m,n ∈N, d ∈R. d=((2−(√3))/m)=(((√8)−2)/n) ⇒2(m+n)=m(√8)+n(√3) ⇒4(m+n)^2 =8m^2 +3n^2 +4mn(√6) ⇒(√6)=((4(m+n)^2 −8m^2 −3n^2 )/(4mn))=((integer)/(integer)) i.e. (√6) is rational, but (√6) is in fact not rational. contradiction! ⇒(√3), 2, (√8) can not be terms of an A.P.!

$$\sqrt{\mathrm{3}}<\mathrm{2}<\sqrt{\mathrm{8}} \\ $$ $${assume}\:{they}\:{are}\:{terms}\:{of}\:{an}\:{A}.{P}., \\ $$ $${then}\:{we}\:{have} \\ $$ $$\mathrm{2}=\sqrt{\mathrm{3}}+{md} \\ $$ $$\sqrt{\mathrm{8}}=\mathrm{2}+{nd} \\ $$ $${with}\:{m},{n}\:\in{N},\:{d}\:\in{R}. \\ $$ $${d}=\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{{m}}=\frac{\sqrt{\mathrm{8}}−\mathrm{2}}{{n}} \\ $$ $$\Rightarrow\mathrm{2}\left({m}+{n}\right)={m}\sqrt{\mathrm{8}}+{n}\sqrt{\mathrm{3}} \\ $$ $$\Rightarrow\mathrm{4}\left({m}+{n}\right)^{\mathrm{2}} =\mathrm{8}{m}^{\mathrm{2}} +\mathrm{3}{n}^{\mathrm{2}} +\mathrm{4}{mn}\sqrt{\mathrm{6}} \\ $$ $$\Rightarrow\sqrt{\mathrm{6}}=\frac{\mathrm{4}\left({m}+{n}\right)^{\mathrm{2}} −\mathrm{8}{m}^{\mathrm{2}} −\mathrm{3}{n}^{\mathrm{2}} }{\mathrm{4}{mn}}=\frac{{integer}}{{integer}} \\ $$ $${i}.{e}.\:\sqrt{\mathrm{6}}\:{is}\:{rational},\:{but}\:\sqrt{\mathrm{6}}\:{is}\:{in}\:{fact}\:{not} \\ $$ $${rational}.\: \\ $$ $${contradiction}! \\ $$ $$\Rightarrow\sqrt{\mathrm{3}},\:\mathrm{2},\:\sqrt{\mathrm{8}}\:{can}\:{not}\:{be}\:{terms}\:{of}\:{an}\:{A}.{P}.! \\ $$

Commented bymehdee42 last updated on 27/Apr/23

beravo sir W the result of contradiction can also be obtained frome the relation ((2−(√3))/m)=(((√8)−2)/n)⇒((2−(√3))/( (√8)−2)) =(m/n) , because we have rational number on one side anf dumb nimber on thr other side

$${beravo}\:{sir}\:{W} \\ $$ $${the}\:{result}\:{of}\:\:{contradiction}\:{can}\:{also}\:{be}\:{obtained}\:{frome}\:{the}\:{relation}\:\:\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{{m}}=\frac{\sqrt{\mathrm{8}}−\mathrm{2}}{{n}}\Rightarrow\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{8}}−\mathrm{2}}\:=\frac{{m}}{{n}}\:, \\ $$ $${because}\:{we}\:{have}\:{rational}\:{number}\:{on}\:{one}\:{side}\:{anf}\:{dumb}\:{nimber}\:{on}\:{thr}\:{other}\:{side} \\ $$

Commented bymehdee42 last updated on 27/Apr/23

anyway , your proof for this contradiction is appriciated

$${anyway}\:,\:{your}\:{proof}\:\:{for}\:{this}\:{contradiction}\:{is}\:{appriciated} \\ $$

Commented bymr W last updated on 27/Apr/23

but we can not generally say that ((irrational number)/(irrational number)) is also irrational.

$${but}\:{we}\:{can}\:{not}\:{generally}\:{say}\:{that} \\ $$ $$\frac{{irrational}\:{number}}{{irrational}\:{number}}\:{is}\:{also}\:{irrational}. \\ $$

Commented bymehdee42 last updated on 28/Apr/23

exactly

$${exactly} \\ $$

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