Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 149567 by mathdanisur last updated on 06/Aug/21

if q is prime number fixed, then solve for natural numbers the equation: (1/q) = (1/x) + (1/y) - (1/z)

$${if}\:\:\boldsymbol{{q}}\:\:{is}\:{prime}\:{number}\:{fixed},\:{then} \\ $$$${solve}\:{for}\:{natural}\:{numbers}\:{the}\:{equation}: \\ $$$$\frac{\mathrm{1}}{{q}}\:=\:\frac{\mathrm{1}}{{x}}\:+\:\frac{\mathrm{1}}{{y}}\:-\:\frac{\mathrm{1}}{{z}} \\ $$

Commented by Rasheed.Sindhi last updated on 07/Aug/21

My answer is too lengthy. The question is waiting for a better solution.

$$\mathrm{My}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{too}\:\mathrm{lengthy}. \\ $$$$\mathrm{The}\:\mathrm{question}\:\mathrm{is}\:\mathrm{waiting}\:\mathrm{for}\:\mathrm{a}\:\mathrm{better} \\ $$$$\mathrm{solution}. \\ $$

Commented by mathdanisur last updated on 07/Aug/21

Dear Ser We hawe: (1/n) = (1/(2(n-1))) + (1/(2(n+1))) - (1/((n-1)(n+1))) ⇒ n=q .... Ser, this true or wrong.?

$$\mathrm{Dear}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$$$\mathrm{We}\:\mathrm{hawe}:\:\frac{\mathrm{1}}{\mathrm{n}}\:=\:\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{n}-\mathrm{1}\right)}\:+\:\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{n}+\mathrm{1}\right)}\:-\:\frac{\mathrm{1}}{\left(\mathrm{n}-\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)} \\ $$$$\Rightarrow\:\mathrm{n}=\mathrm{q}\:.... \\ $$$$\mathrm{Ser},\:\mathrm{this}\:\mathrm{true}\:\mathrm{or}\:\mathrm{wrong}.? \\ $$

Commented by Rasheed.Sindhi last updated on 07/Aug/21

(1/(2(q−1)))+(1/(2(q+1)))−(1/((q−1)(q+1))) =((q+1+q−1−2)/(2(q−1)(q+1)))=((2q−2)/(2(q−1)(q+1))) =((2(q−1))/(2(q−1)(q+1)))=(1/(q+1))≠(1/q) Not correct ser.

$$\frac{\mathrm{1}}{\mathrm{2}\left({q}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}\left({q}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\left({q}−\mathrm{1}\right)\left({q}+\mathrm{1}\right)} \\ $$$$\:\:\:=\frac{{q}+\mathrm{1}+{q}−\mathrm{1}−\mathrm{2}}{\mathrm{2}\left({q}−\mathrm{1}\right)\left({q}+\mathrm{1}\right)}=\frac{\mathrm{2}{q}−\mathrm{2}}{\mathrm{2}\left({q}−\mathrm{1}\right)\left({q}+\mathrm{1}\right)} \\ $$$$=\frac{\cancel{\mathrm{2}\left({q}−\mathrm{1}\right)}}{\cancel{\mathrm{2}\left({q}−\mathrm{1}\right)}\left({q}+\mathrm{1}\right)}=\frac{\mathrm{1}}{{q}+\mathrm{1}}\neq\frac{\mathrm{1}}{{q}} \\ $$$${Not}\:{correct}\:{ser}. \\ $$

Commented by mathdanisur last updated on 07/Aug/21

Sorry Ser, ... (1/(n(n-1)(n+1)))

$$\mathrm{Sorry}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\:...\:\frac{\mathrm{1}}{\mathrm{n}\left(\mathrm{n}-\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)} \\ $$

Commented by Rasheed.Sindhi last updated on 07/Aug/21

(1/(2(n-1))) + (1/(2(n+1))) - (1/(n(n-1)(n+1))) ((n(n+1)+n(n−1)−2)/(2n(n-1)(n+1))) ((n^2 +n+n^2 −n−2)/(2n(n-1)(n+1)))=((2n^2 −2)/(2n(n-1)(n+1))) ((2(n−1)(n+1))/(2n(n-1)(n+1)))=(1/n) ✓ It means that this is the solution.

$$\:\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{n}-\mathrm{1}\right)}\:+\:\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{n}+\mathrm{1}\right)}\:-\:\frac{\mathrm{1}}{\mathrm{n}\left(\mathrm{n}-\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)} \\ $$$$\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)+\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)−\mathrm{2}}{\mathrm{2n}\left(\mathrm{n}-\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)} \\ $$$$\frac{\mathrm{n}^{\mathrm{2}} +\mathrm{n}+\mathrm{n}^{\mathrm{2}} −\mathrm{n}−\mathrm{2}}{\mathrm{2n}\left(\mathrm{n}-\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)}=\frac{\mathrm{2n}^{\mathrm{2}} −\mathrm{2}}{\mathrm{2n}\left(\mathrm{n}-\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)} \\ $$$$\frac{\mathrm{2}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2n}\left(\mathrm{n}-\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{n}}\:\checkmark \\ $$$$\mathrm{It}\:\mathrm{means}\:\mathrm{that}\:\mathrm{this}\:\mathrm{is}\:\mathrm{the}\:\mathrm{solution}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com