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Question Number 26410 by Joel578 last updated on 25/Dec/17

If GCD(a,b) = 1 and GCD(c, d) = 1 a ≠ b ≠ c ≠ d ≠ 1, a < b, c < d is it possible that (a/b) + (c/d) is an integer number?

IfGCD(a,b)=1andGCD(c,d)=1a≠b≠c≠d≠1,a<b,c<disitpossiblethatab+cdisanintegernumber?

Answered by mrW1 last updated on 25/Dec/17

since gcd(c,d)=1, (c/d) is already reduced to lowest terms, i.e. (c/d) can not be simplified further. let′s asume (a/b)+(c/d)=n, then (c/d)=n−(a/b)=((nb−a)/b)=(e/b) with e=nb−a=integer that means (c/d) can be simplified to (e/b) with b<d. but this is a contradiction. ⇒(a/b)+(c/d) can not be an integer number.

sincegcd(c,d)=1,cdisalreadyreducedtolowestterms,i.e.cdcannotbesimplifiedfurther.let′sasumeab+cd=n,thencd=n−ab=nb−ab=ebwithe=nb−a=integerthatmeanscdcanbesimplifiedtoebwithb<d.butthisisacontradiction.⇒ab+cdcannotbeanintegernumber.

Commented by Joel578 last updated on 27/Dec/17

thank you very much

thankyouverymuch

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