Prove that $~a,~b~$ are integers and $~a \ge2~$, then either $~b~$ is not divisible by $~a~$ or $~b+1~$ is not divisible by $~a~$.

Question

Prove that $~a,~b~$ are integers and $~a \ge2~$, then either $~b~$ is not divisible by $~a~$ or $~b+1~$ is not divisible by $~a~$.

Should I use contradiction of uniqueness of prime factorisation to prove or simply prove by mathematical induction?

Special case $,c=1,$ in the dupe. – Bill Dubuque Sep 20 '20 at 04:48 — Bill Dubuque, Sep 20 '20 at 04:48

Kenta S · Answer 1 · 2020-02-09T06:23:35.650

Assume for a contradiction that $a,b$ are integers such that $a\ge 2,$ and both $b$ and $b+1$ is divisible by $a.$ Then, $(b+1)-b=1$ must also be divisible by $a,$ a contradiction. (By definition, there exists an integer $n$ such that $an=1.$ However, since $a,1>0,$ $n$ must be greater than $0,$ or in other words, $n\ge 1.$ Thus, we have $1=an\ge a\ge 2,$ a contradiction.)

Thus, either $b$ or $b+1$ is not divisible by $a.$

score 2 · Answer 2 · answered Feb 09 '20 at 06:47

We can also explain by remainder theorem ,i.e let $b=aq+r$ where $q\in NU\{0\}$. And $r=0,1,2,...(a-1)$ .$$ $$ Case 1:: If r=0 then only b is divisible by a but $ b+1=aq+1$ hence (b+1) is not divisible by a.$$. $$ Case 2:: If. $1\lt r \lt (a-2)$ then neither b , nor b+1 is divisible by a.$$. $$ Case 3. If r=(a-1) the b+1=aq hence only (b+1) is divisible by a .

Prove that $~a,~b~$ are integers and $~a \ge2~$, then either $~b~$ is not divisible by $~a~$ or $~b+1~$ is not divisible by $~a~$.

2 Answers2