I have a question. Suppose that $a$ and $b$ are two natural numbers so that $ a<b$ and $ a\nmid b$. Put $ d=ka$, where $ k\not=0,1,t\dfrac{b}{\gcd(a,b)}$, for $ t\geq 1$. I want to prove that $ b\nmid d$.
my proof: if $ b\mid d$ then $ d=ka=k'b$, for $ k'\in \mathbb{Z}$. Then $ k'=k\dfrac{a}{b}$, as regards $ a\nmid b$ and choice $ k$, to deduce that $ k'\not\in\mathbb{Z}$, which is impossible.
is correct my proof? please give me another proof. I'm in urgent need to find the answer.
Your proof is incorrect, as just because $\frac ab \not\in \mathbb Z$ doesn't mean that $k\frac ab \not\in \mathbb Z$.
You need to use the fact that $\displaystyle k \ne t\frac{b}{gcd(a,b)}$ for $t\ge 1$.
Starting from $ka =k'b$, that means that $\displaystyle k\frac{a}{gcd(a,b)} = k'\frac{b}{gcd(a,b)}$. But $\displaystyle \frac{a}{gcd(a,b)}$ and $\displaystyle \frac{b}{gcd(a,b)}$ are coprime, so that means that we must have $\frac{a}{gcd(a,b)} \mid k'$, so $$k=\left(\frac{k'gcd(a,b)}{a}\right)\left(\frac{b}{gcd(a,b)}\right)=t\frac{b}{gcd(a,b)},$$a contradiction.