Prove $a < (p+1)/n < b$ if $a,b$ are irrationals with $0<a<b$ and $p$ is greatest integer with $p/n < a$

Question

Suppose that $a$ and $b$ are positive irrational numbers, where $a < b$. Choose any positive integer $n$ such that $1/n < b - a$, and let $p$ be the greatest integer such that $p/n < a$.

Prove that the rational number $(p + 1)/n$ lies between $a$ and $b$.

I’ve been stuck on this question, attempting to merge all of the inequalities into one equation, such that $$ \frac{p}{n} < a < \frac{1}{n} + a < b $$ Can anyone advise me on how to properly solve this equation. I’ve reached a dead end and feel like I’ve done something wrong. Any help is appreciated, thanks!

Answer 1

$p$ is defined so $na-1\le p\lt na$,

so $na\le p+1\lt na+1$,

so $a\le \dfrac{p+1}n\lt a+\dfrac1n<b$, since $\dfrac1n<b-a$,

and, since $a$ is irrational, we can't have $a=\dfrac{p+1}n$.

Answer 2

You have

$$\frac{1}{n} \lt b - a \tag{1}\label{eq1A}$$

$$\frac{p}{n} \lt a \tag{2}\label{eq2A}$$

Adding these $2$ inequalities together (since if $c \lt d$ and $e \lt f$, then $c + e \lt d + f$) gives

$$\frac{p + 1}{n} \lt (b - a) + a = b \tag{3}\label{eq3A}$$

Since $p$ is the greatest integer such that $\frac{p}{n} \lt a$, as $a$ is irrational (so since $\frac{p + 1}{n}$ is rational, it can't be equal to $a$), you also have that

$$\frac{p + 1}{n} \gt a \tag{4}\label{eq4A}$$

Putting \eqref{eq3A} and \eqref{eq4A} together gives

$$a \lt \frac{p + 1}{n} \lt b \tag{5}\label{eq5A}$$

as requested.