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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!

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    $\begingroup$ First prove $\frac{p+1}{n}\geq a$ then prove $\frac{p+1}{n}\leq b$. $\endgroup$ – 79037662 Feb 17 at 5:35
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$p$ 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 $a=\dfrac{p+1}n$.

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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.

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  • $\begingroup$ Sorry, I’m just struggling to understand the 4th line, about how to prove than p+1 / n > a, and how that ‘a’ being irrational affects the equation. Thanks. $\endgroup$ – Subbota Feb 17 at 6:12
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    $\begingroup$ @Subbota The question states that since $p$ is the largest integer such that $\frac{p}{n} \lt a$, this means that $\frac{p + 1}{n} \ge a$ (as, otherwise, you have $\frac{p + 1}{n} \lt a$, so $p + 1$ would be a larger integer which works, contradicting that $p$ is the largest). However, note that $p + 1$ and $n$ are integers, so $\frac{p + 1}{n}$ is rational. Since $a$ is irrational, it can't be equal. Thus, you have the strict inequality of $\frac{p + 1}{n} \gt a$. $\endgroup$ – John Omielan Feb 17 at 6:15
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    $\begingroup$ Ohhh I get it now. I was having trouble interpreting what they meant by ‘P was the largest integer’, assuming it mean that its greater than n. But now I understand it. Thank you. $\endgroup$ – Subbota Feb 17 at 6:33

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