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

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!

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

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

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

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

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

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.

• 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. – Subbota Feb 17 at 6:12
• @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 $\frac{p + 1}{n} \gt a$. – John Omielan Feb 17 at 6:15
• 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. – Subbota Feb 17 at 6:33