# Integer within certain interval.

I have to show that there is a unique integer within the interval $$[a,b]$$ where

$$a = -\frac{3}{2} + \sqrt{\frac{9}{4} + 2(n+1)}$$

and

$$b = -\frac{1}{2} + \sqrt{\frac{1}{4} + 2(n+1)}$$

as well as $$n \in \mathbb{N}$$.

Now the uniqueness is easy by showing that there can be one integer within this interval at maximum. This is done by calculating $$(b-a)$$ and finding that $$(b-a)<1$$.

But how do I prove that there indeed exists one?

• It might be worth seeing when $a$ and $b$ are integers. It then becomes easy Apr 28 '20 at 13:59
• But that won't prove for the cases when a and b are not integers. I think we can break the problem into 2 parts.... Apr 28 '20 at 14:01
• Wait, I think you went wrong somewhere, @Octavius. If you can prove 1=|b-a| for all a and b which are not integers, then there will definitely be an integer between a and b when and b are themselves not integers. Apr 28 '20 at 14:10
• If the difference between any 2 real numbers a and b is greater than 1, then there will be at least one integer between [a, b]. Apr 28 '20 at 14:15
• yes, but I proved that the difference (b-a) is smaller than 1, i.e. (b-a) < 1. Apr 28 '20 at 14:22

Assume the contrary - $$\nexists$$ an integer lying in $$[a,b]$$. Then, $$\exists\ m \in \mathbb{N}$$ such that $$[a,b] \subset [m,m+1]$$.

So, \begin{align} m &< a \\ \implies m+\frac{3}{2} &< \sqrt{\frac{9}{4} + 2(n+1)} \\ \implies m^2 + 3m &< 2n+ 2 \end{align} and \begin{align} b &< m+1 \\ \implies \sqrt{\frac{1}{4} + 2(n+1)} &< m+\frac{3}{2} \\ \implies 2n +2 &< m^2 + 3m + 2 \\ \implies 2n &< m^2 + 3m\end{align}

Hence, we must have $$m(m+3) = 2n+1$$. But, $$m(m+3)$$ is an even number, while $$2n+1$$ is odd.

This is a contradiction, and so the result is proved.

• An elegant solution, which I accept gratefully. Thank you. Apr 28 '20 at 18:29

I am down to something so I thought I should share it. It is given (or can be proved) b>a. Let a = m-h, b=m+k, where m is the integer unique integer lying in [a,b] and h and k are real positive numbers. We can (or have already above) prove that h and k are not integers that their sum is less than 1. First let n+1 be k. Reduce a and b to a form of k. Solving that, we get (2a+3)^2 = (2b+1)^2 + 8.

It reduces to (a+b+2)(a-b+1) = 2.

Putting in values for a and b, we get

(2m+h-k+2)(h+k+1) = 2.

This reduces to proving that m is indeed an integer. Hope this can help somehow.

Please tell if any mistakes are there.