Let $l$ be a natural number. Prove that $n\lt\sqrt{n ^ 2 + l}\lt n+1$ for almost every $n$.

In my assignment I have to prove the following statement:

Let $l$ be a natural number. Prove that for almost every $n$ the following inequality is true: $$n\lt\sqrt{n ^ 2 + l}\lt n+1$$

I chose to prove this by contradiction, and I wanted to know if it's correct.

• It is obvious that $n \lt n+1$.
• Assume for the sake of contradiction that $n \ge \sqrt{n ^ 2 + l}$.

Squaring both sides, we get: \begin{align} n^2 &\ge n^2 + l \\ 0 &\ge l \end{align} which contradicts the fact that $l$ is natural. Therefore, $n \lt \sqrt{n ^ 2 + l}$.

• Assume for the sake of contradiction that $\sqrt{n ^ 2 + l} \ge n+1$.

Squaring both sides, we get: \begin{align} n^2+l &\ge n^2+2n+1 \\ l &\ge 2n+1 \end{align} but this is another contradiction, because $l$ is a constant number, and can't be bigger than an infinite amount of numbers.

Therefore the inequality $n\lt\sqrt{n ^ 2 + l}\lt n+1$ is true.

Is my solution correct?

Thank you,

Alan

• `can't be bigger than an infinite amount of numbers.' is not a phrase you want to see in a proof. The idea is there though Apr 28, 2015 at 16:23
• @JackYoon thank you, I am very glad to hear that. Can you be more specific about the idea I have to improve there?
– Alan
Apr 28, 2015 at 16:24
• Search for $n$ such that which the inequality definitely hold. Apr 28, 2015 at 16:26
• Let's say I choose n to be bigger than $2n+1$. Then for almost every n, it's bigger than this $l$, and we have a contradiction?
– Alan
Apr 28, 2015 at 16:40
• Minor point: consider using a letter other than $l$ for a variable, looks too much like $1$.
– user153918
Apr 28, 2015 at 17:53

$$l\geq 2n+1$$ This doesn't contradict a known fact. It just states a boundary for $l$ given your assumption is correct. In a proof by contradiction, you need to arrive at results contradicting a general result for all $n$. Also $\sqrt {n^2+l}<n+1$ isn't true for all $n$, given an $l$. In fact, it is true only for $n>{\dfrac{l-1}{2}}$

To prove the inequality, you can simply say that it holds for all $n>{\dfrac{l-1}{2}}$, which will also satisfy the phrase "for almost every $n$".

• Thank you. I see what is wrong, how do I make it right?
– Alan
Apr 28, 2015 at 16:34
• Let's say I choose n to be bigger than $2n+1$. Then for almost every n, it's bigger than this $l$, and we have a contradiction?
– Alan
Apr 28, 2015 at 16:41
• @Alan, the answer has been edited. Hope it is clear now. (It had a small mistake earlier) Apr 30, 2015 at 3:35
• thank you! I'll have a look.
– Alan
Apr 30, 2015 at 3:50