Prove $\sqrt{n+1} < \sqrt{n} + 1 $

Question

Prove $\sqrt{n+1} < \sqrt{n} + 1 $ for all $n \ge 1$. I have proven the base step for $n = 1$. $\sqrt{2}$ is less than $2$. The inductive hypothesis is $\sqrt{n+1} < \sqrt{n} + 1$. From here, I am unsure of how to proceed. How can I show that the inequality $\sqrt{(k+1)+ 1} < \sqrt{k+1} + 1$? After simplifying the expression $\sqrt{(k+1)+1}$, it becomes $\sqrt{k+2}$. Perhaps I can use this somehow to come up with an expression? The right hand side does not have a radical over the 1, so I do not think it is a duplicate question that has an answer in the suggested question.

Answer 1

To prove by induction: Consider n=1, $\sqrt {1+1} \lt \sqrt{1} + 1$ $$\sqrt{2} \lt 2$$ Assume n=k true: $$\sqrt{k+1}\lt \sqrt{k} + 1$$ Consider n=k+1: If true, this would mean that: $$\sqrt{k+2}\lt\sqrt{k+1} + 1$$ We know that $\sqrt{k+1}\lt \sqrt{k} + 1$, so: $$\sqrt{k+1} + 1 \lt \sqrt{k} + 2$$ Hence: $$\sqrt{k+2}\lt\sqrt{k+1} + 1\lt\sqrt{k} + 2$$ Squaring both sides: $$k+2\lt k + 4 + 4\sqrt{k}$$ This is indeed true, so n=1 true, and n=k true implies that n=k+1 true hence true for all natural numbers (integers greater than or equal to 1).

Answer 2

If we assume $n > 0$ and $\sqrt{n + 1} > 0$ then $\sqrt n > 0$ so $\sqrt{n+1}^2 = n+1 < n + 2 \sqrt{n} + 1 = (\sqrt{n} + 1)^2$

So $\sqrt{n+1} < \sqrt{n}+1$

This relies upon knowing if $a,b > 0$ then $a > b \iff a^2 > b^2$ which is easy to prove and a basic known fact. Oh, and that $\sqrt{pos} > 0$ .

Answer 3

A geometric proof:

Consider two segments of lengths $1$ and $\sqrt{n}$. If you attach these segments by their endpoints at an angle of $90^\circ$ (so they become two legs of a right triangle), then by Pythagorean theorem the hypotenuse of this right triangle is $\sqrt{n+1}$.

Now increase the angle between the two "leg" segments. By the law of cosines, the distance between the "loose" endpoints also increases, until the angle becomes $180^\circ$. But now at $180^\circ$, the distance between the "loose" endpoints is just the sum of the two segments $\sqrt{n}+1$; therefore $$\sqrt{n+1} < \sqrt{n}+1. $$ Q.E.D.