Is it true or false for all instances of $m>0$, $\sqrt{m^2+1}$ is irrational? 
Is it true or false for all instances of $m>0$, $\sqrt{m^2+1}$ is irrational?

My textbook gives an answer

This is true for integers $m$:
$$m>0\implies m^2<m^2+1<m^2+2m+1=(m+1)^2$$
Since $m^2+1$ exists between two consecutive perfect squares it cannot itself be a perfect square and thus its square root is irrational.

but I'm still confused as to the answer. How exactly does one get $m^2+2m+1$?
 A: I'm assuming that you are missing the condition that $m\in\mathbb{N}$. Else, one can see that $m=\frac{3}{4}$ or $m=\sqrt{3}$ are both counter example.
Since $m^2+1\in\mathbb{N}$, it's not hard to see that $\sqrt{m^2+1}$ is either irrational or an integer (proof of this can be found online).
Hence, we want to prove that $m^2+1$ is never a perfect square. A sure condition to prove that an integer is never a perfect square is if there exists $k\in\mathbb{N}$ such that $k^2<m^2+1<(k+1)^2$. This is because $m^2+1$ would be exclusively between two adjacent perfect squares.
We can easily see that $m^2<m^2+1<(m+1)^2$. Hence $m^2+1$ can never be a perfect square.
A: As other responses have indicated, if $m$ is restricted to the positive integers, the result follows from the assertion that given any positive integer $N$, either $\sqrt{N}$ is an integer, or it is irrational.
To prove this assertion:

Preliminary results
(1)
If $N \in \Bbb{Z^+}$ and $p$ is a prime such that $p$ divides $N^2$, then $p$ divides $N$.
Proof: 
By the fundamental theorem of arithmetic, any positive integer $N > 1$ can be uniquely expressed as
$\prod_{i=1}^r (p_i)^{a_i}.$
Here $p_1, \cdots, p_r$ are prime numbers, 
and $a_1, \cdots, a_r$ are positive integers. 
This implies that $N^2$ can be uniquely expressed as 
$\prod_{i=1}^r (p_i)^{2a_i}.$
This means that the only prime numbers that occur in the prime factorization of $N^2$ also occur in the prime factorization of $N$.
(2)
Suppose that you have two positive integers $a,b$ and a prime $q_1$ such that $q_1 ~| ~a^2$ and $q_1 ~| ~b^2$. 
Then $q_1$ is a common factor to both $a$ and $b$.
Proof:
By (1) above, $q_1$ appears in the prime factorization of both $a$ and $b$.

Suppose that this assertion is not true.
Then, $\sqrt{N}$ is not an integer and $\sqrt{N}$ can be expressed as $\frac{p}{q}$, where $p,q$ are in lowest terms (i.e. $p,q$ are relatively prime).  Since $\sqrt{N}$ is not an integer, $q$ is a positive integer greater than $1$.
This implies that there is some prime number $q_1$ that is a factor of $q$.  Since $p,q$ relatively prime, the prime number $q_1$ can not be a factor of $p$.
However, this generates a contradiction, as follows:
Since $\sqrt{N} = \frac{p}{q}$, it is implied that 
$N = \frac{p^2}{q^2}.$
This implies that $q^2 \times N = p^2$.
Therefore, since $(q_1)^2$ divides $(q^2 \times N), ~~ (q_1)^2$ divides $p^2$.
Therefore, prime $q_1$ is a common factor to both $q^2$ and $p^2$.  By preliminary result (2) above, this implies that $q_1$ is a common factor to both $q$ and $p$.  This contradicts the premise that $p,q$ are relatively prime.
This contradiction was generated by the assumption that
$\sqrt{N} = \frac{p}{q}$, with $p,q$ relatively prime and $q > 1$.  Therefore, this assumption must be impossible.
Therefore, if $\sqrt{N}$ is not an integer, it can not be a rational number.
