Step 1: Assumption: Assume $\sqrt{3}$ is a rational number. It can be written as $\frac{p}{q}$, $q \neq 0$ in its simplest form, meaning $p$ and $q$ are co-prime. So,

$$\sqrt{3} = \frac{p}{q}$$

Squaring both sides gives:

$$3 = \frac{p^2}{q^2}$$

or

$$3q^2 = p^2$$

Step 2: Deduction for $p$: Because $p^2$ is 3 times an integer, $p^2$ is divisible by 3. If the square of a number is divisible by 3, the number itself must be divisible by 3. Therefore, $p$ is divisible by 3. Let $p = 3k$. Substituting this gives:

$$3q^2 = (3k)^2 = 9k^2$$

which simplifies to:

$$q^2 = 3k^2$$

Step 3: Contradiction: Because $q^2$ is 3 times an integer, $q$ must also be divisible by 3. This contradicts our initial assumption that $p$ and $q$ share no common factors other than 1. Because our logical steps are flawless, our initial assumption must be wrong.

∴ $\sqrt{3}$ is irrational.

Step 4: Application to $\sqrt{5}$ and $\sqrt{7}$: Yes, the same proof by contradiction approach works for $\sqrt{5}$ and $\sqrt{7}$. Since 5 and 7 are prime numbers, if $p^2$ is divisible by them, $p$ must also be divisible by them, leading to the same contradiction that $p$ and $q$ are not co-prime.

Step 5: Application to $\sqrt{10}$: For $\sqrt{10}$, the approach also works. If $10q^2 = p^2$, $p^2$ is divisible by 10. This implies $p$ is divisible by the prime factors of 10 (2 and 5), which again leads to $q$ being divisible by those factors, contradicting that $p$ and $q$ are co-prime.