# What are three irrational numbers between 2 and 3?

Apr 4, 2016

#### Explanation:

Powers of $2$ are $2 , 4 , 8 , 16 , 32$

and powers of $3$ are $3 , 9 , 27 , 81 , 243$

Hence $\sqrt{7}$, $\sqrt{17}$, $\sqrt{54}$ and $\sqrt{178}$ are all irrational numbers between $2$ and $3$,

as $4 < 7 < 9$; $8 < 17 < 27$; $16 < 54 < 81$ and $32 < 178 < 243$.

For other ways of finding such numbers see What are three numbers between 0.33 and 0.34?

Apr 4, 2016

$\sqrt{2} + 1 , e , \pi - 1$ and many others.

#### Explanation:

Adding on to the other answer, we can easily generate as many such numbers as we'd like by noting that the sum of an irrational with a rational is irrational. For example, we have the well known irrationals $e = 2.7182 \ldots$ and $\pi = 3.1415 \ldots$.

So, without worrying about the exact bounds, we can definitely add any positive number less than $0.2$ to $e$ or subtract a positive number less than $0.7$ and get another irrational in the desired range. Similarly, we can subtract any positive number between $0.2$ and $1.1$ and get an irrational between $2$ and $3$.

$2 < e < e + 0.1 < e + 0.11 < e + 0.111 < \ldots < e + \frac{1}{9} < 3$

$2 < \pi - 1.1 < \pi - 1.01 < \pi - 1.001 < \ldots < \pi - 1 < 3$

This can be done with any irrational for which we have an approximation for at least the integer portion. For example, we know that $1 < \sqrt{2} < \sqrt{3} < 2$. As $\sqrt{2}$ and $\sqrt{3}$ are both irrational, we can add $1$ to either of them to get further irrationals in the desired range:

$2 < \sqrt{2} + 1 < \sqrt{3} + 1 < 3$

Aug 9, 2017

Irrational numbers are those that never give a clear result. Three of those between $2 \mathmr{and} 3$ could be: $\sqrt{5} , \sqrt{6} , \sqrt{7}$, and there are many more that go beyond pre-algebra.

#### Explanation:

Irrational numbers are always approximations of a value, and each one tends to go on forever. Roots of all numbers that are not perfect squares (NPS) are irrational, as are some useful values like $\pi$ and $e$.

To find the irrational numbers between two numbers like $2 \mathmr{and} 3$ we need to first find squares of the two numbers which in this case are ${2}^{2} = 4 \mathmr{and} {3}^{2} = 9$.

Now we know that the start and end points of our set of possible solutions are $4 \mathmr{and} 9$ respectively. We also know that both $4 \mathmr{and} 9$ are perfect squares because squaring is how we found them.

Then using the definition above, we can say that the root of all NPS numbers between the two squares we just found will be irrational numbers between the original numbers. Between $4 \mathmr{and} 9$ we have $5 , 6 , 7 , 8$; whose roots are $\sqrt{5} , \sqrt{6} , \sqrt{7} , \sqrt{8.}$

The roots of these will be irrational numbers between $2 \mathmr{and} 3$.

Eg: $\sqrt{8} \approx 2.82842712474619 \ldots \ldots \ldots \ldots \ldots$ where the wavy lines mean approximately, or, we will never have the exact numerical answer.