Please Help!

Could someone please give me an example of an irrational number and explain why it is irrational?

https://www.mathsisfun.com/irrational-numbers.html

Thank you Ms. Sue!

You're welcome, Bethany.

Certainly! An example of an irrational number is √2 (read as square root of 2). To understand why it is irrational, we need to know the definition of an irrational number.

An irrational number is a real number that cannot be expressed as a fraction of two integers. In other words, it cannot be written as a/b, where a and b are integers and b is not zero.

Now, let's prove that √2 is irrational. We'll use proof by contradiction, which assumes the opposite of what we want to prove and shows that it leads to a contradiction.

Assume that √2 is rational, so it can be expressed as √2 = a/b, where a and b are integers with no common factors (except 1), and b is not zero.

Squaring both sides, we get 2 = (a^2)/(b^2), which implies that 2b^2 = a^2. This means that a^2 is an even number because multiplying any integer by 2 gives an even number.

Now, we notice that if a^2 is even, a must also be even (because the square of an odd number is odd). So, we can write a as a = 2k, where k is an integer.

Substituting a = 2k into our previous equation, we get 2b^2 = (2k)^2, which simplifies to 2b^2 = 4k^2. Dividing both sides by 2, we get b^2 = 2k^2.

The same logic applies here: If b^2 is even, then b must also be even. So we can write b as b = 2m, where m is another integer.

Substituting b = 2m into the equation b^2 = 2k^2, we get (2m)^2 = 2k^2, which simplifies to 4m^2 = 2k^2.

Dividing both sides by 2, we get 2m^2 = k^2. Following the same pattern as before, this equation tells us that k^2 is even, so k must also be even.

By assuming both a and b are even, we have contradicted our initial assumption that a/b has no common factors (except 1). The contradiction disproves our assumption that √2 is rational.

Therefore, we conclude that √2 is an irrational number.