Can you prove that #sqrt(3)# is irrational?

Prove that #sqrt(3)# is irrational by contradiction. E.g. show that it is not rational

2 Answers
Oct 22, 2016

The idea behind the proof is pretty simple, although depending on how rigorous you want to be, the setup can make it a little lengthy. The idea is that you suppose it's of the form #a/b#, meaning #3b^2 = a^2#, and then show that the left hand side is divisible by #3# an odd number of times, whereas the right hand side is divisible by #3# an even number of times.

The same technique can be used to show that #sqrt(n)# is irrational for any integer which is not a perfect square.


Suppose, to the contrary, that #sqrt(3)# is rational. Then there exist integers #a# and #b# such that #sqrt(3) = a/b#. Without loss of generality, we will assume that #a# and #b# share no common factors, as if they did, we could cancel them and obtain two new integers with the desired properties.

#sqrt(3) = a/b#

#=> 3 = (a/b)^2#

#=> 3 = a^2/b^2#

#=> 3b^2 = a^2#

Let #k_a# be the greatest power of #3# by which #a# is divisible. Then #a = 3^(k_a)a_0#, where #a_0 = a/3^(k_a)# is an integer not divisible by #3#. Similarly, let #b = 3^(k_b)b_0# where #b_0 = b/3^(k_b)# is an integer not divisible by #3#.

Substituting those in, we get

#3(3^(k_b)b_0)^2 = (3^(k_a)a_0)^2#

#=> 3^(2k_b+1)b_0^2 = 3^(2k_a)a_0^2#

As neither #a_0# nor #b_0# is divisible by #3#, then by the uniqueness of the prime factorization of integers, we have

#2k_b+1 = 2k_a#.

However, as the left hand side of this is an odd number and the right is even, we have reached a contradiction. Thus our initial premise was false, meaning #sqrt(3)# is not rational, i.e., #sqrt(3)# is irrational. ∎

Oct 22, 2016

Answer:

Well, we'll prove it by proving that it's not a rational number. I'm really sorry this is long but hopefully I explain every step thoroughly.

Explanation:

So, let's start with the definition of a rational number:
Rational Number: a real number that can be expressed through a fraction of REAL, INTEGER, NONZERO numbers. The fraction has to be in it's lowest form (#1/2# instead of #2/4#)

Remember an integer is a whole number (1, 2, 3, etc) and not a fraction of any sort (so not 3.592 or #1/2#)

We know that 5 is rational because it can be expressed as #10/2# or #5/1#. 4.333333333(...) is rational because it can be expressed as #13/3#.

So, let's prove that #sqrt(3) = a/b#. We're doing this because we want to show that it can never be a rational number, so this proof should have a contradiction in it.

Let's start with the statement in the previous paragraph.
#sqrt(3) = a/b#. Let's square both sides to start.
#3 = a^2/b^2#. We just squared each number, getting rid of the root. Next, we'll multiply both sides by #b^2#

#3b^2 = a^2#

And now, we divide both sides by 3.

#b^2 = a^2 /3#

Ok so now we need to use some logic. We know both #a# and #b# are integers (whole numbers) because they have to be for #sqrt(3)# to be a rational number. So, this means that #a^2# is also an integer.

Now, this also means that #b# and #b^2# are both integers. Keep that in mind.

So, since #b#, #b^2#, #a#, and #a^2# are all integers, #a^2/3# should be an integer as well. We know this because #b^2# is an integer, and #a^2/3# is equal to #b^2#, so whatever #a^2/3# is, it has to be an integer since its equal to an integer. Here's a quick visualization of that.

#b^2# is an integer
#b^2 = a^2/3#
an integer #= a^2/3#

We figured that out through simple substitution of "an integer" and #b^2#.


Ok, so back to the problem. We just figured out that #a^2/3# is an integer. So, a number squared divided by three is a whole number. That means that #a^2# is divisible by three. This may not seen revolutionary, but it helps a lot.

Mathematical Rule: A square of an integer is divisible by the same prime numbers that it's square root is divisible by.

This rule means that #a^2# should be divisible by the same prime numbers that #a# is. And, 3 is a prime number. So, #a# should also be divisible by three.
Here's a pretty simple way to look at it. Let's say that #a=14#. This means that #a^2 = 196#. We know that #146# is divisible by #7# and #2# because #14# is divisible by #7# and #2#

So, now we know that #a# and #a^2# are divisible by 3.


Let's keep going here. Since we know that #a# is divisble by #3#, we can make this equation.
#a = 3c#, where #c# is an integer. This makes sense because 3 times some number equals a. But the key here is that C is an integer.
If we square both sides in the equation we made, we get that:
#a^2 = 9c^2#

Now, let's substitute the equation we just made above into the original equation. The original is:
#3b^2 = a^2 #

Our new one is:
#3b^2 =9c^2#

Let's divide both sides by 3 to get:
#b^2 =3c^2#

And divide both sides by 3 again:
#b^2/3 = c^2#

And suddenly we find ourselves repeating ourselves. Remember when we proved that #a# was divisible by 3? We can do that again here.

We can prove that b is divisible by 3 since c is an integer just like we proved that #a# was divisible by 3.

So, I'm going to skip over that step but re-read the part where we proved a was divisible by 3 if you want to jog your memory.


Ok, we're getting close to finishing. Let's examine the definition of a rational number again:
Rational Number: a real number that can be expressed through a fraction of REAL, INTEGER, NONZERO numbers. The fraction has to be in it's lowest form (1/2 instead of 2/4).

Based on what we have proven, #a# and #b# are both divisible by 3. If we had the fraction #a/b#, it would not be in it's simplest form because we can divide both sides by 3. Therefor, #sqrt(3)# cannot be rational since it's proposed fraction isn't in simplest form.

Since it's not rational, it has to be irrational.

Thanks for reading, sorry for it being so long!!!!