Question

Quick tutorial: Another way to prove numbers are irrational.

Answer 1

This is going to be quick like I said:
Suppose you have the following question:

Prove that following:
\[\sqrt{3}\] is irrational

Answer 2

Now normally people would use a proof by contradiction as follows: Suppose it is rational. Thus it can be expressed as the ratio of integers such that they do not have common factors (or in equation form):\[Suppose \sqrt{3} = p/q; p, q \in Z\] with p/q as a simple fraction.
\[Thus 3 = p^2/q^2, 3q^2 = p^2\]

implying p is of the form 3k , k being an integer, suggesting that p has a factor of 3. Substituting 3k as p into the original statement yields \[q^2 = 3k^2\] thus implying that q^2 and thus q has a factor of 3. But we established that p and q don't have common factors and yet they both have 3 as a factor. Thus, \[\sqrt{3} is irrational.\]

Answer 3

Now going on to the method I mentioned actually involves the rational root theorem. As a refresher, the Rational Root Theorem is as follows:

For a polynomial of the form \[ax^n +....................+p\], its roots are of the form \[\pm (p/q)\], with p and q being factors of a and p respectively.

Answer 4

Going back to the original example proving that \[\sqrt{3}\] is irrational,
let x = 3^1/2. Thus, squaring both sides yields\[x^2 = 3 \rightarrow x^2 - 3 = 0\] By the rational root theorem, the possible candidates for solutions of the equation are \[\pm (p/q) = \pm (3/1), \pm (1/1)\] Plugging in each does not yield a solution to the quadratic equation above. Thus, we conclude by the Rational Root Theorem that 3^1/2 is irrational. I find this a more straightforward way for some people when they get confused with proof by contradictions and the problem doesn't require that much "plugging in."

Answer 5

Another example:
Prove that \[\sqrt{3}-\sqrt{2}\] is irrational using the Rational Root Theorem.

Answer 6

Let x= \[\sqrt{3}-\sqrt{2}\]
Squaring both sides yields \[x^2 = 1 -2\sqrt{6}\] or \[x^2 - 1 = 2\sqrt{6}\].

Answer 7

Squaring both sides yields \[x^4 -2x^2 +1 = 24\] or \[x^4 - 2x^2 - 23 = 0\] Using the rational root theorem, the possible roots are \[\pm(23/1), \pm(1/1)\], none of which are solutions. Thus, \[\sqrt{3}-\sqrt{2}\] is irrational.

Answer 8

For now the tutorial posts have ended, but I'll leave you guys a problem for those who want to practice. Prove using the Rational Root Theorem: \[\sqrt{2+\sqrt[3]{5}} \] is irrational.

Answer 9

\[x=\sqrt{2+\sqrt[3]5}\]\[x^2=2+\sqrt[3]5\]\[x^2-2=\sqrt[3]5\]\[(x^2-2)^3=5\]\[x^6-6x^4+12x^2-8=5\]\[x^6-6x^4+12x^2-13=0\]Hence, rational roots must be one of\[x=\pm13\quad x=\pm1\]Neither are equal to our original root, so it must be irrational.

This is an excellent method I had not heard of before. Thanks for the information.

Answer 10

How about \[\sqrt[3]{6+\sqrt{7}}\]For the next person who want to practice a bit.

Answer 11

Here's another fun one for a little extra challenge.

Prove: \(\sqrt{n+\sqrt[3]{n+1}}\) is irrational for all \(n∈\mathbb{N}\).

Answer 12

\[\sqrt{n+(n+1)^{1/3}}\]

Answer 13

\[(n+(n+1)^{1/3})^{1/2}\]

Answer 14

\[x = \sqrt[3]{6+\sqrt{7}} \implies x^3 - 6 = \sqrt{7} \implies x^6 + 36 - 12x^3 = 7\]

\[x^6 - 12x^3 + 29 = 0\]

Possible roots will be: \(\pm 29\) Or \(\pm 1\)

Since none of these are the solutions, So:

\(\sqrt[3]{6+\sqrt{7}}\) is irrational..

Have I tried right ??

Answer 15

Yes you have!

Answer 16

But how we say that these are not the solutions I am not getting that part..

How can we say that \(\pm 29\) is not the solution ??

Answer 17

Since they are not roots to the polynomial provided which was derived by setting x equal to the number . The rational root theorem helps to provide all possible rational roots with which to work with; if none of the candidates are solutions, then we can say that there are no rational roots, and thus the roots that DO satisfy the polynomial are irrational by nature .

Answer 18

It seems like the sum and difference of two irrational numbers (not the same) should be irrational always. Is this true or not, do you know?
Great post!

Answer 19

^In fact it is. A proof by contradiction would prove this. And thanks for the comment!

Answer 20

How about other operations:
add a rational and irrational
multiply a rational and irrational
take the rational root of an irrational
Do you know?

Answer 21

Addition of a rational and irrational is irrational always. Same for multiplying a rational by an irrational. I believe that rational root of an irrational is also irrational. All of these seem to be proven using my method, but I find that using a proof by contradiction is perhaps easier for me in regards to these operations.

Answer 22

It's like the old story about wine. If you put a teaspoon of sewage in a barrel of wine what have you got?
A barrel of sewage.

Answer 23

Lol.