# Math Help - Showing that a number is rational/irrational

1. ## Showing that a number is rational/irrational

I have a bunch of questions that I don't know how to approach about whether a number is rational or irrational

Could someone give me an idea of how to approach questions like the following:

1 + srt(2) + srt(3/2)

Is this rational or irrational (and why)?

thanks

2. ## Re: Showing that a number is rational/irrational

Originally Posted by kinhew93
I have a bunch of questions that I don't know how to approach about whether a number is rational or irrational

Could someone give me an idea of how to approach questions like the following:

1 + srt(2) + srt(3/2)

Is this rational or irrational (and why)?
Can you classify each of $1 ~,~ \sqrt 2 ~,\&~ \sqrt {\frac{3}{2}}$ as rational or irrational ?

3. ## Re: Showing that a number is rational/irrational

Originally Posted by Plato
Can you classify each of $1 ~,~ \sqrt 2 ~,\&~ \sqrt {\frac{3}{2}}$ as rational or irrational ?
Well obviously 1 is rational, srt(2) and srt(3) are irrational, but I don't know about srt(3)/srt(2), or what happens when you sum rational and irrational numbers?

4. ## Re: Showing that a number is rational/irrational

Originally Posted by kinhew93
Well obviously 1 is rational, srt(2) and srt(3) are irrational, but I don't know about srt(3)/srt(2), or what happens when you sum rational and irrational numbers?
If $\sqrt {\frac{3}{2}}$ were rational then what would mean ?

Would that mean $\sqrt 3 = r\sqrt 2$ where $r$ is rational ?

5. ## Re: Showing that a number is rational/irrational

Originally Posted by Plato
If $\sqrt {\frac{3}{2}}$ were rational then what would mean ?

Would that mean $\sqrt 3 = r\sqrt 2$ where $r$ is rational ?
Yes it would but I don't see the issue with that? I know that the product of a rational and irrational number is always an irrational number...

6. ## Re: Showing that a number is rational/irrational

Originally Posted by kinhew93
Yes it would but I don't see the issue with that? I know that the product of a rational and irrational number is always an irrational number...
Well, if $\sqrt 3 = r\sqrt 2$ where $r$ is rational does that not mean that $3 = r\sqrt 6$ where $r$ is rational ?
What is wrong with that?

7. ## Re: Showing that a number is rational/irrational

Originally Posted by Plato
Well, if $\sqrt 3 = r\sqrt 2$ where $r$ is rational does that not mean that $3 = r\sqrt 6$ where $r$ is rational ?
What is wrong with that?
I really don't know. Is it that 3/srt(6) cannot be rational? If it is is still cannot see why

8. ## Re: Showing that a number is rational/irrational

Originally Posted by Plato
Well, if $\sqrt 3 = r\sqrt 2$ where $r$ is rational does that not mean that $3 = r\sqrt 6$ where $r$ is rational ? What is wrong with that?
Originally Posted by kinhew93
I really don't know. Is it that 3/srt(6) cannot be rational? If it is is still cannot see why
You have a huge gape in understanding about rational numbers,
If $K$ is a non-square positive integer then $\sqrt K$ is irrational. Can you prove that?

So if $3 = r\sqrt 6$ where $r$ is rational what in wrong with that?

9. ## Re: Showing that a number is rational/irrational

Originally Posted by Plato
You have a huge gape in understanding about rational numbers,
If $K$ is a non-square positive integer then $\sqrt K$ is irrational. Can you prove that?

So if $3 = r\sqrt 6$ where $r$ is rational what in wrong with that?
Ok gotcha. Ive actually done this question now but I still don't see what I can say about the sum of irrational numbers?

Srt(2) + srt(3) + srt(6)
?

10. ## Re: Showing that a number is rational/irrational

Let's go back to your original question.

If $r = 1+ \sqrt{2} + \sqrt{\frac{3}{2}}$ is rational, surely r - 1 is rational.

So let's focus on s = r - 1 instead. Again, if s is rational, surely s2 is rational as well.

Now: $s^2 = \left(\sqrt{2} + \sqrt{\frac{3}{2}}\right)^2 = 2 + 2\(\sqrt{2})\left(\sqrt{\frac{3}{2}}\right) + \frac{3}{2} = \frac{7}{2} + 2\sqrt{3}$.

If s2 is rational, then t = s2 - 7/2 must also be rational.

Now $t = 2\sqrt{3}$, and if t is rational then t/2 = (1/2)t is also surely rational, which means that the square root of 3 is rational.

But this is absurd.

Working backwards, we see that t cannot be rational, which means that s2 cannot be rational, which means that s itself cannot be rational.

But this means (finally!) that r cannot be rational, for assuming r rational leads to $\sqrt{3} \in \mathbb{Q}$.

11. ## Re: Showing that a number is rational/irrational

The way I would do this would be to note that $(x- 1- \sqrt{2}-\sqrt{3/2})(x-1+ \sqrt{2}+ \sqrt{3/2})= (x- 1)^2- (\sqrt{2}+ \sqrt{3/2})^2= (x- 1)^2- (2+ \sqrt{3}+ 3/2)= (x^2- 2x+ 1- 2- 3/2)- \sqrt{3}= (x^2- 2x- \frac{5}{2})- \sqrt{3}$.

Now multiply that by $(x^2- 2x- \frac{5}{2})+ \sqrt{3}$ to get
$(x^2- 2x-\frac{5}{2})^3- 3= x^4- 4x^3- x^2+ 10x+ \frac{13}{4}$
Finally multiply by 4 to get $4x^5- 16x^3- 4x^2+ 40x+ 13$
a fourth degree polynomial with integer coefficients.

Since that was created by multiplying $x- \sqrt{2}- \sqrt{3/2}$ by other terms, $x- \sqrt{2}- \sqrt{3/2}$ is a factor so $\sqrt{2}- \sqrt{3/2}$ is a zero of the polynomial.

Now, the "rational root theorem" says that if a polynomial with integer has rational zeros then they must be of the form $\frac{m}{n}$ where the denominator, n, evenly divides the leading coefficient, here "4", and the numerator, m, evenly divides the constant term, here 13.

The only integers that evenly divide 4 are 1, -1, 2, -2, 4, and -4. The only integers that evenly divide 13 are 1, -1, 13, and -13. Thus, the only rational number that could possibly satisfy the equation are 1, -1, 13, -13, $\frac{1}{2}$, $-\frac{1}{2}$, $\frac{13}{2}$, $\frac{1}{4}$, $-\frac{1}{4}$, $\frac{13}{4}$, and $-\frac{13}{4}$. But putting those into the polynomial show that, in fact, none of them are actually zeros. Therefore no zeros of that polynomial are rational. Since $\sqrt{2}+ \sqrt{3}{2}$ is a root, it cannot be rational.