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 :)

Re: Showing that a number is rational/irrational

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**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 $\displaystyle 1 ~,~ \sqrt 2 ~,\&~ \sqrt {\frac{3}{2}} $ as rational or irrational ?

Re: Showing that a number is rational/irrational

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**Plato** Can you classify each of $\displaystyle 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?

Re: Showing that a number is rational/irrational

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**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 $\displaystyle \sqrt {\frac{3}{2}} $ were rational then what would mean ?

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

Re: Showing that a number is rational/irrational

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**Plato** If $\displaystyle \sqrt {\frac{3}{2}} $ were rational then what would mean ?

Would that mean $\displaystyle \sqrt 3 = r\sqrt 2 $ where $\displaystyle r$ 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...

Re: Showing that a number is rational/irrational

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**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 $\displaystyle \sqrt 3 = r\sqrt 2 $ where $\displaystyle r$ is rational does that not mean that $\displaystyle 3 = r\sqrt 6 $ where $\displaystyle r$ is rational ?

What is wrong with that?

Re: Showing that a number is rational/irrational

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**Plato** Well, if $\displaystyle \sqrt 3 = r\sqrt 2 $ where $\displaystyle r$ is rational does that not mean that $\displaystyle 3 = r\sqrt 6 $ where $\displaystyle r$ 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

Re: Showing that a number is rational/irrational

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**Plato** Well, if $\displaystyle \sqrt 3 = r\sqrt 2 $ where $\displaystyle r$ is rational does that not mean that $\displaystyle 3 = r\sqrt 6 $ where $\displaystyle r$ is rational ? What is wrong with that?

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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 $\displaystyle K$ is a **non-square** positive integer then $\displaystyle \sqrt K$ is irrational. Can you prove that?

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

Re: Showing that a number is rational/irrational

Quote:

Originally Posted by

**Plato** You have a huge gape in understanding about rational numbers,

If $\displaystyle K$ is a **non-square** positive integer then $\displaystyle \sqrt K$ is irrational. Can you prove that?

So if $\displaystyle 3 = r\sqrt 6 $ where $\displaystyle r$ 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?

What can be said about

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

?

Re: Showing that a number is rational/irrational

Let's go back to your original question.

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

So let's focus on s = r - 1 instead. Again, if s is rational, surely s^{2} is rational as well.

Now: $\displaystyle 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 s^{2} is rational, then t = s^{2} - 7/2 must also be rational.

Now $\displaystyle t = 2\sqrt{3}$, 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 s^{2} 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 $\displaystyle \sqrt{3} \in \mathbb{Q}$.

Re: Showing that a number is rational/irrational

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

Now multiply that by $\displaystyle (x^2- 2x- \frac{5}{2})+ \sqrt{3}$ to get

$\displaystyle (x^2- 2x-\frac{5}{2})^3- 3= x^4- 4x^3- x^2+ 10x+ \frac{13}{4}$

Finally multiply by 4 to get $\displaystyle 4x^5- 16x^3- 4x^2+ 40x+ 13$

a fourth degree polynomial with integer coefficients.

Since that was created by multiplying $\displaystyle x- \sqrt{2}- \sqrt{3/2}$ by other terms, $\displaystyle x- \sqrt{2}- \sqrt{3/2}$ is a factor so $\displaystyle \sqrt{2}- \sqrt{3/2}$ 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 $\displaystyle \frac{m}{n}$ 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, $\displaystyle \frac{1}{2}$, $\displaystyle -\frac{1}{2}$, $\displaystyle \frac{13}{2}$, $\displaystyle \frac{1}{4}$, $\displaystyle -\frac{1}{4}$, $\displaystyle \frac{13}{4}$, and $\displaystyle -\frac{13}{4}$. 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 $\displaystyle \sqrt{2}+ \sqrt{3}{2}$ **is** a root, it cannot be rational.