James Lobo

Posted on Thursday, 08 May, 2003 - 12:49 pm:

I was stuck on the following questions and was wondering if you could help.

1) Let a, b be rational and x irrational. Show that if (x+a)/(x+b) is rational then a=b.

2) Let x, y be rationals such that $(x^{2} + x + \sqrt{2}) / (y^{2} + y + \sqrt{2})$ is also rational. Prove that either $x = y$ or $x + y = - 1$ .

3) Prove that if n is any positive integer then $\sqrt{n} + \sqrt{2}$ is irrational.

Thanks.

Arun Iyer

Posted on Thursday, 08 May, 2003 - 02:57 pm:

For 1>
assume a not equal to b, then take (x+a)/(x+b) = c/d (since (x+a)/(x+b) is rational) then simply try and get a contradiction.

for 2>

let $(x^{2} + x + \sqrt{2}) / (y^{2} + y + \sqrt{2}) = c / d$ performing dividendo we get,

$(x^{2} - y^{2} + x - y) / (y^{2} + y + \sqrt{2}) = (c - d) / d$

now simply finish this one off using some logical arguments.

for 3>

if n is any positive integer then $\sqrt{n}$ is either rational or irrational. Now what do we know about $\sqrt{2}$ ?? and then what can we say about $\sqrt{n} + \sqrt{2}$ ??

love arun
P.S-> if anyone finds any mistake above please do point out and correct them if possible.All comments are welcome.

Demetres Christofides

Posted on Thursday, 08 May, 2003 - 03:14 pm:

For (2) you can also use (1) to show that x² + x = y² + y.

(3) If expression is rational then show that

\sqrt{n} - \sqrt{2}

is also rational. Now show that

\sqrt{2}

must also be rational, a contradiction. [Do you know how to prove that

\sqrt{2}

is irrational ?]

James Lobo

Posted on Thursday, 08 May, 2003 - 04:00 pm:

I know now how to solve 1)
I was wondering if you could show me how to do 2)

For 3) I am thinking that since (2)^0.5 is irrational and (n)^0.5 can be either rational or irrational. Looking at the case when (n)^0.5 is rational, we can prove by contradiction that a rational-irrational=irrational. When (n)^0.5 is irrational we know that in general,irrational-irrational=(rational or irrational). But the only way of obtaining a rational as an answer is if the two irrationals that we are subtracting cancel each other out or a real number occurs. This can't occur for this case because the irrational is of the form (n^0.5) therefore for this question an irrational-irrational=irrational.

Thanks

Marcos

Posted on Thursday, 08 May, 2003 - 04:24 pm:

For (2) did you understand what Demetres said?

We have to show $(x^{2} + x + \sqrt{2}) / (y^{2} + y + \sqrt{2})$ is irrational unless x=y or x+y = -1. From (1) we know that if r is irrational then (r+a)/(r+b) is also irrational unless a=b. In our example in (2) the irrational is $\sqrt{(} 2)$ and $a = x^{2} + x$ , $b = y^{2} + y$ .

Thus, we get to the point, x² + x = y² + y
We can solve this by rearranging:
x² - y² + x - y = 0
(x + y)(x - y) + (x - y) = 0
(x + y + 1)(x - y) = 0, giving the required results...

In Arun's method if we assume the numerator is not zero then it is rational but the denominator is irrational (make sure you can see why) and hence the fraction is irrational (again, make sure you can see why) contradicting the fact that (c - d)/d is rational. Thus, the numerator must be 0 and the result follows as before...

For (3), to be honest I don't quite get your reasoning of why it can't be a case of irrational - irrational = rational...
Try following Demetres hint for what seems to be the simplest solution...

Marcos

P.S. If I'm wrong anywhere please correct me anyone Also, if you need a better explanation don't hesitate to ask...

James Lobo

Posted on Thursday, 08 May, 2003 - 05:39 pm:

Hi Marcos

Question 2 is crystal clear but for question 3, I know how to prove that (2)^0.5 is irrational by contradiction but I was wondering if you could show me how to solve the whole of question 3.

Thanks

Marcos

Posted on Thursday, 08 May, 2003 - 07:41 pm:

Consider

\sqrt{n} + \sqrt{2}) (\sqrt{n} - \sqrt{2}) = n - 2

. Now, if we assume

\sqrt{n} + \sqrt{2}

is rational, it follows (by the rationality of the RHS) that

I hope that clears things up now,
Marcos