February 20th, 2011, 05:31 AM  #1 
Member Joined: Nov 2010 Posts: 48 Thanks: 0  Problem on irrational numbers
If a, b are rational numbers such that , show that a=b=0. I have done , which is valid for rational numbers a and b only when a=b=0. However, or If the question is correct, then a=b=0 then which I know is wrong. Any idea? 
February 20th, 2011, 07:29 AM  #2  
Senior Member Joined: Dec 2007 Posts: 687 Thanks: 47  Re: Problem on irrational numbers Quote:
and so on  
February 20th, 2011, 08:05 AM  #3 
Member Joined: Nov 2010 Posts: 48 Thanks: 0  Re: Problem on irrational numbers
But if a=b=0, the expression becomes .

February 20th, 2011, 08:46 AM  #4  
Senior Member Joined: Dec 2007 Posts: 687 Thanks: 47  Re: Problem on irrational numbers Quote:
you just cannot divide by zero, otherwise yes, you'll get nonsense results  
February 20th, 2011, 08:54 AM  #5 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Problem on irrational numbers
0/0 is what's known as an indeterminate form.

February 20th, 2011, 08:58 AM  #6 
Member Joined: Nov 2010 Posts: 48 Thanks: 0  Re: Problem on irrational numbers
But the expression can also be written as . Then if a=b=0, we get . What then? eta Is the question correct? 
February 20th, 2011, 09:05 AM  #7  
Senior Member Joined: Dec 2007 Posts: 687 Thanks: 47  Re: Problem on irrational numbers Quote:
 
February 20th, 2011, 09:14 AM  #8 
Member Joined: Nov 2010 Posts: 48 Thanks: 0  Re: Problem on irrational numbers
almahed, if I understand correctly, if a=b=0, then would not make any sense if written as . Then would it be safe to say there are no rational numbers a and b that satisfies and the question is wrong?

February 20th, 2011, 09:30 AM  #9 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Problem on irrational numbers
Since ax = by is true for any x,y when a=b=0, you can then go on to show there are no other rational values of a and b that satisfy the equation, which boils down to showing that is irrational, as mentioned above by almahed. Use a proof by contradiction, where you begin by assuming it IS rational: where p and q are coprime, the fraction p/q is fully reduced. In order for to be equal to an even number, then q must be even, so p must be odd since they are coprime, but if q is even, then is divisible by 4, meaning p is even. Since we must have p being both odd and even, we have shown, by reductio ad absurdum, that our initial assumption that is rational is false. Thus, we know a = b = 0 is the only rational solution set to the original equation. 
February 20th, 2011, 09:47 AM  #10 
Member Joined: Nov 2010 Posts: 48 Thanks: 0  Re: Problem on irrational numbers
Thanks Mark, I got the proof of being irrational but I still can't grasp how to resolve the division by 0 problem. If , for rational solutions of a and b, we can't transform the equation? Lets forget about a=b=0 and write the equation as . Now the same equation can have no rational solutions, is that right? Sorry for being thick. 

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