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February 20th, 2011, 05:31 AM   #1
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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?
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February 20th, 2011, 07:29 AM   #2
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Re: Problem on irrational numbers

Quote:
Originally Posted by Deb_D
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?
I think you can use a similar argument to that of the irrationality of

and so on
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February 20th, 2011, 08:05 AM   #3
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Re: Problem on irrational numbers

But if a=b=0, the expression becomes .
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February 20th, 2011, 08:46 AM   #4
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Re: Problem on irrational numbers

Quote:
Originally Posted by Deb_D
But if a=b=0, the expression becomes .
but 0*2=0*5 is no absurd at all!

you just cannot divide by zero, otherwise yes, you'll get non-sense results
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February 20th, 2011, 08:54 AM   #5
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Re: Problem on irrational numbers

0/0 is what's known as an indeterminate form.
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February 20th, 2011, 08:58 AM   #6
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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?
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February 20th, 2011, 09:05 AM   #7
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Re: Problem on irrational numbers

Quote:
Originally Posted by Deb_D
But the expression can also be written as .
Then if a=b=0, we get . What then?

eta Is the question correct?
What you claim that "can be written" actually does not, it can be "physically written", not "mathematically" so to speak
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February 20th, 2011, 09:14 AM   #8
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Re: Problem on irrational numbers

al-mahed, 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?
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February 20th, 2011, 09:30 AM   #9
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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 al-mahed.

Use a proof by contradiction, where you begin by assuming it IS rational:

where p and q are co-prime, 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 co-prime, 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.
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February 20th, 2011, 09:47 AM   #10
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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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