My Math Forum Problem on irrational numbers

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 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 $a\sqrt{2}=b\sqrt{5}$, show that a=b=0. I have done $a\sqrt{2}-b\sqrt{5}=0$, which is valid for rational numbers a and b only when a=b=0. However, $\frac{a}{b}=\frac{\sqrt{5}}{\sqrt{2}}$ or $\frac{a^{2}}{b^{2}}=\frac{5}{2}$ If the question is correct, then a=b=0 then $\frac{0}{0}=\frac{5}{2}$ which I know is wrong. Any idea?
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 $a\sqrt{2}=b\sqrt{5}$, show that a=b=0. I have done $a\sqrt{2}-b\sqrt{5}=0$, which is valid for rational numbers a and b only when a=b=0. However, $\frac{a}{b}=\frac{\sqrt{5}}{\sqrt{2}}$ or $\frac{a^{2}}{b^{2}}=\frac{5}{2}$ If the question is correct, then a=b=0 then $\frac{0}{0}=\frac{5}{2}$ which I know is wrong. Any idea?
I think you can use a similar argument to that of the irrationality of $\sqrt{2}$

$2a^2=5b^2\ ==>\ b=2b#39;$ 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 $\frac{0}{0}=\frac{5}{2}$.
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 $\frac{0}{0}=\frac{5}{2}$.
but 0*2=0*5 is no absurd at all!

you just cannot divide by zero, otherwise yes, you'll get non-sense 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 $a\sqrt{2}=b\sqrt{5}$ can also be written as $\frac{a}{b}=\frac{\sqrt{5}}{\sqrt{2}}$. Then if a=b=0, we get $\frac{0}{0}=\frac{5}{2}$. What then? eta Is the question correct?
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 $a\sqrt{2}=b\sqrt{5}$ can also be written as $\frac{a}{b}=\frac{\sqrt{5}}{\sqrt{2}}$. Then if a=b=0, we get $\frac{0}{0}=\frac{5}{2}$. 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

 February 20th, 2011, 09:14 AM #8 Member   Joined: Nov 2010 Posts: 48 Thanks: 0 Re: Problem on irrational numbers al-mahed, if I understand correctly, if a=b=0, then $a\sqrt{2}=b\sqrt{5}$ would not make any sense if written as $\frac{a}{b}=\frac{\sqrt{5}}{\sqrt{2}}$. Then would it be safe to say there are no rational numbers a and b that satisfies $a\sqrt{2}=b\sqrt{5}$ 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 $\sqrt{\frac{5}{2}}$ is irrational, as mentioned above by al-mahed. Use a proof by contradiction, where you begin by assuming it IS rational: $\frac{p}{q}=\sqrt{\frac{5}{2}}$ where p and q are co-prime, the fraction p/q is fully reduced. $2p^2=5q^2$ In order for $5q^2$ 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 $q^2$ 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 $\sqrt{\frac{5}{2}}$ is rational is false. Thus, we know a = b = 0 is the only rational solution set to the original equation. Thanks from greg1313
 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 $\sqrt{\frac{5}{2}$ being irrational but I still can't grasp how to resolve the division by 0 problem. If $a\sqrt{2}=b\sqrt{5}$, for rational solutions of a and b, we can't transform the equation? Lets forget about a=b=0 and write the equation as $\frac{a}{b}=\frac{\sqrt{5}}{\sqrt{2}$. Now the same equation can have no rational solutions, is that right? Sorry for being thick.

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