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Is there an irrational multiple of another irrational to yield Integer 
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#1
May1714, 11:43 AM

P: 50

This question specifically relates to a numerator of '1'. So if I had the irrational number √75:
1/(x*√75) Could I have some irrational nontransindental value x that would yield a non '1', positive integer while the x value is also less than 1/√75? Caviat being x also can't just be a division of '1/√75', which would just yield the number dividing '1/√75' (canceling out the sqrt(75)) (So x is a number able to be represented by like sqrt(2) or 1/3 i.e not transindental) If there is a way, that anyone knows of how I would go about finding it? Thanks! 


#2
May1714, 11:53 AM

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#3
May1714, 12:11 PM

P: 50

Thanks, I'll elaborate on the question and make it more specific. 


#4
May1714, 01:07 PM

P: 429

Is there an irrational multiple of another irrational to yield Integer
Given ##y=\frac{1}{x\sqrt{75}}##, you have ##x=\frac{1}{y\sqrt{75}}##, yes?



#5
May1714, 01:13 PM

P: 50

Thanks 


#6
May1714, 01:35 PM

P: 429

It's a fact (though not necessarily easy to prove depending on your background) that the product of two algebraic numbers is algebraic. That, combined with the following facts, which are much easier to prove, should be useful in understanding your question.
The following are left as exercises for the reader: If ##x## is irrational, then ##x^{1}## is irrational. If ##x## is rational and ##y## irrational, then ##xy## is irrational. If ##x## is algebraic, then ##x^{1}## is algebraic. 


#7
May1714, 09:38 PM

P: 50

Then can we have y = z*1/x as a rational algebraic integer under the conditions I've stipulated? I don't see why not if x is irrational, and if so, is there a way to find the irrational values for x, acceptable under these conditions? (like not just an x value that's a division of 1/√75, and x < 1/√75) 


#8
May1714, 10:16 PM

P: 429

First off, every integer is rational, and every rational is algebraic. So "rational algebraic integer" is a bit redundant and/or slightly confusing.
Now if you have ##z=\frac{1}{\sqrt{75}}## and ##y=z\cdot\frac{1}{x}## where ##y## is an integer, then per my original post, ##x=\frac{1}{y\sqrt{75}}## must be of a form that you have deemed "not really useful". Furthemore, if ##y>1##, then ##x< z##. I think an issue that you might not have considered is that for real numbers (and rational numbers for that matter) every nonzero number is a divisor/multiple of every other nonzero number. For example, as integers ##3## is neither a divisor of nor a multiple of ##5##. As real numbers (or, again, rationals), ##3## is both a divisor and a multiple of ##5##. So to demand that ##x## not be "just a division of" ##z## is to want something impossible of the solution. Additionally, and I really need you to see this for how obvious it is (no offense intended), if you are demanding that ##y=\frac{z}{x}## and that ##y## is some special brand of number (integer, rational, algebraic, etc.) then it must be true that ##x## is just a division of ##z## by that special brand of number. Just out of curiosity, is your question related to something specifically that you're working on? I.e did this come up in the context of something particular? Or is this just a general query in regards to the relationship between the various "types" of real numbers? 


#9
May1814, 06:12 AM

P: 50

Yes, a "special brand of number" indeed....you have added a lot of perspective. A member of that brand may not exist for the now apparent but always obvious reason you pointed out. I was thinking for instance if x = √q (any irrational): 1/(√q * √75) = Y (an integer) where q is less than 75, perhaps much less, so Y would be a very large number. But that doesn't seem to exist. I had it in my mind that like how you can divide 1 by 1/3 (irrational) and get an integer (3), this would be similar. It's part of a larger theorum I'm working on that involves a number that divides into 1 an integer number of times and also the root of an integer (like √75) an integer number of times. Hence the conditions. I'm assuming this is impossible? 


#10
May1814, 09:11 AM

P: 429

I can't tell if you're making silly errors or if you just haven't thought this through.
##\sqrt{q}## isn't necessarily irrational. For instance ##\sqrt{25}## and ##\sqrt{\frac{16}{49}}## are rational. It strikes me as odd that you would claim ##\frac{1}{3}## as being irrational immediately after expressing it as the ratio of two integers. Perhaps that is a typo. As for your theorem ... to say that a number ##x## divides into ##1## an integer number of times, you presumably mean that ##\frac{1}{x}=n## where ##n## is an integer. Per my previous post, this implies that ##x=\frac{1}{n}##; i.e. ##x## is what is called an Egyptian fraction. If ##x## were also to divide the root of another integer ##k## an integer number of times, then you have ##\frac{\sqrt{k}}{x}=m##, where ##m## is an integer, and thus ##x=\frac{\sqrt{k}}{m}##. So you have ##\frac{\sqrt{k}}{m}=\frac{1}{n}##, which gives ##\sqrt{k}=\frac{m}{n}##. Since ##m## and ##n## are both integers, ##\frac{m}{n}## is rational, as is ##\sqrt{k}##. It can be shown that if the square root of an integer is rational, then it must be an integer. So ##\frac{m}{n}## is an integer, and ##n## divides ##m## as integers. I had a random number generator give a me an integer between ##0## and ##100## for the purposes of demonstration, and it gave me ##33##. ##x=\frac{1}{33}## is a number which divides ##1## an integer number of times. It also divides into ##\sqrt{4356}##, ##\sqrt{9801}##, ##\sqrt{17424}##, and many other square roots of integers an integer number of times. Now you may notice that ##\sqrt{4356}=66=2*33## and ##\sqrt{9801}=99=3*33## and ##\sqrt{17424}=132=4*33## and I think you see the pattern. And it's not by luck that I got an Egyptian fraction which just so happened to divide a bunch of square roots. I hope you see how every Egyptian fraction (i) divides into ##1## (and infinitely many other integers) an integer number of times and (ii) divides into infinitely many other square roots an integer number of times. In fact every rational number divides into infinitely many integers and square roots of integers and cube roots of integers and ##n##th roots of integers an integer number of times. Now I suggest that you read and reread the paragraph with all the ##n##s and ##m##s and ##k##s until you realize that (almost) every single word of it is something that you could have come up with if you had just started writing things down and working things out. I also encourage you to try and see how I used all of my "findings" in that paragraph to construct all of the numbers in the following paragraph and draw the big conclusion at the end of that paragraph (which at first seems amazing, then completely boring, then potentially interesting). 


#11
May1814, 10:00 AM

P: 50

Thanks for picking that up, I'm a bit frazzled atm doing a few things without sleep, last post I started to reply got side tracked and when I resumed I did so without resuming the train of thought.
"Egyption fraction", interesting. The elaboration you made using K and m was helpful but the distinction being that the sqrt(4356) or sqrt(9801) is an integer, not irrational like sqrt(75), never the less it doesn't matter as you say 'the product of an integer and an irrational number is irrational' so I suppose that just means 'm' couldn't be an integer as it is in your example. That is probably the crux I proceded without realising. The thing is, you're correct about writing it down but this is one of those things where you think you can visualise it in your head no problem, like dreaming about walking around an Escher building, but in reality something isn't right. Looking at this through the smaller scope of discussion out of what I've been doing it is much more clear. Thanks for your time, I appreciate it. 


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