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/sci/ - Science & Math


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7495902 No.7495902 [Reply] [Original]

<span class="math">\sqrt{2}[/spoiler] is irrational, and <span class="math">\sqrt{2}\sqrt{2}=\sqrt{4}=2[/spoiler] is rational.

Are there any irrational numbers multiplied by some other irrational number so that you end up with a rational number?

>> No.7495915

>>7495902
sqrt(2) and sqrt(8)

>> No.7495916

>>7495902
yes

>> No.7495930
File: 3 KB, 124x125, 1412901128869s.jpg [View same] [iqdb] [saucenao] [google]
7495930

>>7495902

>> No.7495939

>>7495902
sqrt(3)

>> No.7495946

>>7495930
It was an honest question.

I'm a CS major so I might be retarded though. Working through Rudin atm.

>> No.7495956

>>7495946
>Working through Rudin atm
there's still hope

>> No.7495984

>>7495946
<div class="math">a^{\frac{n}{2}}\cdot a^{\frac{m}{2}}</div> for m,n odd

>> No.7496014
File: 17 KB, 555x555, chart.png [View same] [iqdb] [saucenao] [google]
7496014

>>7495902

The general question (except for OP's case!) is of course simple, and I had occasion to describe the general case recently, for the reals.

In the picture, we describe multiplication of 0, some random real nonzero rational r, and some random real, irrational i (not to be confused with the imaginary unit of complex numbers). Because multiplication of real numbers is commutative (order of elements doesn't matter), our chart is symmetric about the main diagonal. We include the trivial 0 cases in the interest of simple completeness, since 0 is an obvious "spoiler" for various versions of OP's question, which bears mentioning.

0 times any real number is 0, and five of our nine cases are done. 5/9

Any rational number (zero or otherwise, but we already covered that case) times any other rational number (zero etc.), always gives a rational number. A little check is all that's needed to know this. 6/9.

Meanwhile, any NON-ZERO rational number, times any NON-ZERO irrational number, always returns an irrational number. A proof involves supposing the opposite (that their product is rational) and showing that this is absurd. 8/9.

It's only the latter case, which OP is wondering about, which is interesting, since the situation can go either way. But OP shouldn't feel too bad (nor should he be scolded too harshly), for there are elementary products of irrational numbers which are not known to be rational or irrational e.g. π x e ! Similar open questions entail other elementary arithmetic and functional operations involving pi and e, viz.

https://en.wikipedia.org/wiki/Irrational_number#Open_questions

>> No.7496025

>>7495902
Yes.
Let x be our irrational (and nonzero) number.
Then 1/x is also irrational, and x * (1/x) = 1, a rational number.

>> No.7496051

>>7496025
sorry, when <span class="math">x[/spoiler] is irrational, <span class="math">x \cdot \frac{1}{x} = 0.\overline{9}[/spoiler].

>> No.7496056

>>7496014
I think that your diagram is overkill.
For 8/9, a simple look at the fundamental theorem of arithmetic shows immediately that the number is irrational.
For 9/9, see >>7496025. What you're creating though is a nice discussion about >>7496014
transcendental numbers, which are big bitches.

>> No.7497899

>>7495902
(3-sqrt(2))*(3+sqrt(2))=5

>> No.7498962

>>7495902
OP, you just did multiple a irrational number by some other irrational number and got a rational number as a result. I don't understand what you are asking

>> No.7498970

>>7495902
Think about the strategy you used to generate that example.

>Oh, hey, the square root of this prime number is irrational.
>Let's look at the square roots of other prime numbers.
>Let's see what happens when we take square roots of different prime numbers and multiply them together.

>> No.7499283
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7499283

>>7495946
>>I'm a CS major so I might be retarded though
>might be

>> No.7500190

>>7495902
so the answer is yes, see >>7496025

So if you choose an irrational number x you can obtain every rational number : if q is a rational number then q/x is irrational and x(q/x)=q. Conversely if you multiply x by an irrational number y and the product is a rational number q then y=q/x. So this are the only irrational that will give a rational product.

Because there are infinitely more irrational numbers than rational numbers (in a sense i won't explain) the product of two random irrational numbers is almost certainly (in a sense i won't explain) an irrational number. So you can considere that the product of two irrational giving a rational number is an exception and not the usual case.