Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am really curious about the Vedanta behind the arithmetic operations on irrational numbers. It still got aggrevated after the productive discussions with my friend. So I decided to ask it here. Basically there is some confusion with some 4 operators.

  1. Let me start with $+$ operator. There is no confusion. Suppose the "$+$" in $\sqrt{3} +\sqrt{2}$ just add the decimal part of the both numbers. Its like adding $ 1+0.73205... + 1 +0.41421... $ . So it makes sense when we add them linearly .
  2. The problem arises with the multiplication. Can some one explain how is $\sqrt{3}*\sqrt{3} = 3$ ?. How can a product of two irrational numbers turns out to be a rational number ? . Its like $1.73205.....*1.73205.....$ , so the multiplication operator just multiplies the decimal part too. It must give rise to the infinite decimal part in the output too. But in converse we are getting a rational number ( specifically an integer ) $3$. How can one make sense out of this contradiction ? .
  3. Similarly coming to Division the same problem arises . But once we have well defined multiplication, division may make some sense because we can always rationalize numerator and denominator. So the problem is just finding the explanation behind the multiplication.
  4. Similarly how about the exponents ? . Raising an irrational number to the power of another irrational number ? . For example take $\sqrt{3}^{\sqrt{2}}$ . How can one raise the irrational number which has infinite precision to another number which has infinite precision ? .

Thank you. Awaiting for your responses.

share|improve this question
1  
Thank you chandu for fixing the typo. –  Iyengar Jun 3 '12 at 6:27
4  
Don't forget that you can add two irrational numbers to get a rational one. :) You have two problems, I think: (1) you are letting your concept of "decimal expansion" is get in the way of developing a concept of "number", and (2) you have some finitist notion of "able to compute" which you are letting get in the way of developing a concept of "able to define something". –  Hurkyl Jun 3 '12 at 6:28
1  
@Iyengar: I'm going to be signing off in a bit, but I'm sure someone will be able to provide a great answer. Also, could you clarify what Vedanta is and why it is relevant to your mathematical question? –  Zev Chonoles Jun 3 '12 at 6:38
5  
@ZevChonoles : Vedanta, should be taken according to the context. It means the quintessence, the true meaning etc.. which can be applied here. Your wiki-link refers to the actual meaning of veda-anta... I hope you are clear now . –  Iyengar Jun 3 '12 at 6:42
3  
It's an interesting word (and a fine question!), but in the future I would not assume that others will know what it means. –  Dylan Moreland Jun 3 '12 at 17:10

5 Answers 5

up vote 12 down vote accepted

$1.732 \times 1.732 = 2.999824$

$1.73205 \times 1.73205 = 2.9999972025$

$1.7320508 \times 1.7320508 = 2.99999997378064$

$...$

The truncation of the decimal expansion of $\sqrt{3}$ to $n$ decimal digits is, in fact, the largest number $x$ with $n$ decimal digits such that $x^2 \le 3$. As you take more and more digits, the square gets closer and closer to $3$. And so the limit of these decimal approximations, which is $\sqrt{3}$, satisfies $\sqrt{3} \times \sqrt{3} = 3$.

share|improve this answer
1  
+1, good and neat explanation. –  Iyengar Jun 3 '12 at 6:55
    
For a useful discussion about the foundational issues involved in this topic, Iyengar may want to look at: David Fowler, Dedekind's theorem: $\sqrt{2} \times \sqrt{3} = \sqrt{6}$, American Mathematical Monthly 99 #8 (October 1992), 725-233. –  Dave L. Renfro Jun 4 '12 at 15:26
    
@DaveL.Renfro : Nice paper sir. But unfortunately I cant find a copy. It would be very great if you can give me the link the PDF file sir. –  Iyengar Jun 4 '12 at 16:35
    
@Iyengar: When I made my comment I looked on the internet for a freely available copy but couldn't find one. It's at JSTOR of course, being an MAA journal, but JSTOR is not freely available (to me also, although FWIW I do happen to have a hard copy of the paper filed away somewhere at home). –  Dave L. Renfro Jun 4 '12 at 19:16
    
@DaveL.Renfro : Thank you sir for your response. No problem sir, sometime we need to have fate to read something.. –  Iyengar Jun 5 '12 at 5:21

To get an answer to your question, you first need to understand what a real number is, and what does multiplying two real numbers mean exactly. Every real number can be thought of as an equivalence class of cauchy sequence of ratonal numbers. For example, $\sqrt 3$ can be thought of as the following cauchy sequence of rational numbers

$$1, \frac{17}{10}, \frac{173}{100}, \frac{1732}{1000}, \frac{173205}{100000} \ldots$$

Similarly, the rational number $3$ can though of the cauchy sequence $3,3,3, \ldots$.Same can be done with any other rational number. Now, two different cauchy sequence can represent the same real number, that's why every real number should be thought as an equivalence class of cauchy sequence of ratonal number, and element cauchy sequence belonging to that equivalence class can be used to represent that real number. For example, the rational number $3$ can also be though as the as the cauchy sequence $\{3+ \frac{1}{n}\}_{n=1}^{\infty}$. In fact any sequence of rational numbers converging to $3$ can be used to represent $3$.Now, it's time for some formal definitions.

A cauchy sequences is a sequence $x_1,x_2,x_3,...$ of rational numbers such that for every rational $\epsilon > 0$, there exists an integer $N$ such that for all natural numbers $m,n > N, |x_m-x_n|<\epsilon$.

Two cauchy sequences $(a_n)$ and $(b_n)$ are equivalent if the cauchy sequence $(a_n-b_n)$ has the limit zero. Check that this relation is indeed an euivalence relation. The set $\mathbb R$ is then the set of equivalance classes of $R$ under this relation.

Now, we can define the binary operation multiplication on $\mathbb R$.Let $(a_n)$ and $(b_n)$ be two cauchy sequences , $(a_n) \times (b_n)$ is the cauchy sequence $(a_n \times b_n)$. Let $a=[(a_n)],b=[(b_n)] \in \mathbb R$, $a \times b = [(a_n \times b_n)]$.Here $[(a_n)]$ denotes the equivalence class containing $(a_n)$. You should check that the operation multiplication nf the equivalence classes is independent of the respresentative we use to denote the equivalence class.

Now we are ready to answer your first question.Let $x=[(1, \frac{17}{10}, \frac{173}{100}, \frac{1732}{1000}, \frac{173205}{100000} \ldots)]$. So $x^2=[(1,\frac{289}{100},\frac{173}{100},\frac{29929}{10000}, \ldots)]$. But $(1,\frac{289}{100},\frac{173}{100},\frac{29929}{10000}, \ldots)$ converges to $3$. So, $3=[(3,3,3,3, \ldots)]=[(1,\frac{289}{100},\frac{173}{100},\frac{29929}{10000}, \ldots)]=x^2$. So, $x=\sqrt 3$.

share|improve this answer

Lets say that you multiply two numbers $a * b$. This product will have the digit representation $c.c_1c_2c_3....$. There are infinitely many digits in this product.

Now, wen you start multiplying lots of numbers at random, $c_1$ could be any of the digits, and so can be $c_2$ and $c_3$ and so on....

It means that sometime, $c_1=0$, and that sometimes you also get $c_2=0$, and that sometimes you also get $c_3=0$, and so on....

Basically, extremely rarely it will happen that all the $c_i$ digits will be zero, and then your product is an integer.

Basically

$$1.7320508..... \times 1.7320508.... = 3.0000.....$$ so we still get infinitely many digits, we just don't write them since they are 0.

share|improve this answer
    
Good one to understand.. +1. But they are not going to be zeroes as per the Robert's answer above. They turn out to be all $9'$s that will give rise to a round-up $3$. –  Iyengar Jun 3 '12 at 17:25
1  
Well, not exactly, but even if they turn out to be all 9's you get an integer too... Robert calculated the product $\sqrt{3}*\sqrt{3}$ by taking a certain approximation (truncation), and in this approximation, you could see the digits turn into 9's one by one... BUT, while this is the "standard" way of doing it, you could approximate $\sqrt{3}$ in different ways, and for some of them you could see all $0$'s... Basically this is what the above posters mean by the fact that the product of two irationals is a number, not a digit representation... –  N. S. Jun 3 '12 at 18:01
    
Keep in mind that $\sqrt{3}=1.7320508...$. So, if I ask you to approximate it to 6 digits, which approximation would you take? $1.732050$ or $1.7320514? ;) –  N. S. Jun 3 '12 at 18:04
    
Claps , very well said. @N.S –  Iyengar Jun 3 '12 at 18:13

I am little baffled that you seem to be totally happy with addition of two irrationals, but not with multiplication.

If you take $a = \sqrt{2} = 1.4142135...$ and $b = 2-\sqrt{2} = 0.5857864...$, then adding them digit by digit is always going to give you something like $0.999...$, and I would have thought that would be as objectionable as the similar sort of infinite decimal when multiplying?

share|improve this answer
1  
But when you add them linearly there is no confusion. The terms may be linearly added. But coming to multiplication the sequence still gets doubled... –  Iyengar Jun 3 '12 at 17:08

If it doesn't bother you that $${3\over7}\times{7\over3}=.428571...\times2.333...=1$$ the product of two non-terminating decimals giving a whole number, then it shouldn't bother you if the non-terminating decimals happen to be irrational.

share|improve this answer
    
Yes , really a great explanation. But why does that happen ?. Why does the infinite decimal series go to give a whole number when multiplied. Does the same principle of Robert Israel apply there ? –  Iyengar Jun 3 '12 at 15:51
4  
@Iyengar: Because the decimal representation is just a representation of a number, not the number itself. –  Asaf Karagila Jun 3 '12 at 17:01
    
It's a good exercise to see if you can adapt Robert Israel's discussion to my example. –  Gerry Myerson Jun 4 '12 at 1:18

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.