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Irrational , Rational Doubts

Hey frndzz, my doubt is dat we know dat 2/4 is not a rational no. because it's not co-prime. But if we consider another case dat in proving root 2 as irrational. We assume it as rational in the form a/b. And at last its irrational coz they both are not co prime. But that a/b can be 2/4 also............................... THINK OVER IT......

Note by Aßhĩмanyu Singh
4 years ago

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\(\frac {2}{4}\) is a rational number. A rational number is a number which can be expressed through the ratio of two integers (whole numbers). Another way of saying that is, a rational number can be expressed as \(\frac {a}{b}\) where \(a,b \epsilon Z\) and \(b\neq0\). So the numerator and the denominator don't have to be co-prime. Another important thing is that a same rational number can be expressed in many ways. As long as the numerator and the denominator are integers, it is a rational number. For example: \(\frac {2}{4}\) can also be written as \(\frac {1}{2}\), \(\frac {5}{10}\), \(\frac {17}{34}\) etc. Proving the irrationality of \(\sqrt2\) is a different case entirely. We call this a proof by contradiction. In this case, we start by assuming the exact opposite of what we are trying to prove. Then we show that our assumption is self-contradictory. In the case of the proof of the irrationality of \(\sqrt2\), We start by saying it is rational (the opposite of what we are trying to prove) and it can be expressed as \(\frac {a}{b}\) where \(a\) and \(b\) are co-prime integers [\(b\neq0\)]. Then we proceed and find that \(a\) and \(b\) are both even numbers and therefore not co-prime : a blatant contradiction. So, saying \(\sqrt2\) is rational is so crazy that it contradicts itself. So it can't be rational. So, long story short, a rational number is the ratio of two integers which are not necessarily co-prime. However all rational numbers can be expressed through the ratio of two co-prime integers. Hope this helps! Mursalin Habib · 4 years ago

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@Mursalin Habib No, according to the real definition, they must be co-prime, if we think that it is not rational, then it must be irrational, but according to me can't it be a no. like 2/4 Aßhĩмanyu Singh · 4 years ago

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@Aßhĩмanyu Singh well, they are co-primes when reduced to lowest form, in your case it is 1/2 where 1 and 2 are co primes Sudha Parimala · 4 years ago

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@Aßhĩмanyu Singh I'd like to know where you found your "real" definition. From what I understand (from the internet right now!) that the co-primality of the denominator and the numerator is not a necessity. If it were, it should be mentioned in the definitions. I'm giving you a couple of links that gives the definition of rational numbers. They nowhere mention the necessity of the co-primality of the denominator and the numerator. http://en.wikipedia.org/wiki/Rational_number http://www.mathsisfun.com/rational-numbers.html http://mathworld.wolfram.com/RationalNumber.html The website below also gives a couple of examples where the rational number is not reduced (the numerator and the denominator aren't co-prime). http://www.eduplace.com/math/mathsteps/7/a/ (Notice the instance when they say \(3.75\) is rational. As far as I'm concerned, I believe \(\frac{1}{2}\) and \(\frac{2}{4}\) are the same number as they represent the same point on the real number line. So why should one be considered rational while the other isn't? I'd like to see where you got that definition from and then we can come to a conclusion. Mursalin Habib · 4 years ago

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The proof of irrationality of root2 is based on the assumption that a and b are co-prime and then the contradiction, but here as I already know that 2 and 4 are not co-prime, even if I show in the proof that they are having a common factor won't be a contradiction as it agrees to the observation that 2 and 4 are not co-prime. A twisty answer to a twisty question. Siddharth Kumar · 4 years ago

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@Siddharth Kumar Can u prove that by solving? Kashish Goel · 1 week, 3 days ago

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is pie(3.14) is an irrational number or not? Yousuf Mahtab · 4 years ago

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@Yousuf Mahtab \(\pi\) is not a rational number simply because it can't be expressed through the ratio of two integers. If you have a circle with an integer circumference, its diameter's going to be irrational and vice versa. \(\pi\) is not equal to \(3.14\). It is an approximate value. The decimal expansion of \(\pi\) never ends and it is therefore an irrational number. However the proof(s) of the irrationality of \(\pi\) are a bit more complicated than the proof of the irrationality of \(\sqrt2\). Hope this helps! Mursalin Habib · 4 years ago

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Everything which can be expressed in fractions is a real number and which cannot be expressed in fractions is an irrational number, so what's the point of discussion? 2/4 is a fraction and hence a real number and not an irrational number. The point of co-prime numbers must have been provided by the mathematicians to make the analysis easier, however the "co-prime numbers" element is not a part of definition. Siddharth Kumar · 4 years ago

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@Siddharth Kumar I agree with you. There is no point in discussion. A rational number is the ratio of two integers. It's as simple as that (we use the co-prime point in the proof of irrationality without the loss of generality). Mursalin Habib · 4 years ago

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@Aßhĩмanyu: Another point : every rational number has an irreducible form where the numerator and denominator are co-prime. Reducing a number doesn't change its value. So numbers that are expressed through the ratios of two integers but are not reduced are also rational numbers. Would you consider \(0.5\) to be rational? By definition, \(0.5=\frac{5}{10}\) Mursalin Habib · 4 years ago

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@Mursalin Habib yes, but the fact is in proving irrationality of a number, we say that if it is not co-prime , therefore it is iraational. Aßhĩмanyu Singh · 4 years ago

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@Aßhĩмanyu Singh Let me give you an example -

Suppose we need to prove that Root 2 is irrational - Then we say let Root 2 be in the form of a/b or let it be rational then - therefore Root 2 = a/b Root 2 * b = a 2b^2 = a^2 (Squaring on both the sides) - 1 Therefore, according to a theorem that if a prime no. is a factor of square of no. then it is also a factor of that no. Therefore 2 is a factor of a.
Therefore a can be written as - 2 * p (Where p is a positive integer)

a = 2p a^2 = 4p^2 (Squaring on both the sides) Putting this in Eq. 1 4p^2 = 2b^2 or, 2p^2 = b^2 therefore 2 is a factor of b since, a and b are not coprime, therefore Root2 is irrational. Here we see that we prove Root 2 as irrational on the basis of its co-primity..

And, thus we conclude that A no. is rational if it is no co-prime... Aßhĩмanyu Singh · 4 years ago

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@Aßhĩмanyu Singh There is a slight misconception there. We don't say \(\sqrt2\) is irrational on the basis of the co-primality of the numerator and the denominator. We say it is irrational because we reach a contradiction when we assume it is rational (in the reduced form). I hope the statements below will clear things up: Let's say \(\sqrt2=\frac{a}{b}\) where \(a\) and \(b\) are not co-prime. So there exists a number \(\frac{m}{n}\) such that \(\frac{m}{n}=\frac{a}{b}\) and \(m\) and \(n\) are co-prime. This is true for all \(\frac{a}{b}\). So we are doing this without the loss of generality. Because it can be done for all rational numbers that are not reduced. So we can say, \(\sqrt2=\frac{a}{b}=\frac{m}{n}\) [ \(\frac{m}{n}\) is the reduced form of \(\frac{a}{b}\)]. Now notice that \(m\) and \(n\) are co-prime (even though \(a\) and \(b\) are not). But if we proceed like you did, we'll find that \(m\) and \(n\) are both even and therefore not co-prime : a contradiction.So our previous assumption was incorrect. So even if a rational number is not reduced, we can reduce it to have our numerator and denominator co-prime, and still reach our contradiction. We are using the principle that something that contradicts itself can't be true. I hope this clears things up! Mursalin Habib · 4 years ago

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@Mursalin Habib Oh, i see, it means that at the time of assuming only, we think that it is no. in reduced form. Thnxxx frnd.. A lot lot thanxx to u Aßhĩмanyu Singh · 4 years ago

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@Aßhĩмanyu Singh You're welcome! Mursalin Habib · 4 years ago

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