How come the set of real numbers is group under addition when set of irrational number is not?
We say that set of real numbers is union of sets of rational and irrational numbers.Now set of irrational numbers is not a group under addition.Then how is it possible that its union set can be a group under addition.
In short I mean to say that is it possible always that a union set possess a property that isn’t in one of its subset?
As hemu said, the irrationals lack an identity element.
Your question about unions and subsets is worth addressing. Yes, a property of a set will be a property of any subset IF the property is stated in terms of individual elements. so for example, "for any two real numbers x and y, then x+y=y+x." If you accept this, then the property is inherited by any subset such as the irrationals or the integers, because they are special cases of the general statement.
but the property "there exists an identity" will not be inherited by any subset that omits the identity! there is no property of the real numbers that says "for any real numbers x1,x2,……., there exists j such that xj = 0."
September 11th, 2009 at 3:59 am
Yes, easily.
Natural numbers are not a group under subtraction; however, natural numbers are a subset of integers, which are a group under subtraction.
Because new elements are available to union sets that were not available in subsets, union sets can have different properties.
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September 11th, 2009 at 4:18 am
simple example: the square root of two plus minus squre root of two is zero…zero is rational
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September 11th, 2009 at 4:36 am
There is no irrational identity of addition.
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September 11th, 2009 at 5:25 am
It’s always possible to define a function such that when it is performed on members of a set the result is outside that set. By adding additional members, you may be able to close the function. That’s essentially what’s going on here.
If A and B are members of set X and A+B is not a member of set X then it’s not a group under addition. Let A be some arbitrary irrational number. Let B be 1-A. B is also an irrational number because if there is no pattern to the digits in A, then there can be no pattern to the digits in B either. Yet A+B=1, which is not in the set of irrational numbers.
Some irrational numbers therefore add up to rational numbers and some add up to irrational numbers. All rational numbers add up to rational numbers. Therefore, the union set of rational and irrational numbers (the real numbers) is a group under addition while irrational numbers are not.
Here’s a similar example to illustrate that this is not so unusual:
Look at the square root function. The set of real numbers is not a group on this function because the square roots of negative numbers are not real numbers, they are imaginary. The square roots of imaginary numbers can be real or imaginary, so if you take the union of the two sets, you get a closed group under the square root function. Note, however, that the set of positive real numbers is closed under the square root function, although the set of positive rational numbers is not (you need to have irrational numbers in there too).
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September 11th, 2009 at 6:02 am
As hemu said, the irrationals lack an identity element.
Your question about unions and subsets is worth addressing. Yes, a property of a set will be a property of any subset IF the property is stated in terms of individual elements. so for example, "for any two real numbers x and y, then x+y=y+x." If you accept this, then the property is inherited by any subset such as the irrationals or the integers, because they are special cases of the general statement.
but the property "there exists an identity" will not be inherited by any subset that omits the identity! there is no property of the real numbers that says "for any real numbers x1,x2,……., there exists j such that xj = 0."
References :