Let x and y be irrational numbers, is it always the case that x * y are irrational?How can I prove or disprove
357184
No. It can be easily be disproved by simply providing a counterexample.\n\nAll surds are irrational (root 4 is not a surd because it simplifies to an integer). Root 2 is irrational, but\nroot 2 * root 2 = 2.\n\nTherefore the statement is false.
451479
hav u ever heard of the Principle of Mthematical Induction ?\nthat could solve ur problem in a jiffy!!\n\nHowever if u havn't, dont worry!!\n\nNote that, the sum in ur series(i.e. 1+1/2+1/3+...+1/n),where n>=2, is always less than 2, but greater than 1. so it can't b an integer.\nwanna know why?\ntry cramming up some theorems about Sequence & series of real numbers!! its a regulation part of any under-graduate level Real analysis course.
328117
Yes. This is a Frequently Asked Question. Following is my previous answer:\n\nThe real numbers can be defined using Cauchy sequences. Basically, the real numbers make a number line continuous by filling in all possible "holes". If you have a set of real numbers A, and that set is bounded above by some real number (there is a real number that is greater than every real number in A), then we define the set to be a real number that is the smallest real number greater than or equal to every real number in A. This is how one constructs the real number system, by defining every point on a line segment to be a real distance from the start of the line segment (thus all square roots, corresponding to hypotenuses of right triangles, have been incorporated into our number system). \nIn other words, every bounded (from above) subset of rational numbers corresponds to a real number. Now, rational numbers are defined as the set of all ratios of integers. Just like two ratios of integers can correspond to the same rational number (for example, 2/4 and 500/1000 are the same rational number), so can two bounded subsets of rational numbers correspond to the same real number. \nA common way of representing these sets is to use decimal notation. Decimal notation is an infinite series in powers of ten. Each partial sum of the series is taken to be an element of the set in study. A series is said to converge if the set of partial sums is bounded as described above.\nFor example, the number 0.333... is the decimal way of writing the series 0 + 3/10 + 3/(10^2) + 3/(10^3) + ... which corresponds to the set of partial sums {0, 3/10, 33/100, 333/1000, ...}. It is trivial to show that this set is bounded by 1, so this set corresponds to a real number. It can be shown by algebraic manipulation that this set actually represents the rational number 1/3.\nLet us look at the set corresponding to 0.999..., or {0, 9/10, 99/100, 999/1000, ...}. This set corresponds to a real number because each element is of the type (1 - (9/10)^n)/(1/10) = 10(1 - (9/10)^n) (it is a geometric sum; the expression can be derived after a moment's thought). Obviously, no power of any fraction less than 1 is greater than 1 since all such powers are less than the original number, so this is bounded by 1.\nWe hypothesize that 1 is the smallest number that bounds this sequence. Consider the difference 1 - L where L is the smallest number that bounds this sequence. By the field axioms (common algebra), if 1 - L = 0, then 1 = L. Now L is greater than every element of the set in question by definition. Thus, consider 1 - 10(1 - (9/10)^n) = 9^n/10^(n - 1) - 9 = 9((9/10)^(n - 1) - 1). Suppose 1 - L = d for some positive number d. Then (9/10)^N - 1 > D, or (9/10)^N > E ( = d/9 + 1 ) for some constant E (E is not a function of N, only of the fixed difference d between 1 and L) for every possible choice of N. This is absurd, as it implies that after some N, (9/10)^(N + 1) either increases or remains the same as (9/10)^N which is nonsense. Thus, our assumption that there exists a positive difference (and thus any difference) between 1 and L must have been false, so the difference must be 0. \nBy the construction of real numbers, that makes 1 = 0.999... . This is a bit more tedious than proving two rational numbers equivalent (and thus why there exist many hand-waving arguments or misunderstandings of the issue), but is conceptually the same.
912338
Neither, it is undefined, i.e. it has none of the properties of all rational or whole numbers.\n\nI might be wrong, but I think this is one of those trick questions that are so common in math courses.
664717
I'm not sure what you are asking, but yes, there are an infinite number of numbers between 1 and 0.  \n\nIn fact, the whole numbers are the smallest group of infinities on the number line.  The infinite set of irrational numbers far exceed the whole numbers.\n\nTherefore, if you were to have all of the numbers on a number line, and you were to throw a dart, the probability of you hitting a whole number with that dart is basically zero!
1151541
This is impossible:\nProof. Let the area of each identical square equal n^2 (so each side is n). Since the large square is has 3 of these pieces, its area is 3n^2. Thus, each of its side is sqrt(3)n. However, you can never divide the square to do this (since sqrt(3) is irrational).
1092202
looks like you've defined e using a series.  I think you need some periods of ellipsis to show that that series is infinite, since e is an irrational number
624409
Clearly false. Let K={i}.\n\nDid you mean for K to be a subfield? In that case, K contains the rationals and so is dense in R.
981709
To extend the above two discussions and bring the matter to a conclusion, more or less.\n\nLet us define a sequence by T0 = 1, Tn = x^Tn-1.  Call the limit of this sequence as n -> infinity  as f(x).\n\nFor any given y = f(x), we see that:\n\n     lim Tn = lim Tn-1 = y\n>>\n    lim Tn = lim(x^Tn-1)=x^limTn-1\n>>\n     y = x^y\n>> \n    x = y ^ (1/y)\n\nFor y >= 1:\nThis can easily be seen to satisfy the  special cases of y = 1, 2 and 4.\n\nx, as a function of y has a maximum at y = e.~ 2.718, which leads to x~1.447.\n\nFor values of  x > e^(1/e),  f(x) does not exist; and the sequence T cannot have a limit > e.\n\nFor y < 1, we can similarly see that limits < 1/e are not possible.  The smallest x for which the limit exists is (1/e)^e ~ 0.066, giving a limit 1/e ~ 0.368
472136
False. For example: F(x)=x^2 increases for x>0.\n\nF(x)=x^3 always increases.
904022
Surprisingly, yes.  This bugged the heck out of me in middle school, but it's a consequence of the way infinity works.\n\nThe classic proof is:\n\n    x = 0.999999....\n10x = 9.999999....\n\nSubtract the first line from the second, and you get\n\n9x = 9, therefore x = 1.
57970
does 1+1=2 is actually a really good question because there is really no way of proving it. Because numbers are abstract(they are not real) they represent other things. But if you look at the equation 1+1=2 you will actually see that by itself it makes no sense because one plus one does not equal a two symbol. so without subsituiting the number symbols for actual things the equation is irrational and cannot be proven.
198310
It depends on what you mean by "flanked". When you are talking about the rationals and the irrationals, there is no thing as "next to". \n\nWhat you can show is the following:\n\nGiven a rational number q, and ε > 0, you can find an infinite number of irrationals numbers x such that |x-q| < ε. That is now matter how small an interval around q you pick, it will contain irrationals.\n\nSimilarly, given a rational number x, and ε > 0, you can find an infinite number of rationals numbers q such that |x-q| < ε. That is now matter how small an interval around x you pick, it will contain rationals.\n\nBetween any two rational numbers, there is an irrational number.\n\nBetween any two irrational numbers, there is a rational.\n\n\nBy the way, .99999... is not next to 1, it is 1. If you want to know about that, go back through older answers. It's been answered dozens of times.
935707
x<0\nso for all values of x, x will be a negative. when you absolute x, the positive absolute will always be bigger than the negative x
252300
I think this is otherwise known as recurring numbers. This occurs when you have an irrational number. The number is usually a fraction that does always has a remainder, but this remainder when written in the decimal is somewhat repeated. In other words thre is a sequence. An example is 0.722372237223. The numbers 7223 are recurrent.A number which does have a lot of numbers in its decimal but doesnt have recurrent numbers is e
759729
I think, therefore I am.\n\nAssume 1 does not exist.\n\nLet x = 0\nLet y = 2\n\nAny pair of numbers have a number between them, known as the arithmetic mean.\n\nTake the arithmetic mean of x and y:\n(x + y)/2 = (0 + 2)/2 = 1\n\nBut we now have a result of a number that shouldn't exist.\n\nSince we have a contradiction, this proves that our initial assumption was incorrect.\n\nTherefore 1 exists -- proof by contradiction.
535772
If squares of two numbers are equal the two numbers need not be equal always i.e. there is no such axiom or rule. \n\nLet us explain. \nSuppose x^2=y^2. \nThen x^2-y^2=0 [Transposition rule]\nor (x-y)*(x+y)=0 [Factorising]\n\nAs product of two numbers is zero so at least one of them must be zero [since anything multiplied by zero is zero] so\neither (x-y)=0 or (x+y)=0.\n\nWhen x-y=0, x=y but when x+y=0, x=-y.\nNow 'x' and 'y' are such numbers that there squares are equal but that not imply x = y because x = -y is also possible.\n\nSo your step "rooting both sides" is wrong leading to the wrong result.\n\nUNDERSTAND?