Report post Posted October 16, 2017 Somehow I ended up with xx5 gold. I think it was during the double gold event? I'm not sure. But, now I have 5 gold I can never use. Anyone else? I searched for an old thread regarding this but didn't find anything. Share this post Link to post Share on other sites
Report post Posted October 16, 2017 If you play arena, you can get another gold amount ending with 5 (such as 45) that will help you with returning the gold number back to increments of 10. 1 Share this post Link to post Share on other sites
Report post Posted October 16, 2017 37 minutes ago, Tupi said: Somehow I ended up with xx5 gold. I think it was during the double gold event? I'm not sure. But, now I have 5 gold I can never use. Anyone else? I searched for an old thread regarding this but didn't find anything. Like positiv2 said, it is from arena. nerdmode: Double-Gold can have nothing to do with it. If you double a natural number (1, 2, 3...) (all rewards are natural numbers) you can never get gold ending on the digit 5. You would need a rational number like to get a number ending on the digit "5" ;) e.g. 2.5 = resulting in 5 or 17.5 = resulting in 35 Share this post Link to post Share on other sites
Report post Posted October 17, 2017 12 hours ago, WedgeAntilles said: Like positiv2 said, it is from arena. nerdmode: Double-Gold can have nothing to do with it. If you double a natural number (1, 2, 3...) (all rewards are natural numbers) you can never get gold ending on the digit 5. You would need a rational number like to get a number ending on the digit "5" ;) e.g. 2.5 = resulting in 5 or 17.5 = resulting in 35 Needing a rational number doesn't, by necessity, exclude natural numbers, since the set of all natural numbers are a subset of the rational numbers. Share this post Link to post Share on other sites
Report post Posted October 17, 2017 3 hours ago, VaraTreledees said: Needing a rational number doesn't, by necessity, exclude natural numbers, since the set of all natural numbers are a subset of the rational numbers. Yes, but it is impossible to multiply any natural number with the factor "2" and get as an result an odd-number like 5. To get such a result you need to increase the base of numbers from natural to rational. That rational numbers include natural numbers is correct. My statement would only be incorrect or incomplete if I had said that you would need a real number. To be more precise the defined numbers in my statement would then have been unnessesarily broad. Because it does not need to be a real number, a rational number is sufficient. Even more, the statement would be more confusing: Real numbers are the sum of rational numbers (like 2.5 e.g.) and irrational numbes (like squareroot of 2 e.g.). But there is per definition no irrational number that you can multiply with two and get any natural number (like e.g. 5). That's why speaking of real numbers and by this way increasing the base from rational numbers only to rational AND irrational numbers wouldn't be useful, on the contrary. Share this post Link to post Share on other sites
Report post Posted October 17, 2017 The set of numbers that could be the base reward are numbers ending with 7.5 or 2.5, ie. {2.5(2n-1)} for n∈N. The "smallest" number system that contains this set are the rational numbers. It is also a subset of Q\N, and there are no rewards in Hearthstone that are members of that set. If I am wrong, please do correct me. 1 Share this post Link to post Share on other sites
Report post Posted October 18, 2017 (edited) 11 hours ago, positiv2 said: The set of numbers that could be the base reward are numbers ending with 7.5 or 2.5, ie. {2.5(2n-1)} for n∈N. The "smallest" number system that contains this set are the rational numbers. It is also a subset of Q\N, and there are no rewards in Hearthstone that are members of that set. If I am wrong, please do correct me. Kind of. The set is correct, and that can easily be proven by induction, but it would be wrong to use the term smallest. The set of natural numbers, and the set of natural numbers divisible by 5 but not 10, and the set of rational numbers are all the same size. Edited October 18, 2017 by VaraTreledees Share this post Link to post Share on other sites
Report post Posted October 18, 2017 11 hours ago, WedgeAntilles said: Yes, but it is impossible to multiply any natural number with the factor "2" and get as an result an odd-number like 5. To get such a result you need to increase the base of numbers from natural to rational. That rational numbers include natural numbers is correct. My statement would only be incorrect or incomplete if I had said that you would need a real number. To be more precise the defined numbers in my statement would then have been unnessesarily broad. Because it does not need to be a real number, a rational number is sufficient. Even more, the statement would be more confusing: Real numbers are the sum of rational numbers (like 2.5 e.g.) and irrational numbes (like squareroot of 2 e.g.). But there is per definition no irrational number that you can multiply with two and get any natural number (like e.g. 5). That's why speaking of real numbers and by this way increasing the base from rational numbers only to rational AND irrational numbers wouldn't be useful, on the contrary. Rational numbers are real numbers. The rational numbers are a subset of the real numbers. While saying you need a real number is not incorrect, it is not very useful. Similarly, saying you need a rational number is also not very useful. Again, not incorrect, just not useful. If you want to define a useful set, as positiv did you can, and that set is a subset of the rational numbers but not a subset of the natural numbers, you can then use that to prove your point. Or you can just prove that you cannot double a natural number and have it divisible by 5, but not 10, the direct proof is fairly trivial, since doubling any natural number will result in an even number, ie one that can be written in the form of 2n, where as any natural number divisible by 5 and not 10, would be in the form 2n+1, an odd number, where n is a natural number. Share this post Link to post Share on other sites
Report post Posted October 18, 2017 7 hours ago, VaraTreledees said: Rational numbers are real numbers. The rational numbers are a subset of the real numbers. While saying you need a real number is not incorrect, it is not very useful. Similarly, saying you need a rational number is also not very useful. Again, not incorrect, just not useful. If you want to define a useful set, as positiv did you can, and that set is a subset of the rational numbers but not a subset of the natural numbers, you can then use that to prove your point. Or you can just prove that you cannot double a natural number and have it divisible by 5, but not 10, the direct proof is fairly trivial, since doubling any natural number will result in an even number, ie one that can be written in the form of 2n, where as any natural number divisible by 5 and not 10, would be in the form 2n+1, an odd number, where n is a natural number. I don't get your point. I just defined which kind of numbers would be needed to reach "5" with a multiplication by two. And I said that you need rational number to do that. A rational number is the smallest part of existing numbers that can achieve that goal. EVERY number that can be doubled and result in "5" is a rational numbers. (With rational number as the smallest defined number system avaiable) I just defined the name of the pool of numbers where you have to choose from. I did not intend to proof that a natural number can not be mulitplied with 2 and result in 5. I did not intend to use a mathematical function to define the numbers more closely. Yes, you obviously can do that. Like positiv2 did, thank from my part for that! But I didn't choose to do that. You have several groups of numbers: Natural numbers Integers Rational Numbers Transcendental Numbers Irrational numbers Real Numbers (did I miss one?) I just defined the group, nothing more, nothing less. And while a mathematical function defines a set of numbers way more precicsly (no question about that!), a mathematical function has no name for the resulting numbers themselves. The existing definition of groups is very limited - there exist just a handful. With functions you can define numbers more closely - resulting in infinite avaiable functions. But again, I just wanted to define the group of numbers, not a function. Thankfully positiv2 answered the question of the thread in his first answer, because this topic has definitly become quite unreadable and hasn't anylonger much to do with the topic itself *g* Share this post Link to post Share on other sites
Report post Posted October 25, 2017 lol, nice divert from OT. Ah, I don't play Arena hardly at all...but that must have been where it came from. Thanks for the reply. Share this post Link to post Share on other sites