Originally posted by royalchickenI see. Does this mean that for you puzzle all numbers are written as fractions. eg. 1 = 1/1
No problem. I'll try and be clear. A ratioinal number is a fraction in lowest terms with whole numbers as the numerator and denominator. For example, 3/2 is a rational number.
Originally posted by royalchickenOk...here goes...The number you have to choose from are 1/1 to infinity over 1 or infinity. Due to another thread I know that all sets of infinity are equal so the choices 1/2, 3/2, 5/2.... are infinite So the chances theoretically are infinity over infinity or one. Did I get it or am I missing something?
No...of all the rational numbers, what is the probability that a randomly selected one has an even denominator?
Originally posted by royalchickenI'm afraid I am still missing something....if you count even denominators...1/2, 1/4, 1/6, etc....you still come up with an infinite set of numbers, right. So therefore my answer would stand. What am I missing?
I'm afraid not sir. Think of all of the rationals less than, say, 10, with denominators less than, say, 10. Count how many have even denominators. Then make your parameters bigger and see what patterns emerge.
Originally posted by royalchickenAlas, that gets back into the infinity question....it would still be a 1 to 1 ratio due to the properties of infinity. Any other puzzles?
I'm trying not to be too technical here. Thing is, you're right: there is an infinity with even denominators, and also an infinity with odd denominators. My question could be phrased "For each rational with an even denominaotr, how many rationals with odd denominators are there?"
Originally posted by royalchickenThis can't be solved.
What is the probability that a randomly selected rational number (fraction in lowest terms) has an even denominator?