Prove the following using Proof techniques: Are the numbers, i.e., Irrational, c

Question

Prove the following using Proof techniques:

Are the numbers, i.e., Irrational, countable of uncountable?
(Saying "the set is easy to list" or "not easy to list" is not a sufficient answer for this question.)

Explanation / Answer

let A be the biggest rational number(this is our hypothese) we know that the real numbers are uncountable A is rational than A is real than there is an other real bigger than A like A + 1/2 1/2 is rational & A is rational then [A+ 1/2]is rational (cuz the sum of 2 rationals is rational) [A +1/2]>A & it is a rational so A isnt the biggest rational (contradiction with the hypothese) then there is no "biggest rational number" so there is an infinite rational number The direct answer to your question is "Because they are". If you are asking for a proof, the outline goes like this: Assume the irrationals are countable. Then there is a 1-1 mapping from the natural numbers onto the irrationals. Given that, you can construct an number that can't be on the list, a contradiction. The simple example goes like this: Just consider all the decimals between 0 and 1. They can be represented in the form .dddddddd..... Now suppose they are countable, then there has to be a mapping, f(n), from the natural numbers onto them. Construct the following number X: If the nth digit of f(n) is 0, the nth digit of X will be 1. If the nth digit of f(n) isn't 0, make the nth digit of X = 0. For each n, X is different from f(n) in the nth decimal place, so X is not equal to f(n) for an n. That says X isn't in the list, which was assumed to be all inclusive. Contradiction.

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