Let α∈[0,9]. Is the set of real numbers such that the average of the digits of each of these numbers =α, countable or uncountable?

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Declan Cameron · Accepted Answer

Step 1Given:&amp;nbsp;α∈[0,9], for&amp;nbsp;n∈N&amp;nbsp;letxn=⌊nα⌋−⌊(n−1)α⌋.Note that&amp;nbsp;xn&amp;nbsp;is a non-negative integer and&amp;nbsp;≤9, so can be used as a decimal digit. Let&amp;nbsp;A⊆BN&amp;nbsp;be an arbitrary set (so there are continuum-many choices for A). Then consider the number&amp;nbsp;Xα,A=0.d1d2d3…&amp;nbsp;that has (no integer part and) as nth decimal digitdn={1n=2k,k∈A0n=2k,k∉Axn−k2k&amp;lt;n&amp;lt;2k+1&amp;nbsp;Then one verifies quickly that the average digit is&amp;nbsp;α

Let \(\displaystyle\alpha\in{\left[{0},{9}\right]}\). Is the set of real

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