Let {xn} be a Cauchy sequence such that every term xn is an integer. Show that {xn} is ""eventually constant"" – i. e. there exist and N > 0 such that xn = xm for all n > m ≥ N

Given epsilon equal to 1/2, then should be n0 such that, for any values n and m greater than or equal to n0, we have that |an-am| < 1/2

Since the difference between 2 integers is an integer, and the module may not be negative, we have no other choice than |an-am| = 0. Thus, an = am. This means that for any k bigger than n0, we have that ak = an0, as a result, the sequence is eventually constant.

in problems involving proportions, we can use cross products to test whether two ratios are equal and form a proportion. to find the cross products of a proportion, we multiply the outer terms, called the extremes, and the middle terms, called the means.

hope it

the aswere is the one with all 8's and the second one is with the 10,8 and 6

step-by-step explanation: