Prove that there is a strictly decreasing sequence

[Alexmahone]

Given any irrational number c > 0, prove that there is a strictly decreasing sequence of rational numbers that converges to c.

 

[ThePerfectHacker]

Given any irrational number c > 0, prove that there is a strictly decreasing sequence of rational numbers that converges to c.

Let n be a positive integer. Then there is rational number q_n such that c < q_n < c+1/n (denseness of rationals).

Now q_n is sequeezed between c and c+1/n. As c and c+1/n go to c in the limit it follows by the squeeze theorem that q_n goes to c also. However, perhaps q_n might not be strictly decreasing. We need to adjust for that. But this is very simple. Consider the sequence q_1,q_2,q_3,... and so forth. Start with q_1. Then keep on going into the sequence until you find the next rational number which is strictly less than q_1 (this must happen), by relabeling we call this q_2. Then keep on going past q_2 into the sequence until you find the next rational number which is strictly less than q_2 (this again must happen), by relabeling we call this q_3. Now by repeating this process we produced a sequence of rational numbers that are strictly deceasing and converges to c.

 

[HallsofIvy]

Another way of thinking about this. Every real number, including the irrationals and, in particular, c, has a decimal expansion. Let a_0 be the c rounded to the next larger integer (the smallest integer larger than c). Let a_1 be c rounded to the next higher first decimal place. In general, let a_n be c rounded to the next higher nth decimal place. (If the n decimal digit is 9, skip that term.)

For example, if c= pi which is 3.1415926..., a_0= 4, a_1= 3.2, a_2= 3.15, a_3= 3.142, a_4= 3.1416, we skip the "9" so a_5= 3.141593, a_6= 3.1415927, etc.