Prove that There exists a sequence of rationals approaching any real number

[iceblits]

Homework Statement

Let x be any real number. Prove that there exists a sequence {Rn} of rationals different from x such that {Rn} converges to x.

Use the Archimedian property, the fact that the rationals are dense in the reals, and the squeeze principle.

Homework Equations

The Archimedian property states that given any real number x, we may find a natural number n such that n>x.

The squeeze principle states that if the sequence {an} converges to L, the sequence {bn} converges to L, {an}<{bn} for all n, then, if {an}≤{cn}≤{bn}, then cn converges to L.

The Attempt at a Solution

Proof:
If x is rational, then let {an}=x+1/n which is a sequence of rationals that converges to x.
Assume x is irrational and without loss of generality assume x>0. Using the Archimedian property we can find a natural number n such that 0<x<n. We will now construct two sequences {an}, {bn} which approach x from below and above respectively. Take the interval [0,n] and divide it into two equal parts. Then look at the interval [a1,b1] containing x. We know that X cannot be halfway point because x is irrational. Divide this interval into two again and consider the interval [a2,b2] containing x. Continue in this manner. Take {an} to be the sequence of left endpoints:{a1,a2...} and {bn} to be the sequence of right endpoints: {b1,b2...}. Both of these sequences are monotone and both are bounded by x. Thus {an} and {bn} both converge to x. Note that {an}<{(an+bn)/2}<{bn} and by the squeeze principle, {(an+bn)/2} converges to x. #

I feel that this is the correct way to go about it but it seems to me that it is not necessary to use the squeeze principle when I've already found two sequences {an},{bn} that converge to x. However, my instructor insists that I have to use the squeeze principle so I feel as if my proof can be greatly shortened somehow and I'm just not seeing how...

 

[SammyS]

Homework Statement

Let x be any real number. Prove that there exists a sequence {Rn} of rationals different from x such that {Rn} converges to x.

Use the Archimedian property, the fact that the rationals are dense in the reals, and the squeeze principle.

Homework Equations

The Archimedian property states that given any real number x, we may find a natural number n such that n>x.

The squeeze principle states that if the sequence {an} converges to L, the sequence {bn} converges to L, {an}<{bn} for all n, then, if {an}≤{cn}≤{bn}, then cn converges to L.

The Attempt at a Solution

Proof:
If x is rational, then let {an}=x+1/n which is a sequence of rationals that converges to x.
Assume x is irrational and without loss of generality assume x>0. Using the Archimedian property we can find a natural number n such that 0<x<n. We will now construct two sequences {an}, {bn} which approach x from below and above respectively. Take the interval [0,n] and divide it into two equal parts. Then look at the interval [a1,b1] containing x. We know that X cannot be halfway point because x is irrational. Divide this interval into two again and consider the interval [a2,b2] containing x. Continue in this manner. Take {an} to be the sequence of left endpoints:{a1,a2...} and {bn} to be the sequence of right endpoints: {b1,b2...}. Both of these sequences are monotone and both are bounded by x. Thus {an} and {bn} both converge to x. Note that {an}<{(an+bn)/2}<{bn} and by the squeeze principle, {(an+bn)/2} converges to x. #

I feel that this is the correct way to go about it but it seems to me that it is not necessary to use the squeeze principle when I've already found two sequences {an},{bn} that converge to x. However, my instructor insists that I have to use the squeeze principle so I feel as if my proof can be greatly shortened somehow and I'm just not seeing how...

Each of the sequences, {a_n} and {b_n}, do appear to converge to x.

However, the reason you give,

"Both of these sequences are monotone and both are bounded by x. Thus {a_n} and {b_n} both converge to x."​

does not guarantee that either converges to x.

Added in Edit:

Suggestions for constructing the sequence, {R_n}:

The sequences, {a_n} and {b_n}, do not need to be rational. As long as each converges to x and a_n < b_n, for each n, then density of rationals in the reals tells you that between each pair, a_n and b_n, there is some rational number --- call it R_n .

​

 

[iceblits]

Ah so...
Given a real number x, I may use the sequences, for example, x+1/n and x-1/n both of which approach x. Then, I can use density to say that between these two possibly irrational sequences I can find a rational sequence {Rn} so by the squeeze principle Rn converges to x?

 

[SammyS]

Ah so...
Given a real number x, I may use the sequences, for example, x+1/n and x-1/n both of which approach x. Then, I can use density to say that between these two possibly irrational sequences I can find a rational sequence {Rn} so by the squeeze principle Rn converges to x?

Looks good to me.