Between 1/3 and 1/2 lies 2/5. Finding one example is straightforward, but density asks for a construction that works between any two distinct real numbers, including irrational endpoints and very small gaps. A dense set meets every nonempty open interval. Density does not mean equality with the whole line: the rational numbers are dense even though many real numbers are irrational.
Predict
Predict how a grid of spacing 1/n can enter an interval (a,b). If its spacing is smaller than b−a, at least one grid point should fit strictly between the endpoints. The Archimedean property lets us choose n with n(b−a)>1. Scaling the interval by n turns the task into locating an integer in an interval longer than one. The argument needs to work for negative a as well as positive a.
Experiment
Trace the construction for a=1.41 and b=1.42.
Step
Choice
Check
Choose denominator
n=1000
n(b−a)=10>1
Next integer above na
m=1411
1410<m≤1411
Divide back
m/n=1.411
a<m/n<b
The displayed decimals here are exact terminating rationals. Repeat the construction for an interval with negative endpoints. The least integer strictly above a real number can be justified from the Archimedean property and well-ordering after shifting into positive integers; it is not an assumption that the number itself is integral.
Understand
The general rational-density proof chooses n as above and an integer m with na<m≤na+1. Then na<m<nb, so a<m/n<b. Irrational density follows too: choose a nonzero rational q in (a/√2,b/√2), using a smaller subinterval avoiding zero if necessary. The number q√2 lies between a and b and is irrational, since otherwise division by nonzero rational q would make √2 rational.
For nested nonempty closed intervals I_n=[a_n,b_n] inside one bounded interval, let s=sup a_n. Every b_n bounds all left endpoints: earlier ones are no larger than a_n, and later ones remain inside I_n. Hence a_n≤s≤b_n for every n, placing s in their intersection. If lengths tend to zero, two distinct common points cannot exist because their positive distance would be bounded by every shrinking length.
Apply
Repair “nested intervals always have one common point.” Closedness and boundedness matter for existence, and shrinking lengths matter for uniqueness. The open intervals (0,1/n) are nested and bounded but have empty intersection. The closed intervals [0,1] repeated forever have many common points. The closed unbounded intervals [n,∞) have no common real point.
In each counterexample identify the exact hypothesis removed. Merely drawing a few intervals does not settle the infinite intersection; use the Archimedean property to exclude any proposed common positive point in the first example.
Fractional parts: bounded samples versus an all-index claim
For a real number x, its floor is the greatest integer at most x; its fractional part is x minus that floor, a number in [0,1). Fix an irrational real α and put a_n equal to the fractional part of nα for positive integers n. No a_n is zero, and different indices give different values: equality would make a nonzero integer multiple of α an integer. Thus every a_n lies strictly between zero and one. Boundedness gives a convergent subsequence, but alone does not identify its limit or prove the values approach one.
For α=√2, the sixth fractional part is about0.485281 and the twelfth about0.970563. A finite window with no points near zero does not prove that the whole sequence avoids zero. Contrast the rational step7/5, whose fractional parts repeat0,1/5,2/5,3/5,4/5.
Inspect a window without confusing it with density
A zero count is a finite-search result. The proof below decides the irrational orbit's eventual behavior; no displayed hit frequency proves equidistribution.
Plot positions and decimals are approximations. Hits for √2 use exact integer squared inequalities; rational hits use integer cross-products. Endpoints are excluded.
Inspect exact expressions and interval decisions
n
Floor of nα
Exact fractional expression
Approximation
Strictly inside?
1
1
1√2 − 1
0.414214
no
2
2
2√2 − 2
0.828427
no
3
4
3√2 − 4
0.242641
no
4
5
4√2 − 5
0.656854
no
5
7
5√2 − 7
0.071068
no
6
8
6√2 − 8
0.485281
yes
7
9
7√2 − 9
0.899495
no
8
11
8√2 − 11
0.313708
no
9
12
9√2 − 12
0.727922
no
10
14
10√2 − 14
0.142136
no
11
15
11√2 − 15
0.556349
no
12
16
12√2 − 16
0.970563
no
13
18
13√2 − 18
0.384776
no
14
19
14√2 − 19
0.798990
no
15
21
15√2 − 21
0.213203
no
16
22
16√2 − 22
0.627417
no
At most128points are computed, with largest index256. Settings are temporary and do not award mastery.
Pigeonhole closeness requires a sign repair
Partition [0,1) into N half-open intervals of length1/N, for a positive integer N. The N+1 fractional parts at indices0 through N place two distinct values in the same interval. Subtract the earlier index from the later to get a positive integer q and an integer p with a signed difference ε=qα−p, where0<|ε|<1/N. Irrationality prevents ε=0. A small absolute difference does not yet say the positive-index fractional part is small: if ε is negative, that fractional part is close to one.
In fact either sign gives a positive-index value near one. If ε<0, the fractional part of qα is1+ε, strictly above1−1/N. If ε>0, set k to the floor of1/ε. Since ε is irrational,1/ε is not an integer. Hence1−ε<kε<1, and the fractional part of kqα is kε. It too is strictly above1−1/N. This proves positive-index values arbitrarily close to one, without assuming the sign supplied by pigeonholes. No bound on the final index kq in terms of N has been claimed.
Turn values approaching one into a genuine subsequence
Choosing one such value for every N creates a sequence of values in the range, but the chosen indices need not increase. We must justify that additional requirement. After an index m has been chosen, all finitely many distances1−a1 through1−a_m are positive. Choose a tolerance smaller than their minimum and smaller than1/(j+1), where j is the current subsequence stage. The near-one result supplies a value at a later index: none of the first m values is close enough. Repeating gives strictly increasing indices n_j with a(n_j) tending to one.
More generally, if values from the range of a sequence approach L and L is not itself in the range, the positive minimum distance from any finite prefix forces a closer value to occur beyond it. This is the same increasing-index construction. If L is in the range, the conclusion need not hold: a sequence with first value0 and all later values1 has range values constantly equal to0 but no subsequence converging to0. The absent-limit hypothesis is a precise tool for this lemma, not a claim that every sequence containing its limit lacks a convergent subsequence.
Continue the mechanism to density and every tail
A positive-index value close to one also produces one close to zero. Write the fractional part of qα as1−δ with irrational0<δ<η, where η is any prescribed positive tolerance. Put k=floor(1/δ). Then0<1−kδ<δ, and the fractional part of kqα is1−kδ. Call this small positive fractional part β.
To hit an arbitrary open interval(a,b) with0≤a<b≤1, choose0<β<b−a and set r=floor(a/β)+1. Then a<rβ≤a+β<b≤1. Because rβ has not wrapped past one, it is the fractional part of rqα. Thus the positive-index fractional parts are dense in [0,1] in the sense of meeting every such open interval.
Every tail is dense too. Inside any target interval choose a smaller open interval avoiding the finitely many values at indices1 through M. Density supplies a value in this smaller interval, and its index must exceed M. Repeatedly shrink target intervals about any t in [0,1], always requiring a later index; this constructs a subsequence converging to t, including endpoints via one-sided intervals. The statement is stronger than merely finding arbitrarily many displayed dots. Density and tail density do not by themselves prove a particular limiting visit frequency; equidistribution is a separate theorem.
The MIT18.100B Spring2025 Problem Set4, Problem5 motivates the fractional-part and increasing-subsequence action. The positive-index sign cases and finite-prefix exclusion above make the mechanism explicit. The bounded inspector compares two fixed steps; it does not numerically decide whether an arbitrary decimal input is irrational.
Continue
Optional depth relates interval bisection to completeness. Repeatedly retaining a closed half creates nested intervals whose lengths tend to zero, giving a unique real point. A finite computation outputs only an interval enclosing that point; it does not generally produce its exact decimal expansion.
Keep a proof ledger with one existence argument, one uniqueness argument and one missing-hypothesis counterexample. These will reappear in Bolzano–Weierstrass and compactness. Density will also supply sequences that witness discontinuity, showing why arbitrarily fine sampling along just one type of input can miss a function’s behavior.
Apply · explain · revise
Practice and reasoning
Drafts save locally as you type. Reveal worked solutions when useful; the site remembers that exposure. Automatic matches check the requested answer, while written arguments require review against the rubric.
Let α be irrational and a_n be the fractional part of nα. Use N equal half-open intervals and indices0 through N to find a positive q and integer p with0<|qα−p|<1/N. Treat both signs to prove that some positive-index a_n exceeds1−1/N. Explain why simply reversing a negative index would not finish the positive-index claim.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Suppose values from the range of a real sequence approach L and L is not in that range. Prove that a subsequence with strictly increasing original indices approaches L. Apply this to the irrational fractional parts approaching1. Give a counterexample when the absent-limit hypothesis is dropped, and explain how small positive fractional parts imply density.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Give the midpoint of 2/5 and 3/5.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
Nested nonempty closed bounded intervals with lengths tending to zero have:
Prove a rational lies between arbitrary real a<b.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Using rational density, prove every open interval contains an irrational.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Prove nested intervals [a_n,b_n] have a common point when nonempty and all lie in [a₁,b₁].
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Give nested bounded nonempty open intervals whose intersection is empty and prove it.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Reflect and continue
Before marking this lesson complete, explain one key idea without the worked solution and identify a condition that matters. Marking complete records participation, not independent mastery.
