The usual order on ℝ: translation, cross-multiplication with positive denominators, finite extrema and closed intervals

On this page
  1. Goal and scope
  2. Givens and symbols
  3. Rules and sources
  4. Main derivation
  5. Substeps and lemmas
  6. Reverse audits
  7. Error controls and negative controls
  8. Result and limitations
  9. Verification and source binding

← Guide index · Original MKO

derivation-mko-order-relation-01 · production-20261019-adopted-96

Goal and scope

Prove for all reals a, b, c that a<b ⇔ a+c<b+c; that p/q<r/s ⇔ ps<rq when q, s>0; and that a finite nonempty set of reals has a minimum, which is unique. Use these to compare the elements of M={−5/9, 1/3, −7/12, −1/2} (listed in this order), find the minimum and maximum of M, and decide whether the two endpoints and 1/2 belong to the closed interval [min M, max M].

Only the usual total order on the reals is considered (MKO a1), with the order axioms of order_basics as prerequisites; the reals are not constructed. Comparisons use rationals with positive denominators (a2); extrema are for nonempty finite sets, and a closed interval needs its lower endpoint no greater than its upper endpoint (a3). Existence of a maximum is covered by 'the same argument with ≥ in place of ≤'. Complex numbers, general partial orders and suprema of infinite sets are not treated; the canonical MKO and its English edition are not changed.

Givens and symbols

  • order_basics
  • frac_rules
  • add_inverse
  • nat_induction
  • numeric_evaluation
  • G1 — The usual order ≤ on ℝ is reflexive, antisymmetric, transitive and total; a<b means a≤b and a≠b.
  • G2 — M={−5/9, 1/3, −7/12, −1/2}, listed in this order; in positive-denominator form: (−5)/9, 1/3, (−7)/12, (−1)/2.
  • G3 — Definitions: m is a minimum of a set M when m∈M and m≤x for all x∈M; a maximum likewise with ≥. For l≤u, the closed interval [l,u]={x∈ℝ ∣ l≤x and x≤u}.
  • a\le b,\ a<b — the usual order and the strict order
  • \frac{p}{q} — a rational with denominator q>0
  • M — the finite set being compared
  • \min M,\ \max M — the minimum and maximum elements
  • [l,u] — the closed interval {x ∣ l≤x≤u}

Rules and sources

  • order_basics · prerequisite — Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw.
  • frac_rules · prerequisite — Fraction rules (all denominators nonzero): a/c+b/c=(a+b)/c; (a/b)(c/d)=(ac)/(bd); (a/b)/(c/d)=(ad)/(bc) (also needing c≠0); a/b=(ak)/(bk) for k≠0.
  • add_inverse · prerequisite — Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v).
  • nat_induction · prerequisite — Principle of induction: for a family P(n) starting at n₀, if P(n₀) holds and P(n)⇒P(n+1) for every n≥n₀, then P(n) holds for every n≥n₀.
  • numeric_evaluation · prerequisite — Exact evaluation of an explicit integer or rational expression; independently recomputable.
  • add_assoc · prerequisite — Associativity of addition: (u+v)+w=u+(v+w).
  • add_zero · prerequisite — Additive identity: u+0=u.
  • mul_inverse_nonzero · prerequisite — Inverse of a nonzero number: in ℚ, ℝ or ℂ, if u≠0 there is 1/u with u·(1/u)=1, and v/u means v·(1/u). 0 has no inverse, so v/0 is undefined.
  • mul_zero · prerequisite — Multiplying by zero: u·0=0. Reason: u·0=u·(0+0)=u·0+u·0; subtracting u·0 from both sides gives 0=u·0.
  • contradiction · prerequisite — Proof by contradiction: if assuming ¬P leads to a contradiction (a statement and its negation both hold), then P holds.
  • mul_comm · prerequisite — Commutativity of multiplication: uv=vu (in commutative number systems such as ℤ, ℚ, ℝ, ℂ).
  • mul_assoc · prerequisite — Associativity of multiplication: (uv)w=u(vw).
  • logic_chain · prerequisite — Chaining: from P⇒Q and Q⇒R infer P⇒R; likewise for ⇔.
  • case_analysis · prerequisite — Proof by cases: if P∨Q holds and both P⇒R and Q⇒R are proved, then R holds. The cases must cover every possibility.
  • and_intro_elim · prerequisite — Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q.
  • mko_order · cited — On R, the usual total order ≤ is reflexive, antisymmetric and transitive; any two real numbers are comparable. Write a<b when a≤b and a≠b. For real a,b,c, a<b exactly when a+c<b+c. A finite nonempty real set has a minimum and a maximum; a closed interval [l,u] with l≤u includes both endpoints. · mko-order-relation
  • min_max_def · prerequisite — Minimum: m∈M and m≤x for all x∈M; maximum: m∈M and m≥x for all x∈M.
  • interval_def · prerequisite — Closed interval: for l≤u, x∈[l,u] ⇔ l≤x and x≤u.
  • L_translate · lemma — For all reals a, b, c: a<b ⇔ a+c<b+c.
  • L_cross · lemma — For reals p, r and q>0, s>0: p/q<r/s ⇔ ps<rq.
  • L_finite_min · lemma — Every nonempty finite set of reals has a minimum.
  • L_min_unique · lemma — A set of reals has at most one minimum (and likewise at most one maximum).

Main derivation

S01 · definition

Before

a,b\in\mathbb{R}

After

a<b\ :\iff\ a\le b\land a\neq b;\ \text{exactly one of}\ a<b,\ a=b,\ b<a

Conditions: none

Rule: mko_order · Uses: G1, mko_order

Leaf basis: mko_order + order_basics

Define the strict order as in the MKO. Totality (a≤b or b≤a) and antisymmetry (both together force a=b) give trichotomy: exactly one of a<b, a=b, b<a holds. This is the MKO's three-way comparison with outputs <, = or >.

S02 · equivalent

Before

a<b

After

a+c<b+c

Conditions: none

Rule: L_translate · Uses: S01

Leaf basis: order_basics + add_assoc + add_inverse + add_zero

By Lemma L_translate, adding c to both numbers does not change the comparison. ⇒ is S02.1 and ⇐ is S02.2.

S03 · equivalent

Before

\frac{p}{q}<\frac{r}{s}

After

ps<rq

Conditions: q>0; s>0

Rule: L_cross · Uses: S01

Leaf basis: order_basics + frac_rules + mul_inverse_nonzero

By Lemma L_cross: when both denominators are positive, comparing two fractions is the same as comparing the cross products. The proof is in S03.1 to S03.3; the positive-denominator condition cannot be dropped.

S04 · equivalent

Before

-\tfrac{7}{12}\ ?\ -\tfrac{5}{9}

After

(-7)\cdot9=-63<-60=(-5)\cdot12\ \Rightarrow\ -\tfrac{7}{12}<-\tfrac{5}{9}

Conditions: 12>0; 9>0

Rule: L_cross · Uses: S03, G2

Leaf basis: order_basics + numeric_evaluation

Take p=−7, q=12, r=−5, s=9; both denominators are positive. ps=−63 and rq=−60, and −60−(−63)=3>0, so −63<−60; by L_cross, −7/12<−5/9.

S05 · equivalent

Before

-\tfrac{5}{9}\ ?\ -\tfrac{1}{2}

After

(-5)\cdot2=-10<-9=(-1)\cdot9\ \Rightarrow\ -\tfrac{5}{9}<-\tfrac{1}{2}

Conditions: 9>0; 2>0

Rule: L_cross · Uses: S03, G2

Leaf basis: order_basics + numeric_evaluation

Take p=−5, q=9, r=−1, s=2: ps=−10, rq=−9 and −10<−9, so −5/9<−1/2.

S06 · equivalent

Before

-\tfrac{1}{2}\ ?\ \tfrac{1}{3}

After

(-1)\cdot3=-3<2=1\cdot2\ \Rightarrow\ -\tfrac{1}{2}<\tfrac{1}{3}

Conditions: 2>0; 3>0

Rule: L_cross · Uses: S03, G2

Leaf basis: order_basics + numeric_evaluation

Take p=−1, q=2, r=1, s=3: ps=−3, rq=2 and −3<2, so −1/2<1/3.

S07 · implies

Before

-\tfrac{7}{12}<-\tfrac{5}{9},\ -\tfrac{5}{9}<-\tfrac{1}{2},\ -\tfrac{1}{2}<\tfrac{1}{3}

After

-\tfrac{7}{12}<-\tfrac{5}{9}<-\tfrac{1}{2}<\tfrac{1}{3}

Conditions: none

Rule: order_basics · Uses: S04, S05, S06

Leaf basis: order_basics + logic_chain

The three comparisons form a chain; by transitivity of <, each number in the chain is less than every number to its right, for example −7/12<−1/2, −7/12<1/3 and −5/9<1/3. All four elements of M are now ordered.

S08 · implies

Before

M\subseteq\mathbb{R}\ \text{finite},\ M\neq\varnothing

After

M\ \text{has a minimum}

Conditions: M\neq\varnothing

Rule: L_finite_min · Uses: S01

Leaf basis: nat_induction + order_basics + case_analysis

The MKO asserts that a nonempty finite set of reals has a minimum and a maximum. By Lemma L_finite_min, the minimum exists, proved by induction on the number of elements n≥1 (S08.1, S08.2); the argument for the maximum is the same with ≥ in place of ≤.

S09 · implies

Before

-\tfrac{7}{12}\in M,\ -\tfrac{7}{12}\le x\ \forall x\in M

After

\min M=-\tfrac{7}{12}

Conditions: none

Rule: min_max_def · Uses: S07, G2, min_max_def

Leaf basis: min_max_def + order_basics

−7/12 belongs to M (listed third), by S07 it is less than the other three elements, and −7/12≤−7/12 for itself, so it is the minimum. M lists −5/9 first, but extrema are decided by value, not by the order of listing.

S10 · implies

Before

\tfrac{1}{3}\in M,\ \tfrac{1}{3}\ge x\ \forall x\in M

After

\max M=\tfrac{1}{3}

Conditions: none

Rule: min_max_def · Uses: S07, G2, min_max_def

Leaf basis: min_max_def + order_basics

1/3 belongs to M and by S07 the other three elements are all less than 1/3, so 1/3 is the maximum.

S11 · implies

Before

m,m'\ \text{both minima of}\ M

After

m=m'

Conditions: none

Rule: L_min_unique · Uses: S09, S10

Leaf basis: min_max_def + order_basics

By Lemma L_min_unique the minimum is unique, so one may speak of 'the' minimum −7/12; the maximum 1/3 is likewise unique. The proof is in S11.1.

S12 · implies

Before

l=-\tfrac{7}{12},\ u=\tfrac{1}{3}

After

l\le u

Conditions: none

Rule: order_basics · Uses: S07, S09, S10, G3

Leaf basis: order_basics + mko_order

MKO assumption a3 asks first to check that the lower endpoint of a closed interval is no greater than the upper one. By S07, l=−7/12<1/3=u, so l≤u and [l,u] is a valid closed interval.

S13 · implies

Before

l\le l\le u,\ l\le u\le u

After

l\in[l,u],\ u\in[l,u],\ M\subseteq[l,u]

Conditions: l\le u

Rule: interval_def · Uses: S12, S09, S10, interval_def

Leaf basis: interval_def + mko_order

Membership in a closed interval is l≤x≤u. For x=l: l≤l (reflexivity) and l≤u (S12), so the left endpoint belongs to the interval; the right endpoint likewise. Every element x of M satisfies min M≤x≤max M (S09, S10), so M⊆[l,u].

S14 · implies

Before

\tfrac{1}{3}<\tfrac{1}{2}\ \ (1\cdot2<1\cdot3)

After

\tfrac12\notin[l,u]

Conditions: none

Rule: interval_def · Uses: S03, S13, interval_def

Leaf basis: interval_def + contradiction + order_basics

By L_cross (p=1, q=3, r=1, s=2): 1·2=2<3=1·3, so 1/3<1/2. If 1/2≤1/3, antisymmetry together with 1/3≤1/2 would give 1/2=1/3, contradicting 1/3≠1/2; so 1/2≤u fails and 1/2 is not in the interval.

S15 · equivalent

Before

-\tfrac{7}{12}<-\tfrac{5}{9}

After

-\tfrac{7}{12}+\tfrac{7}{12}<-\tfrac{5}{9}+\tfrac{7}{12},\ \text{i.e.}\ 0<\tfrac{1}{36}

Conditions: none

Rule: L_translate · Uses: S02, S04

Leaf basis: order_basics + frac_rules + numeric_evaluation

Apply L_translate with c=7/12: −7/12<−5/9 ⇔ 0<−5/9+7/12. Compute −5/9+7/12=−20/36+21/36=1/36>0, consistent with S04. Translation does not change the comparison, so this is a rewriting, not a new kind of exercise.

S16 · implies

Before

\text{S02 to S15}

After

\min M=-\tfrac{7}{12},\ \max M=\tfrac13,\ l,u\in[l,u],\ \tfrac12\notin[l,u]

Conditions: none

Rule: and_intro_elim · Uses: S02, S03, S08, S09, S10, S11, S13, S14

Leaf basis: and_intro_elim

Summary: translation invariance and positive-denominator cross-comparison are proved; a nonempty finite set has a unique minimum; M is ordered −7/12<−5/9<−1/2<1/3 with minimum −7/12 and maximum 1/3; the closed interval [−7/12,1/3] contains both endpoints and all of M, but not 1/2.

Substeps and lemmas

S02.1 · implies

Before

a<b

After

a+c<b+c

Conditions: none

Rule: order_basics · Uses: order_basics

Leaf basis: order_basics

(⇒) This is exactly the order axiom 'u<v ⇒ u+w<v+w' with u=a, v=b, w=c.

S02.2 · implies

Before

a+c<b+c

After

(a+c)+(-c)<(b+c)+(-c),\ \text{i.e.}\ a<b

Conditions: none

Rule: order_basics · Uses: order_basics, add_inverse

Leaf basis: order_basics + add_assoc + add_inverse + add_zero

(⇐) Apply the same axiom to a+c<b+c with w=−c to get (a+c)+(−c)<(b+c)+(−c); simplifying both sides (associativity, c+(−c)=0, adding 0) gives a<b. S02.1 and S02.2 together prove Lemma L_translate.

S03.1 · implies

Before

q>0,\ s>0

After

qs>0,\ \tfrac{1}{qs}>0

Conditions: q>0; s>0

Rule: order_basics · Uses: order_basics, mul_inverse_nonzero

Leaf basis: order_basics + mul_zero + mul_inverse_nonzero + contradiction

Multiply 0<q by s>0 to get 0·s<q·s, i.e. qs>0. Then qs≠0, so 1/(qs) exists; if 1/(qs)<0, multiplying by qs>0 would give 1<0, a contradiction, and 1/(qs)≠0 since (1/(qs))·qs=1, so 1/(qs)>0.

S03.2 · implies

Before

\frac{p}{q}<\frac{r}{s}

After

\frac{p}{q}\cdot qs<\frac{r}{s}\cdot qs,\ \text{i.e.}\ ps<rq

Conditions: qs>0

Rule: order_basics · Uses: S03.1, frac_rules

Leaf basis: order_basics + frac_rules + mul_comm + mul_assoc

(⇒) Multiply both sides by the positive number qs, which keeps the order; simplify (p/q)·qs=ps and (r/s)·qs=rq (fraction rules, q, s≠0).

S03.3 · implies

Before

ps<rq

After

\frac{ps}{qs}<\frac{rq}{qs},\ \text{i.e.}\ \frac{p}{q}<\frac{r}{s}

Conditions: \tfrac{1}{qs}>0

Rule: order_basics · Uses: S03.1, frac_rules

Leaf basis: order_basics + frac_rules

(⇐) Multiply both sides by the positive number 1/(qs) (S03.1), which keeps the order; cancel ps/(qs)=p/q and rq/(qs)=r/s. S03.1 to S03.3 together prove Lemma L_cross.

S08.1 · induction_base

Before

M=\{a\}

After

\min M=a

Conditions: none

Rule: nat_induction · Uses: G3, min_max_def

Leaf basis: min_max_def + order_basics

Base n=1: M={a}. a∈M, and the only element x=a satisfies a≤a (reflexivity), so a is the minimum (and also the maximum).

S08.2 · induction_step

Before

|M|=n+1,\ a\in M,\ m'=\min(M\setminus\{a\})

After

\min M=m'\ \text{if}\ m'\le a,\ \ \min M=a\ \text{if}\ a<m'

Conditions: n\ge1

Rule: nat_induction · Uses: S08.1, min_max_def, case_analysis

Leaf basis: nat_induction + case_analysis + order_basics + min_max_def

Inductive step: assume every nonempty set with n elements has a minimum (n≥1). Take |M|=n+1 and a∈M; M′=M∖{a} has n elements and, by the induction hypothesis, a minimum m′. By trichotomy there are two cases: if m′≤a, then m′ is no greater than every element of M′ and than a, so m′ is the minimum of M; if a<m′, then for x∈M′ we have a<m′≤x, so a≤x by transitivity, and a≤a, so a is the minimum. S08.1 and S08.2 together prove Lemma L_finite_min.

S11.1 · implies

Before

m,m'\in M,\ m\le x,\ m'\le x\ \forall x\in M

After

m\le m',\ m'\le m\ \Rightarrow\ m=m'

Conditions: none

Rule: mko_order · Uses: min_max_def, G1

Leaf basis: min_max_def + mko_order

m is a minimum and m′∈M, so m≤m′; likewise m′≤m. By antisymmetry m=m′. The maximum is handled the same way with ≥. This step proves Lemma L_min_unique.

Reverse audits

  • S16 · sufficient_goal_reduction · S02, S03, S08, S09, S10, S11, S13, S14, and_intro_elim — Goal G is the proofs and decisions in the goal; the sufficient condition H is the conjunction of the step results, all proved forwards, and H⇒G uses only conjunction introduction.
  • S15 · inverse_equivalence · L_translate, frac_rules, numeric_evaluation — L_translate is an equivalence; adding 7/12 and subtracting 7/12 undo each other.
  • S14 · premise_audit · L_cross, interval_def, contradiction — Membership is a conjunction of two inequalities; refuting one of them suffices to exclude the point.
  • S13 · premise_audit · interval_def, S12, mko_order — Endpoint membership uses only reflexivity and l≤u; open intervals are a different object and are not discussed here.
  • S12 · premise_audit · S07, order_basics — From l<u, l≤u follows by the definition of the strict order.
  • S11.1 · premise_audit · min_max_def, mko_order — Only that both minima belong to M and antisymmetry are used.
  • S11 · premise_audit · L_min_unique — Uniqueness relies on antisymmetry.
  • S10 · premise_audit · S07, min_max_def — Symmetric to S09.
  • S09 · premise_audit · S07, min_max_def, order_basics — Both conditions (membership in M and being no greater than every element) are checked; x<y gives x≤y by the definition of the strict order.
  • S08.2 · premise_audit · nat_induction, case_analysis, order_basics, min_max_def — It needs n≥1 so that M′ is nonempty and the induction hypothesis applies; the two cases are exhaustive by totality.
  • S08.1 · premise_audit · min_max_def, order_basics — Only reflexivity is used; the two extrema of a singleton coincide.
  • S08 · premise_audit · L_finite_min — Nonemptiness is essential: the induction starts at n=1, the empty set is outside its range, and it has no minimum.
  • S07 · premise_audit · S04, S05, S06, order_basics — What is needed is the three adjacent comparisons and transitivity; the other three pairs need no further cross-multiplication.
  • S06 · inverse_equivalence · L_cross, numeric_evaluation — As in S04, the denominators are positive and the equivalence applies.
  • S05 · inverse_equivalence · L_cross, numeric_evaluation — As in S04, the denominators are positive and the equivalence applies.
  • S04 · inverse_equivalence · L_cross, numeric_evaluation — L_cross is an equivalence and its conditions 12, 9>0 hold, so the fraction comparison and the integer comparison are true or false together.
  • S03.3 · premise_audit · S03.1, order_basics, frac_rules — It needs 1/(qs)>0, supplied by S03.1.
  • S03.2 · premise_audit · S03.1, order_basics, frac_rules — It needs qs>0; if qs<0, multiplication would reverse the inequality.
  • S03.1 · premise_audit · order_basics, mul_inverse_nonzero, contradiction — A product of two positives is positive, and the reciprocal of a positive is positive; both depend on q, s>0.
  • S03 · inverse_equivalence · L_cross — ⇒ and ⇐ are proved in the sub-steps, and both use the condition q, s>0.
  • S02.2 · premise_audit · order_basics, add_assoc, add_inverse, add_zero — Every real c has an additive inverse −c, so this direction needs no extra condition.
  • S02.1 · premise_audit · order_basics — A direct application of the axiom, valid for every real c.
  • S02 · inverse_equivalence · L_translate — Both directions are proved, so this is an equivalence.
  • S01 · premise_audit · G1, mko_order, order_basics — This only restates a definition and the axioms; trichotomy is also listed in order_basics as a prerequisite.

Error controls and negative controls

Denominators must be positive before cross-multiplying

One misconception identified in the MKO is ignoring the sign of denominators. Cross-multiplying 1/(−2) and 1/3 directly gives 1·3=3>−2=1·(−2) and wrongly suggests 1/(−2)>1/3; in fact 1/(−2)=−1/2<1/3. The cause is that multiplying by the negative (−2)·3 reverses the inequality. Rewriting with a positive denominator (−1)/2 first gives (−1)·3=−3<2 and the correct conclusion.

A closed interval contains its endpoints

The other misconception is to leave out the endpoints of a closed interval. l≤l≤u holds, so l∈[l,u] (S13). When l=u the interval has a single value, for example [2,2]={2}; it is the open interval that excludes endpoints, and that is a different object.

Extrema go by value, not order, and the empty set has none

M lists −5/9 first, but the minimum is −7/12 (S09). The empty set has no minimum because there is no m∈∅; the induction in S08 starts at n=1 and does not cover the empty set, so no extremum may be assigned to it.

  • Cross-multiplying 1/(−2) and 1/3 gives 1·3=3>−2=1·(−2), so 1/(−2)>1/3 → reject — The denominator −2 is negative, so L_cross does not apply; 1/(−2)=−1/2<1/3.
  • −7/12 is an endpoint, so it is not in [−7/12,1/3] → reject — −7/12≤−7/12≤1/3 holds; a closed interval contains its endpoints.
  • The minimum of M is −5/9 because it is listed first → reject — −7/12<−5/9 (S04); the minimum is decided by value and is −7/12.
  • The minimum of the empty set is 0 → reject — A minimum must belong to the set, and 0∉∅; the empty set has no minimum.

Result and limitations

For all real a, b and c, a<b ⇔ a+c<b+c; for q, s>0, p/q<r/s ⇔ ps<rq; a nonempty finite set of reals has a unique minimum (and likewise maximum). The elements of M are ordered −7/12<−5/9<−1/2<1/3, so min M=−7/12 and max M=1/3 (regardless of the listing order); [−7/12,1/3] is a valid closed interval containing both endpoints and all of M, but not 1/2.

  • The order axioms (order_basics) are used as prerequisites; they are not derived from a construction of the reals.
  • Existence of a maximum is covered only by 'the same argument with ≥ in place of ≤' and is not written out step by step.

Verification and source binding

14 declared checks for this guide; only the listed expressions are covered.

  • source commit: b28a5edbd46279b1db2a3f3f48bb3f447e18899f
  • adopted source SHA-256: 2cbe9fb81fea4fdafc3a0280d691e7cbfb470c6ed78948d6657b4f589caa62a8
  • original source SHA-256: 2cbe9fb81fea4fdafc3a0280d691e7cbfb470c6ed78948d6657b4f589caa62a8
  • Lean: NOT_RUN · independent AI: NOT_MEASURED

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