Main derivation
S01 · definition
Before
x\in\mathbb{Q}After
\exists p,q\in\mathbb{Z},\ q\neq 0:\ x=\frac{p}{q}Conditions: none
Rule: mko_rat_def · Uses: G1, G2, mko_rat_def
Leaf basis: mko_rat_def + mul_inverse_nonzero
By the MKO, deciding that x is rational means finding a pair of integers (p,q) with q≠0 and x=p/q. The definition only asks that such a pair exist and does not say it is unique; different integer pairs can represent the same rational, which is what the following steps handle.
S02 · equivalent
Before
\frac{p}{q}=\frac{r}{s}\quad(q\neq0,\ s\neq0)After
ps=rq
Conditions: q\neq 0; s\neq 0
Rule: L_cross · Uses: G1, L_cross
Leaf basis: L_cross
The tool for deciding whether two fractions are equal is Lemma L_cross: when both denominators are nonzero, p/q=r/s ⇔ ps=rq. The lemma is proved in S02.1 and S02.2 and used in S03, S04, S07 and S09.
S03 · equivalent
Before
6\cdot 4=24=(-3)\cdot(-8)
After
\frac{6}{-8}=\frac{-3}{4}Conditions: -8\neq 0; 4\neq 0
Rule: L_cross · Uses: S02, G4
Leaf basis: L_cross + numeric_evaluation + neg_mul
Take p=6, q=−8, r=−3, s=4: ps=6·4=24 and rq=(−3)·(−8)=24. They are equal, so by L_cross, 6/(−8)=−3/4.
S04 · equivalent
Before
6\cdot(-12)=-72=9\cdot(-8)
After
\frac{6}{-8}=\frac{9}{-12}=\frac{-3}{4}Conditions: -8\neq 0; -12\neq 0
Rule: L_cross · Uses: S02, S03, G4
Leaf basis: L_cross + numeric_evaluation + eq_refl_symm_trans
Take p=6, q=−8, r=9, s=−12: ps=6·(−12)=−72 and rq=9·(−8)=−72, so 6/(−8)=9/(−12); with S03, the three different integer pairs (6,−8), (−3,4) and (9,−12) represent the same rational number. This is exactly the MKO's 'different integer pairs can represent the same rational number'.
S05 · equivalent
Before
\frac{6}{-8}After
\frac{6\cdot(-\frac12)}{(-8)\cdot(-\frac12)}=\frac{-3}{4}Conditions: -\tfrac{1}{2}\neq 0
Rule: frac_rules · Uses: G1, frac_rules
Leaf basis: frac_rules + numeric_evaluation
Reduce: gcd(6,8)=2 (N20), so multiply numerator and denominator by k=−1/2 (that is, divide both by −2, which also moves the sign to the numerator). By the fraction rule a/b=(ak)/(bk) for k≠0, 6/(−8)=(6·(−1/2))/((−8)·(−1/2))=(−3)/4 (N7, N8). This is an exact equality, not an approximation.
S06 · implies
Before
\frac{-3}{4}After
\gcd(3,4)=1,\ 4>0:\ \text{canonical form}Conditions: none
Rule: canonical_def · Uses: S05, G3, gcd_def
Leaf basis: canonical_def + gcd_def + divides_def + numeric_evaluation
The positive divisors of 4 are 1, 2 and 4; neither 2 nor 4 divides 3, so gcd(3,4)=1 (N9); and the denominator 4>0. So −3/4 meets the canonical display convention of G3. Normalizing changes the representation, not the value (S05 is an equality).
S07 · equivalent
Before
2\cdot5=10\neq 9=3\cdot3
After
\frac{2}{3}\neq\frac{3}{5}Conditions: 3\neq 0; 5\neq 0
Rule: L_cross · Uses: S02, G4
Leaf basis: L_cross + numeric_evaluation
Take p=2, q=3, r=3, s=5: ps=10 and rq=9, and 10≠9. L_cross is a biconditional, so negating both sides keeps an equivalence: ps≠rq ⇔ p/q≠r/s, hence 2/3≠3/5.
S08 · implies
Before
x\cdot 0=5\ \text{for some}\ xAfter
0=5,\ \text{a contradiction}Conditions: none
Rule: contradiction · Uses: G1, mul_zero
Leaf basis: mul_zero + contradiction + mul_inverse_nonzero
If '5/0' denoted some number x, then x would satisfy x·0=5. But x·0=0 for every x (multiplying by zero), so 0=5, a contradiction. So no number is 5/0; the pair (5,0) violates q≠0 (assumption a2) and represents no rational. By contrast 0/5=0·(1/5)=0 is a legitimate rational; the numerator may be 0 (N11).
S09 · implies
Before
-1.25
After
-\frac{125}{100}=-\frac{5}{4}\in\mathbb{Q}Conditions: 100\neq 0
Rule: exists_intro · Uses: S02, G4, decimal_def
Leaf basis: decimal_def + exists_intro + L_cross + numeric_evaluation
By the meaning of a finite decimal, −1.25=−(1+2/10+5/100)=−125/100 (N12). Taking p=−125 and q=100≠0, −1.25 is rational by definition. Reducing: (−125)·4=−500=(−5)·100, so by L_cross, −125/100=−5/4 (N13, N14). A finite decimal is an exact fraction, not a different kind of number.
S10 · implies
Before
n\in\mathbb{Z}After
n=\frac{n}{1}\in\mathbb{Q}Conditions: 1\neq 0
Rule: exists_intro · Uses: S01, mul_one
Leaf basis: exists_intro + mul_one + forall_intro
For any integer n: 1·1=1, so 1/1=1 and n/1=n·1=n. Taking p=n and q=1≠0, n∈ℚ by definition. In particular 7=7/1 is rational (N15). n was arbitrary, so every integer is rational.
S11 · equivalent
Before
(-3)\cdot3=-9<-8=(-2)\cdot4
After
\frac{-3}{4}<\frac{-2}{3}Conditions: 4>0; 3>0
Rule: L_cross_order · Uses: G4, L_cross_order
Leaf basis: L_cross_order + numeric_evaluation
Both denominators 4 and 3 are positive. By Lemma L_cross_order (p=−3, q=4, r=−2, s=3): ps=(−3)·3=−9 and rq=(−2)·4=−8, and −9<−8, so −3/4<−2/3 (N16, N17). The lemma is proved in S11.1; positive denominators are sufficient hypotheses used by this lemma, not a claim of logically minimal conditions, which is one benefit of the canonical form's positive denominator.
S12 · implies
Before
\text{S04, S06 to S11}After
\tfrac{6}{-8}=\tfrac{9}{-12}=\tfrac{-3}{4};\ \tfrac{2}{3}\neq\tfrac{3}{5};\ -1.25,\ 7\in\mathbb{Q};\ \tfrac{-3}{4}<\tfrac{-2}{3}Conditions: none
Rule: and_intro_elim · Uses: S04, S06, S07, S08, S09, S10, S11
Leaf basis: and_intro_elim
Three integer pairs represent the same rational, whose canonical form is −3/4; 2/3 and 3/5 differ; '5/0' denotes no number; −1.25=−5/4 and 7=7/1 are rational; and −3/4<−2/3.
Substeps and lemmas
S02.1 · implies
Before
q\neq 0,\ s\neq 0
After
qs\neq 0,\ \frac{p}{q}\cdot qs=ps,\ \frac{r}{s}\cdot qs=rqConditions: q\neq 0; s\neq 0
Rule: zero_product · Uses: G1, zero_product
Leaf basis: zero_product + mul_inverse_nonzero + mul_assoc + mul_comm + mul_one
By the zero-product property, qs=0 ⇔ q=0 or s=0; neither holds, so qs≠0. Also (p/q)·qs=p·(1/q)·q·s=p·1·s=ps (associativity, commutativity and q·(1/q)=1); likewise (r/s)·qs=r·(1/s)·s·q=rq.
S02.2 · equivalent
Before
\frac{p}{q}=\frac{r}{s}After
ps=rq
Conditions: qs\neq 0
Rule: eq_congruence · Uses: S02.1, eq_congruence
Leaf basis: eq_congruence + mul_inverse_nonzero + frac_rules
⇒: multiply both sides by qs; by S02.1 this gives ps=rq. ⇐: multiply both sides by 1/(qs), which exists since qs≠0; ps·(1/(qs))=p/q and rq·(1/(qs))=r/s, giving p/q=r/s. S02.1 and S02.2 together prove Lemma L_cross.
S11.1 · equivalent
Before
q>0,\ s>0,\ \frac{p}{q}<\frac{r}{s}After
ps<rq
Conditions: q>0; s>0
Rule: order_basics · Uses: S02.1, order_basics, mul_inverse_nonzero
Leaf basis: order_basics + mul_inverse_nonzero + frac_rules
q>0 and s>0 give qs>0 and also 1/(qs)>0. ⇒: multiplying both sides by the positive number qs keeps the direction, and the simplification in S02.1 gives ps<rq. ⇐: multiplying both sides by the positive number 1/(qs) gives p/q<r/s. This proves Lemma L_cross_order. If a denominator were negative, multiplying by qs could reverse the inequality, and the lemma would not apply.