Goal and scope
In a fixed Cartesian coordinate system, let A=(−3,2), B=(2,−3), B′=(6/3,−3), C=(0,−4), O=(0,0) and E=(−1/2,−5). First prove that the axes meet only at the origin and that every point either lies on an axis or lies in exactly one quadrant; then decide for each point its axis or quadrant, and decide whether A and B, and B and B′, are the same point.
Uses only the MKO's definitions: a fixed origin, perpendicular oriented axes and a unit on each, coordinates as ordered pairs of reals, the x-axis given by y=0 and the y-axis by x=0, and the quadrants off the axes given by the signs of the two coordinates. It infers no distance, angle or figure size (MKO assumption a2) and reads no coordinates from a sketch; the canonical MKO and its English edition are not changed.
Main derivation
S01 · definition
Before
\text{a point in the fixed coordinate system}After
\text{point}\leftrightarrow(x,y)\in\mathbb{R}^2,\quad (x,y)=(x',y')\iff x=x'\land y=y'Conditions: none
Rule: mko_coord_plane · Uses: G1
Leaf basis: mko_coord_plane + cartesian_product_def
First fix the coordinate system (G1); from then on a point is represented by an ordered pair of reals, the first coordinate read along the x-axis and the second along the y-axis. By the equality condition for ordered pairs, two points are the same exactly when both corresponding coordinates are equal; this is the comparison described in MKO para2, and it is meaningful only within one coordinate system.
S02 · definition
Before
\text{axes and quadrants}After
\mathcal{X}=\{y=0\},\ \mathcal{Y}=\{x=0\},\ Q_{\mathrm{I}}=\{x>0,y>0\},\ Q_{\mathrm{II}}=\{x<0,y>0\},\ Q_{\mathrm{III}}=\{x<0,y<0\},\ Q_{\mathrm{IV}}=\{x>0,y<0\}Conditions: none
Rule: def_intro · Uses: G2, S01
Leaf basis: def_intro + mko_quadrant_signs + set_builder
Following the MKO, write the two axes and four quadrants as sets. Note that the quadrants use strict inequalities: Q_I needs x>0 and y>0, not x≥0 and y≥0; this is exactly why points on an axis are excluded from the quadrants.
S03 · implies
Before
(x,y)\in\mathbb{R}^2After
(x=0\lor y=0\Rightarrow\text{no quadrant})\ \land\ (x\neq 0\land y\neq 0\Rightarrow\text{exactly one quadrant})Conditions: none
Rule: L_classify · Uses: S02, order_basics
Leaf basis: L_classify + order_basics + case_analysis
By Lemma L_classify, every point of the plane is classified exactly once by the union of the two axes and the four quadrants: points on an axis are in no quadrant, and points off the axes are in exactly one quadrant. The proof is S03.1 (on an axis), S03.2 (off the axes, at least one) and S03.3 (at most one).
S04 · implies
Before
\mathcal{X}\cap\mathcal{Y}After
\{(0,0)\}Conditions: none
Rule: set_extensionality · Uses: S02
Leaf basis: set_extensionality + set_operations_def
Prove the MKO's statement that the axes meet at (0,0), and only there. The two inclusions are S04.1 and S04.2.
S05 · implies
Before
A=(-3,2)
After
A\in Q_{\mathrm{II}},\ A\notin\mathcal{X}\cup\mathcal{Y}Conditions: -3<0; 2>0
Rule: mko_quadrant_signs · Uses: G3, S02, S03
Leaf basis: mko_quadrant_signs + order_basics + numeric_evaluation
The first coordinate of A is −3<0 and the second is 2>0, sign pattern (−,+), which is the definition of Q_II. Since −3≠0, A is not on the y-axis; since 2≠0, A is not on the x-axis. By L_classify, A is in no other quadrant.
S06 · implies
Before
B=(2,-3)
After
B\in Q_{\mathrm{IV}},\ B\notin\mathcal{X}\cup\mathcal{Y}Conditions: 2>0; -3<0
Rule: mko_quadrant_signs · Uses: G3, S02, S03
Leaf basis: mko_quadrant_signs + order_basics + numeric_evaluation
The first coordinate of B is 2>0 and the second is −3<0, sign pattern (+,−), so B∈Q_IV. Both coordinates are nonzero, so B is on neither axis.
S07 · implies
Before
(-3,2)\ \text{and}\ (2,-3)After
A\neq B,\quad\text{while}\quad\{-3,2\}=\{2,-3\}Conditions: none
Rule: cartesian_product_def · Uses: S01, S05, S06
Leaf basis: cartesian_product_def + set_extensionality + numeric_evaluation
Equal ordered pairs need equal first coordinates, and −3≠2, so A≠B. By contrast the unordered sets {−3,2} and {2,−3} have the same elements and are equal by set_extensionality. Swapping the coordinates changed the point and even its quadrant (Q_II became Q_IV), so a point cannot be treated as an unordered set.
S08 · equivalent
Before
B'=\left(\tfrac{6}{3},-3\right)After
B'=(2,-3)=B
Conditions: none
Rule: cartesian_product_def · Uses: G3, S01, S06
Leaf basis: cartesian_product_def + numeric_evaluation + eq_replace
6/3=2 and −3=−3, so both corresponding coordinates are equal and B′=B by the equality condition for ordered pairs; hence B′ is also in Q_IV. Whether points coincide is decided by the coordinate values, not by how they are written.
S09 · implies
Before
C=(0,-4)
After
C\in\mathcal{Y},\ C\notin\mathcal{X},\ C\ \text{in no quadrant}Conditions: 0=0; -4\neq 0
Rule: L_classify · Uses: G3, S02, S03
Leaf basis: L_classify + set_builder + numeric_evaluation
The first coordinate of C is 0, so C∈𝒴; the second coordinate is −4≠0, so C∉𝒳. By L_classify (the case x=0), C lies in no quadrant, even though its second coordinate is negative.
S10 · implies
Before
O=(0,0)
After
O\in\mathcal{X}\cap\mathcal{Y},\ O\ \text{in no quadrant}Conditions: none
Rule: L_classify · Uses: G3, S03, S04
Leaf basis: L_classify + set_operations_def
By S04, O lies on both axes; by L_classify (x=0), O lies in no quadrant. The origin is the only common point of the axes, not a corner of the first quadrant.
S11 · implies
Before
E=\left(-\tfrac{1}{2},-5\right)After
E\in Q_{\mathrm{III}},\ E\notin\mathcal{X}\cup\mathcal{Y}Conditions: -\tfrac12<0; -5<0
Rule: mko_quadrant_signs · Uses: G3, S02, S03
Leaf basis: mko_quadrant_signs + order_basics + numeric_evaluation
The coordinates of E are −1/2<0 and −5<0, sign pattern (−,−), so E∈Q_III. The coordinates need not be integers; only their signs matter. Both are nonzero, so E is on no axis.
S12 · implies
Before
A,B,B',C,O,E
After
A\in Q_{\mathrm{II}},\ B=B'\in Q_{\mathrm{IV}},\ E\in Q_{\mathrm{III}},\ C\in\mathcal{Y}\setminus\mathcal{X},\ O\in\mathcal{X}\cap\mathcal{Y},\ A\neq BConditions: none
Rule: and_intro_elim · Uses: S05, S06, S07, S08, S09, S10, S11
Leaf basis: and_intro_elim
Collect the decisions: A is in the second quadrant, B and B′ are the same point in the fourth quadrant, E is in the third quadrant, C is on the y-axis but not the x-axis, and O is on both axes; C and O are in no quadrant; A≠B. Every conclusion follows from the coordinate values and the definitions of S02 alone.
Substeps and lemmas
S03.1 · implies
Before
x=0\ \lor\ y=0
After
(x,y)\notin Q_{\mathrm{I}}\cup Q_{\mathrm{II}}\cup Q_{\mathrm{III}}\cup Q_{\mathrm{IV}}Conditions: none
Rule: order_basics · Uses: S02, order_basics
Leaf basis: order_basics + set_operations_def
Every quadrant requires x>0 or x<0, and also y>0 or y<0. By trichotomy, x=0 makes both x>0 and x<0 false, and y=0 makes both y>0 and y<0 false. So whenever x=0 or y=0, none of the four quadrant conditions holds and the point lies in no quadrant.
S03.2 · case_split
Before
x\neq 0\ \land\ y\neq 0
After
(x,y)\in Q_{\mathrm{I}}\cup Q_{\mathrm{II}}\cup Q_{\mathrm{III}}\cup Q_{\mathrm{IV}}Conditions: x\neq 0; y\neq 0
Rule: case_analysis · Uses: S02, order_basics
Leaf basis: case_analysis + order_basics
By trichotomy, x≠0 gives x>0 or x<0, and y≠0 gives y>0 or y<0. The four combinations (+,+), (−,+), (−,−) and (+,−) cover every possibility and are exactly the defining conditions of Q_I, Q_II, Q_III and Q_IV, so the point lies in at least one quadrant.
S03.3 · implies
Before
(x,y)\in Q_i\cap Q_j,\ i\neq j
After
\bot
Conditions: none
Rule: contradiction · Uses: S02, order_basics
Leaf basis: contradiction + order_basics
Any two different quadrants demand opposite signs of x or of y: Q_I and Q_II, and Q_III and Q_IV, differ in x; Q_I and Q_IV, and Q_II and Q_III, differ in y; Q_I and Q_III, and Q_II and Q_IV, differ in both. A point in two different quadrants would need, for example, both x>0 and x<0, contradicting trichotomy. So a point lies in at most one quadrant, and with S03.2 in exactly one. S03.1 to S03.3 together prove Lemma L_classify.
S04.1 · implies
Before
(x,y)\in\mathcal{X}\cap\mathcal{Y}After
(x,y)=(0,0)
Conditions: none
Rule: set_operations_def · Uses: S02, set_operations_def
Leaf basis: set_operations_def + cartesian_product_def
(x,y)∈𝒳∩𝒴 means (x,y)∈𝒳 and (x,y)∈𝒴, that is y=0 and x=0; by the equality condition for ordered pairs, (x,y)=(0,0).
S04.2 · implies
Before
(0,0)
After
(0,0)\in\mathcal{X}\cap\mathcal{Y}Conditions: none
Rule: set_operations_def · Uses: S02, S04.1
Leaf basis: set_builder + set_operations_def + set_extensionality
The second coordinate of (0,0) is 0, so it lies in 𝒳; the first coordinate is 0, so it lies in 𝒴; hence it lies in the intersection. S04.1 and S04.2 together give 𝒳∩𝒴={(0,0)} by set_extensionality.
Error controls and negative controls
Points on an axis are in no quadrant
The first error named in the MKO is to put points on an axis, or the origin, into a quadrant. The quadrants are defined with strict inequalities (S02), so C=(0,−4), although its second coordinate is negative, is in neither the third nor the fourth quadrant, and the origin is not a point of the first quadrant (S03.1, S09, S10).
Ordered pairs do not commute
The second error named in the MKO is to treat an ordered pair as an unordered set. {−3,2}={2,−3}, but (−3,2)≠(2,−3): the first is in the second quadrant and the second in the fourth (S07). Whether two points coincide is decided by comparing corresponding coordinates one by one.
Coordinates are data, not read from a sketch
B′=(6/3,−3) is written differently from B but is the same point (S08), and E with first coordinate −1/2 might look close to the y-axis in a rough sketch. Decisions rest only on the stated coordinate values and the definitions; this guide also infers no distance between points and no figure size (MKO assumption a2).
- The origin (0,0) is in the first quadrant because both its coordinates are ≥0 →
reject — Q_I requires x>0 and y>0; 0>0 is false, so the origin is in no quadrant. - (−3,2)=(2,−3) because both use the same two numbers →
reject — Equal ordered pairs need equal first coordinates, and −3≠2; the two points are even in different quadrants. - C=(0,−4) has a negative second coordinate, so C is in the third or fourth quadrant →
reject — The third and fourth quadrants both require x≠0 (x<0 or x>0), while C has x=0, so C is on the y-axis. - B′=(6/3,−3) and B=(2,−3) are written differently, so they are different points →
reject — 6/3=2, so both corresponding coordinates agree and B′=B.