PROOF 116

Prove.

The graph of inverse proportion never touches the axes

On the coordinate plane.

A way to read this proof

First separate what is assumed from what is to be proved. At each step, ask which definition or earlier result permits the move. If it is unclear, return to that premise rather than skipping to the conclusion.

What you may use

A1∀r, there is n ∈ ℕ with r < n
Archimedean property
A2n ∈ ℕ ⟹ 0 < n
natural numbers are positive
A3(ab)c = a(bc)
associativity of multiplication
A4ab = ba
commutativity of multiplication
A5a × 1 = a
multiplicative identity
A6if a ≠ 0, then a·a⁻¹ = 1
existence of multiplicative inverses
A70 < c, a < b ⟹ ca < cb
multiplication and order
A8|ab| = |a||b|
absolute value and multiplication
A9a ≠ 0 ⟺ 0 < |a|
absolute value and 0
A100 < a ⟹ |a| = a
absolute value and positives
A110 < a ⟹ 0 < a⁻¹
inverses and order
Pa × 0 = 0
obtained in Proof 79 — cited only

D1G = {(x,y) : xy = k}
definition of the inverse-proportion graph (k ≠ 0)
D2x-axis is y=0, y-axis is x=0
definition of the coordinate axes
T1no point of the axes is in G
Theorem 1
T2∀M>0, there is n with |kn|>M
Theorem 2

Nothing outside this list may be cited.

Definition D1

For k ≠ 0, the inverse-proportion graph G is the set of all points (x, y) satisfying xy = k.

Proof.

A graph is not a picture but a set of points. Whether a point belongs to G is decided by xy = k alone.convention

∎

Definition D2

The x-axis is the set of all points with y = 0, and the y-axis is the set of all points with x = 0.

Proof.

An axis is also a set of points. 'Touching' means the two sets share a common point.convention

∎

Theorem T1

G contains no point of the x-axis and no point of the y-axis.

Proof.

(1)Suppose there is a y with (0, y) ∈ G.assumption

(2)0 × y = k(1), D1

(3)0 × y = y × 0 = 0A4, P

(4)k = 0(2), (3)

(5)This contradicts k ≠ 0. Hence no point with x = 0 is in G.(4), D1

(6)If (x, 0) ∈ G, then x × 0 = k, and in the same way k = 0, a contradiction.D1, P

(7)Hence G shares no common point with either axis.(5), (6), D2

∎

Theorem T2

For any M > 0, there is a natural number n such that the point (n⁻¹, kn) belongs to G and |kn| > M.

Proof.

(1)Suppose M > 0.assumption

(2)Since k ≠ 0, we have 0 < |k|, and |k|⁻¹ exists with 0 < |k|⁻¹.D1, A9, A11

(3)There is a natural number n with M|k|⁻¹ < n.(1), (2), A1

(4)Since 0 < n, n⁻¹ exists. Let x = n⁻¹.(3), A2, A11

(5)Letting y = kn, we get xy = n⁻¹(kn) = k(n⁻¹n) = k × 1 = kA3, A4, A6, A5

(6)Hence (x, y) ∈ G.(4), (5), D1

(7)|y| = |kn| = |k||n| = |k|nA8, A2, A10

(8)Since 0 < |k| and M|k|⁻¹ < n, we get |k|(M|k|⁻¹) < |k|n(2), (3), A7

(9)|k|(M|k|⁻¹) = M(|k||k|⁻¹) = M × 1 = MA3, A4, A6, A5

(10)M < |k|n = |y|(7), (8), (9)

∎

Corollary C

The point given by T2 satisfies |x| < |k|M⁻¹. M was arbitrary, so among points of the form x = n⁻¹ there are those with |x| as small as desired, and yet none with |x| = 0.

Proof.

(1)Take the point (x, y) = (n⁻¹, kn) given by T2. Then |y| > M.T2

(2)Since x ≠ 0, we have 0 < |x|.T1, A9

(3)|x|M < |x||y|(1), (2), A7

(4)|x||y| = |xy| = |k|A8, D1

(5)|x|M < |k|(3), (4)

(6)Since 0 < M, M⁻¹ exists and 0 < M⁻¹.(1), A11

(7)M⁻¹(|x|M) = |x|(MM⁻¹) = |x| × 1 = |x|A3, A4, A6, A5

(8)|x| < M⁻¹|k| = |k|M⁻¹(5), (6), (7), A7, A4

(9)M was arbitrary, so among points of the form x = n⁻¹ there are those with |x| as small as desired, and by T1 there is no point with |x| = 0.T2, (8), T1

∎

Remark

Can A1 be removed from this proof?

T1 says only that no point lies on the axes. Check what the lines saying 'approaches without bound' rest on.