PROOF 73

Prove.

Shifting the decimal point multiplies by 10 or by 1/10

On place-value notation.

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.

Definition D1

The place-value notation aₘ…a₁a₀.a₋₁…a₋ₙ means aₘ·10ᵐ + … + a₀·10⁰ + a₋₁·10⁻¹ + … + a₋ₙ·10⁻ⁿ.

Proof.

(1)Each aₖ is an integer from 0 to 9, and k is the position that digit occupies.convention

(2)10⁰ = 1 and 10¹ = 10.convention

(3)For positions with k < 0, 10ᵏ is the multiplicative inverse of 10⁻ᵏ, and A2 guarantees its existence.A2

∎

Definition D2

When b has a multiplicative inverse, a/b means a · b⁻¹.

Proof.

What multiplying by 1/10 means is fixed by this definition.convention

∎

Lemma L1

10 × 10⁻¹ = 1. That is, 10⁻¹ is the multiplicative inverse of 10.

Proof.

(1)10 × 10⁻¹ = 10¹ · 10⁻¹D1

(2)　　　　 = 10⁰A1

(3)　　　　 = 1(2), D1

∎

Theorem T1

10x = aₘ·10ᵐ⁺¹ + … + a₋ₙ·10⁻ⁿ⁺¹. Every digit moves up one position.

Proof.

(1)x = Σ aₖ·10ᵏD1

(2)10x = Σ 10(aₖ·10ᵏ)(1), A3, E

(3)10(aₖ·10ᵏ) = aₖ(10 · 10ᵏ)A4, A5

(4)10 · 10ᵏ = 10¹ · 10ᵏ = 10ᵏ⁺¹D1, A1

(5)10x = Σ aₖ·10ᵏ⁺¹(2), (3), (4), E

(6)The digit aₖ that was at position k goes to position k+1. The order of the digits is unchanged.(5), D1

∎

Theorem T2

10⁻¹x = aₘ·10ᵐ⁻¹ + … + a₋ₙ·10⁻ⁿ⁻¹. Every digit moves down one position.

Proof.

(1)10⁻¹x = Σ 10⁻¹(aₖ·10ᵏ)D1, A3, E

(2)10⁻¹(aₖ·10ᵏ) = aₖ(10⁻¹ · 10ᵏ)A4, A5

(3)10⁻¹ · 10ᵏ = 10ᵏ⁻¹A1

(4)10⁻¹x = Σ aₖ·10ᵏ⁻¹(1), (2), (3), E

(5)The digit aₖ that was at position k goes to position k−1.(4), D1

∎

Corollary C

The number with the decimal point moved one place to the right is 10x, and the one with it moved one place to the left is 1/10 of x.

Proof.

(1)The notation with every digit moved up one position is the notation with the decimal point moved one place to the right, and its value is 10x.T1

(2)1/10 = 1 × 10⁻¹ = 10⁻¹D2, L1, A6

(3)The value of the notation with every digit moved down one position is 10⁻¹x, that is, 1/10 of x.T2, (2)

(4)For j places, the same argument gives 10ʲ and 10⁻ʲ.T1, T2, A1

∎

Remark

Can A1 be removed from this proof?

Check whether moving the decimal point is a rule of notation or a property of exponents.