Uniqueness Proof Technique
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Uniqueness Proof Technique

Suppose If A then B is a proposition involving statements A and B. The existence of B does not necessarily establish the uniqueness of B. The Uniqueness Proof Technique can be used to establish the uniqueness of B.

Prove, by the direct uniqueness method, that if a, b, c, d, e, and f are real numbers such that (ad - bc) ≠ 0, then there are unique real numbers x and y such that (ax + by) = e and (cx + dy) = f

The strings: S7P2A21 (Identity - Physical Property).

The math:
Pj Problem of Interest is of type identity (physical property). Proofs establish truths. So they are identity problems.

The conceptual steps used in the Uniqueness Proof Technique is illustrated in figure 121.4.
Existence is usually explicitly indicated with the quantifiers: there is, there are, for all, for each, etc. These quantifiers can be assigned to two main groups:existential quantifiers(e.g. there is, there are) and universal quantifiers (e.g. for all).

Establish the existence of x and y in the given problem:
The existential quantifier, "there are" is implicit in the problem. In order words, we can restate B as:

There are real numbers x and y such that (ax + by) = e and (cx + dy) = f.

If we multiply (ax + by) = e by d, we have:
d(ax + by) = de ------(1)
If we multiply (cx + dy) by b, we have:
b(cx + dy) = bf ------(2)
subtracting equation (2) from (1), we have:
(dax + dby - bcx -bdy) = de - bf--------(3)
So, (da - bc)x + (db - bd)y = de - bf
So, x = (de - bf)/(da - bc) ------------(4)
Similarly, if we multiply (ax + by) = e by c and (cx + dy) by a, we have:
y = (af - ce)/(ad - bc) ----------------(5)

The values of x and y exist for the given problem if they satisfy:
(ax + by) = e and (cx + dy)
[a(de-bf)/(da-bc) + b(af-ce)/(ad-bc)] = [ade -abf + baf - bce]/(da-bc)
= e(ad-bc)/(da-bc) = e.
[c(de-bf)/(da-bc) + d(af-ce)/(ad-bc)] = [cde -cbf + daf - dce]/(da-bc)
= f(ad-bc)/(da-bc) = f.
So, existence of x and y established.

Establish the uniqueness of x and y in the given problem:
Assume two different x values (x1, x2), and two different y values(y1, y2) then:

ax1 + by1 = e-------(1)
cx1 + dy1 = f-------(2)
ax2 + by2 = e-------(3)
cx2 + dy2 = f-------(4)

equation(1) - Equation(3) gives:
[a(x1-x2) + b(y1-y2)] = 0 -------(5)

equation(2) - Equation(4) gives:
[c(x1-x2) + d(y1-y2)] = 0---------(6)

Multiplying equation(5) by d and equation (6) by b, then subtracting (6) from (5) gives:
[(ad - bc)(x1-x2)] = 0.
So, (x1-x2) = 0 since (ad - bc) ≠ 0.
So, x1 = x2
Similarly, y1 = y2
So, uniqueness is proved.

Blessed are they that have not seen, and yet have believed. John 20:29

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