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# Transitive Property in Proofs

Reference > Mathematics > Introduction to Proofs

The transitive property states that if a = b and b = c, then a = c. This seems fairly obvious, but it's also very important. It's similar to the substitution property we looked at earlier, but not exactly the same. Here's an example of how we might use this property.

Prove: x2 + (a + b)x + ab = (x + a)(x + b)

Note that we don't have an "if - then" format, which is something new. Because of this, we don't have a starting equation that we can assume is true; the only equation we have is the one we are trying to prove, so we can't use that as a given. It would be nice if we had an equation we could start with as our first step, but the only way we can do that is to introduce a new variable and assign it a value. We'll use "variable assignment" as our reason.

```STATEMENT                                           REASON
1. Let y = (x + a)(x + b)                           Variable assignment
2. y = x(x + b) + a(x + b)                          Distributive property
3. y = x2 + bx + ax + ab                            Distributive property
4. y = x2 + (a + b)x + ab                           Distributive property
5. ∴ x2 + (a + b)x + ab = (x + a)(x + b)           Transitive property (1, 4)
```

You see how we did that? Since y = (x + a)(x + b), and y also equals x2 + (a + b)x + ab, then those two quantities must be equal to each other! That's a nice result, and I think we might make use of it later, so I'm going to give it a name, so we can use it as a reason in another proof:

Sum and Product Theorem: x2 + (a + b)x + ab = (x + a)(x + b)

This means that if we see x2 + (a + b)x + ab, we can replace it with  (x + a)(x + b), and if we see  (x + a)(x + b), we can replace it with x2 + (a + b)x + ab. That's a very handy theorem, as we'll see below:

Prove: If x2 + 3x + 2 = 0, then x = -1 or x = -2.

Note: for this proof we're going to use another algebraic property - the "Zero Property of Multiplication," which says that if two or more terms multiply to make zero, one or more of those terms must be zero.

```STATEMENT                                     REASON
1. x2 + 3x + 2 = 0                            Given
2. x2 + (2 + 1)x + 2(1) = 0                   Arithmetic
3. (x + 2)(x + 1) = 0                         Sum and Product Theorem
4. x + 2 = 0 or x + 1 = 0                     Zero Property of Multiplication
5. x = -2 or x = -1                           Mr. T's Linear Equation Theorem```

Here's a more complex problem:

Prove: if (x - 1)(x + 2) = 4 then x = 2 or x = -3

```STATEMENT                                     REASON
1. (x - 1)(x + 2) = 4                         Given
2. x2 + x - 2 = 4                             Sum and Product Theorem
3. x2 + x - 2 - 4 = 4 - 4                     Subtraction property of equality
4. x2 + x - 6 = 0                             Combining like terms
5. (x + 3)(x - 2) = 0                         Sum and Product Theorem
6. x + 3 = 0 or x - 2 = 0                     Zero property of multiplication
7. x = -3 or x = 2                            Mr. T's Linear Equation Theorem
```

As you consider that last problem, you should note that we used two theorems, and those two theorems, combined, saved us about ten extra steps, because without them, we would have to include all the steps in the proof of my Linear Equation Theorem (TWICE!) plus all the steps contained in the proof of the Sum and Product Theorem.

## Questions

1.
Prove: x2 + 2ax + a2 = (x + a)2
2.
Prove: a2 - b2 = (a - b)(a + b)
3.
Prove: a3 - b3 = (a - b)(a2 + ab + b2)
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