Before diving headfirst into geometrical proofs, it's a good idea to revisit algebra. We've already learned how to solve equations for a variable. Now we'll do algebra in the format of the two-column proof.
Sample Problem
Show that if 3x – 7 = 5, then x = 4.
Here, our given statement is 3x – 7 = 5, and we're asked to prove x = 4.
Statements | Reasons |
1. 3x – 7 = 5 | Given |
2. 3x – 7 + 7 = 5 + 7 | Addition of 7 to equation (1) |
3. 3x + 0 = 5 + 7 | Substitution of –7 + 7 = 0 into (2) |
4. 3x = 5 + 7 | Substitution of 3x + 0 = 3x into (3) |
5. 3x = 12 | Substitution of 5 + 7 = 12 into (4) |
6. 3x⁄3 = 12⁄3 | Dividing equation (5) by 3 |
7. x = 12⁄3 | Substitution of 3x⁄3 = x into (6) |
8. x = 4 | Substitution of 12⁄3 = 4 into (7) |
Is there such a thing as being too descriptive? Yep, and that was it, since over half the proof was devoted to telling the reader how to do arithmetic. We'll typically take numerical computation for granted, and write proofs like this:
Statements | Reasons |
1. 3x – 7 = 5 | Given |
2. 3x = 12 | Add 7 to both sides of equation (1) |
3. x = 4 | Divide equation (2) by 3 |
See? That proof looks a lot like how we'd write it in algebra. The only difference is that you give reasons as you go, convincing the readers (like your math teacher) that you know what you're doing. You got this.
Sample Problem
Show that if 5(x + 12) = 30 and x + y = 100, then y = 106.
This time, our two given statements are 5(x + 12) = 30 and x + y = 100. We're supposed to prove y = 106. Here we go.
Statements | Reasons |
1. 5(x + 12) = 30 | Given |
2. x + y = 100 | Given |
3. 5x + 60 = 30 | Distributive property (1) |
4. 5x = -30 | Subtract 60 from both sides of (3) |
5. x = -6 | Divide both sides of (4) by 5 |
6. -6 + y = 100 | Substitute x = -6 into (2) |
7. y = 106 | Add 6 to both sides of (6) |
As you can see, there are lots of ways of phrasing your reasons. The important part is that you justify each step with why your statement is true. Of course, if your "reader" prefers it to be written a certain way, it's probably a good idea to follow his or her suggestions. Just saying.
Example 1
Write a proof for the statement "If 2x – 7 = 13, then x = 10." Which properties are used the proof? |
Example 2
Find the error in the following "proof" of the statement, "If 3 + z2 = 28, then z = 5." Then, find a counterexample to the statement itself.
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Example 3
Write a proof for the statement "If x3 + y = 10 and 3(3x – 3) = 9, then x = y." |
Exercise 1
If 4x + 12 = 0, then prove that x2 + 2 = 11 using an algebraic proof.
Statements | Reasons |
1. 4x + 12 = 0 | ? |
Exercise 2
If 4x + 12 = 0, then prove that x2 + 2 = 11 using an algebraic proof.
Statements | Reasons |
1. 4x + 12 = 0 | Given |
2. 4x = -12 | ? |
Exercise 3
If 4x + 12 = 0, then prove that x2 + 2 = 11 using an algebraic proof.
Statements | Reasons |
1. 4x + 12 = 0 | Given |
2. 4x = -12 | Subtract 12 from (1) |
3. x = -3 | ? |
Exercise 4
If 4x + 12 = 0, then prove that x2 + 2 = 11 using an algebraic proof.
Statements | Reasons |
1. 4x + 12 = 0 | Given |
2. 4x = -12 | Subtract 12 from (1) |
3. x = -3 | Divide (2) by 4 |
4. x2 = 9 | ? |
Exercise 5
If 4x + 12 = 0, then prove that x2 + 2 = 11 using an algebraic proof.
Statements | Reasons |
1. 4x + 12 = 0 | Given |
2. 4x = -12 | Subtract 12 from (1) |
3. x = -3 | Divide (2) by 4 |
4. x2 = 9 | Square (3) |
5. x2 + 2 = 11 | ? |
Exercise 6
Give the statements and the reasons that prove that if 5 – x2 = 1 and , then y = 28.
Statements | Reasons |
1. 5 – x2 = 1 | Given |
2. ![]() | Given |
3. ? | ? |
Exercise 7
Give the statements and the reasons that prove that if 5 – x2 = 1 and , then y = 28.
Statements | Reasons |
1. 5 – x2 = 1 | Given |
2. ![]() | Given |
3. -x2 = -4 | Subtract 5 from both sides of (1) |
4. ? | ? |
Exercise 8
Give the statements and the reasons that prove that if 5 – x2 = 1 and , then y = 28.
Statements | Reasons |
1. 5 – x2 = 1 | Given |
2. ![]() | Given |
3. -x2 = -4 | Subtract 5 from both sides of (1) |
4. x2 = 4 | Multiply (3) by -1 |
5. ? | ? |
Exercise 9
Give the statements and the reasons that prove that if 5 – x2 = 1 and , then y = 28.
Statements | Reasons |
1. 5 – x2 = 1 | Given |
2. ![]() | Given |
3. -x2 = -4 | Subtract 5 from both sides of (1) |
4. x2 = 4 | Multiply (3) by -1 |
5. ![]() | Substitute (4) into (2) |
6. ? | ? |