Assumptions
In our first example we only needed one chain of derivations, which proved that the first term was equal to the last term. The problem to be solved is sometimes more difficult, we might first need to come up with our strategy to solve the problem. We need to consider how to use all the information we get from the assignment.
In these cases we use a more general structure. We start with a description of the task. Then we make a list of all information we are given; the assumptions.
We denote the task by the \(\bullet\) sign (bullet) and we denote every assumption by either an id in parentheses (\((a), (b), ...; (ass1), (ass2), ...\)) or by the \(-\) sign (minus). When we have made the possible assumptions, we start the actual derivation, and we denote this by the \(\Vdash\) sign ("proof"). We conclude the derivation with the \(\square\) sign ("which was to be demonstrated"). This shows that the task has been successfully completed. As before, we use two columns for the derivation: In the first column we have our relation signs (in the previous example \(=\)) and other special signs (\(\bullet\), \(\Vdash\)). In the second column we have the terms and the motivations.
Let us consider the following assignment:
Assignment:
Step 1:
Structured derivation:
We start by describing the task and write task to structured derivation.
| \(\bullet\) | How much paint is needed to cover the floor? |
We then write down the given information we need to solve the problem. We know
- the floor is to be painted twice,
- the dimensions of the rooms
- the average paint consumption.
This information is included in the derivation as assumptions. In this way we have all the information in the same place.
| \(\bullet\) | How much paint is needed to cover the floor when |
| (a) | the floor is painted twice and |
| (b) | the living room is \(5m\) long and \(3m\) wide and |
| (c) | the kitchen is \(2m\) long and \(1m\) wide and |
| (d) | the average paint consumption is \(\frac{1l}{5m^{2}}\) |
After listing the assumptions we are ready for the solution. We use the \(\Vdash\) sign to mark the start of the solution (or proof) and write down what we want to solve. Usually the first step of solving a problem is to understand what needs to be calculated. We see from our assignment that the problem was to calculate "How much paint Pelle needs to buy when ...". In other words, we want to know how much paint is needed.
| \(\bullet\) | How much paint is needed to cover the floor when |
| (a) | the floor is painted twice and |
| (b) | the living room is \(5m\) long and \(3m\) wide and |
| (c) | the kitchen is \(2m\) long and \(1m\) wide and |
| (d) | the average paint consumption is \(\frac{1l}{5m^{2}}\) |
| \(\Vdash \) | The amount of paint needed |
Next we solve the problem step by step. How can we use this information to find out the needed amount of paint? The needed amount of paint can be calculated based on the area to be painted and the average paint consumption.
| \(\bullet\) | How much paint is needed to cover the floor when |
| (a) | the floor is painted twice and |
| (b) | the living room is \(5m\) long and \(3m\) wide and |
| (c) | the kitchen is \(2m\) in length and \(1m\) in width and |
| (d) | the average paint consumption is \(\frac{1l}{5m^{2}}\) |
| \(\Vdash\) | The amount of paint needed |
| \(=\) | {The amount of paint needed (\(l\) ) = 2 \(\cdot\) floor area (\(m^{2}\) ) . average paint consumption (\(\frac{l}{m^{2}}\) ), these are written down as assumptions (a), (d)} |
| \(2\cdot\) floor area \((m^{2})\cdot\frac{1l}{5m^{2}}\) |
The floor area can be calculated based on the dimensions of the floors.
| \(\bullet\) | How much paint is needed to cover the floor when |
| (a) | the floor is painted twice and |
| (b) | the living room is \(5m\) long and \(3m\) wide and |
| (c) | the kitchen is \(2m\) long and \(1m\) wide and |
| (d) | the average paint consumption is \(\frac{1l}{5m^{2}}\) |
| \(\Vdash\) | The amount of paint needed |
| \(=\) | {The amount of paint needed (\(l\) ) = 2 \(\cdot\) floor area (\(m^{2}\) ) . average paint consumption (\(\frac{l}{m^{2}}\) ), these are written as assumptions (a), (d)} |
| \(2\cdot\) floor area \( (m^{2})\cdot\frac{1l}{5m^{2}}\) | |
| \(=\) | {floor area = floor area of kitchen + floor area of living room. The formula for area is length \(\cdot\) width, we get the dimensions from assumptions (b), (c)} |
| \(2\cdot(5m\cdot3m+2m\cdot1m)\cdot\frac{1l}{5m^{2}}\) |
Now all we need to do is simplify the answer.
| \(\bullet\) | How much paint is needed to cover the floor when |
| (a) | the floor is painted twice and |
| (b) | the living room is \(5m\) long and \(3m\) wide and |
| (c) | the kitchen is \(2m\) long and \(1m\) wide and |
| (d) | the average paint consumption is \(\frac{1l}{5m^{2}}\) |
| \(\Vdash\) | The amount of paint needed |
| \(=\) | {The amount of paint needed (\(l\) ) = 2 \(\cdot\) floor area (\(m^{2}\) ) . average paint consumption (\(\frac{l}{m^{2}}\) ), these are written as assumptions (a), (d)} |
| \(2\cdot \) floor area \( (m^{2})\cdot\frac{1l}{5m^{2}}\) | |
| \(=\) | {floor area = floor area of kitchen + floor area of living room. The formula for area is length \(\cdot\) width, we get the dimensions from assumptions (b), (c)} |
| \(2\cdot(5m\cdot3m+2m\cdot1m)\cdot\frac{1l}{5m^{2}}\) | |
| \(=\) | {We do the calculations} |
| \(2\cdot17m^{2}\cdot\frac{1l}{5m^{2}}\) | |
| \(\approx\) | {We simplify \(m^{2}\) } |
| \(7l\) | |
| \(\square\) |
Instead of long variable names we can use letters. If we denote the needed amount of paint as \(x\) and the average paint consumption as \(k\), the derivation goes as follows:
| \(\bullet\) | How much paint (\(x\)) is needed to cover the floor when |
| (a) | the floor is painted twice and |
| (b) | the living room is \(5m\) long and \(3m\) wide and |
| (c) | the kitchen is \(2m\) long and \(1m\) in width and |
| (d) | the average paint consumption (\(k\)) is \(\frac{1l}{5m^{2}}\) |
| \(\Vdash\) | \(x\) |
| \(=\) | {\(x\) (\(l\)) = 2 \(\cdot\) floor area (\(m^{2}\) ) \(\cdot\) \(k\) (\(\frac{l}{m^{2}}\) ), these are written down as assumptions (a), (d)} |
| \(2\cdot \) floor area \( (m^{2})\cdot\frac{1l}{5m^{2}}\) | |
| \(=\) | {floor area = floor area of kitchen + floor area of living room. The formula for area is length \(\cdot\) width, we get the dimensions from assumptions (b), (c)} |
| \(2\cdot(5m\cdot3m+2m\cdot1m)\cdot\frac{1l}{5m^{2}}\) | |
| \(=\) | {We do the calculations} |
| \(2\cdot17m^{2}\cdot\frac{1l}{5m^{2}}\) | |
| \(\approx\) | {We simplify \(m^{2}\) } |
| \(7l\) | |
| \(\square\) |
