Mathematical reasoning

Mathematical reasoning isn’t explicitly taught the same way that addition or multiplication is taught. Mathematical reasoning develops after plenty of experience using numbers, quantity, numerical relationships and problem solving. Mathematical reasoning allows people to solve a math problem without algorithms, or a set process. Classroom is the only place where math looks like a bunch of problems on a sheet of paper. In life, and in careers, math springs up in context with bigger issues. Mathematical reasoning is a critical skill in these situations.

In a few words, we can say that mathematical reasoning is about various data representation, its interpretation and use. We’ll explain how to solve problems using simple math.

In every aspect of life we are dealing with massive amount of information and data. We always strive to read these information as easily as we can. That is why we are labeling them, drawing all kinds of graphs or pictures that show us everything we need to know without wasting any time on searching for information in messy lists.

The most popular way of interpretation data is the bar chart. It consists of two axes with values of elements we are observing.

For example we’ll make a simple chart. For the start we’ll need some kind of information and values in areas we are interested in.

Let’s say every day you take apples to school. On Monday you take 1 apple, on Tuesday you take 2 apples, on Wednesday you take 5 apples, on Thursday you take 3 apple and on Friday you take 4 apples. And you want to make a chart of it.

First you draw two perpendicular lines like as shown in a picture below. You want to observe the number of apples a day you took through 5 days. You have 5 days, which means that you’ll draw 5 segments representing your days onto horizontal axis. On the vertical axis you’ll also draw 5 segments because 5 is the maximum number of apples you took.

Then you should get something like this: Let’s see how can we mark  how many apples you took on each day using rectangles.

On Monday you took 1 apple. This means that 1 will be the height of your rectangle. On Tuesday you took 2 apples, so the height of the rectangle is 2: We can continue this process until we are done. This is the final result: From this chart you can easily read out the information you need, no matter how complex and messy it might sometimes be.

Next thing is to show how to make a pie chart. A pie chart or pie graph is a circular chart divided into wedge-like sectors, illustrating proportion. Each wedge represents a proportionate part of the whole, and the total value of the pie is always 100 percent. It is named for its resemblance to a pie which has been sliced.

For example if you want to draw a pie chart of the things you do in your free time, and you know that 25% of the time you read, 25% you play with your friends and 50% you play video games, this is how you would do it. One circle is one whole, you know that 25% is $\frac{1}{4}$, and 50% is $\frac{2}{4}$ which means that you’ll divide a circle into four parts. One part will be for reading, one for playing with friends and two for playing video games (two because  50% = $\ 2 \cdot \frac{1}{4}$).

Also, a pie chart should always have a legend beside it. A legend is a list of your information and matching colors so you can easily read your pie chart. From every of these charts you can also retrieve information.

There are many more charts that are used.
The most common chart is known as the line chart. If we imagine our information and its values as points in a plane and connect them with a line. We get a function – like chart.

For example, let’s say you are selling chocolate bars to your neighbors. This would be the line chart of your sales: What is the name of the month in which you have sold the most cookies? You simply look at the growth of the function, if your sales grows, so does your function. This means that all you have to do is find the highest point which would be in December where you sold 15 cookies. The lowest point is in October where you sold only one cookie.

Constructing numerical expressions

In this lesson you’ll learn how to transform simple math problems from words into numerical expressions.

First we’ll start with the easy one. What is the quotient of ten minus two and two?

This, written as a fraction is:

$\frac{(10-2)}{2} = \frac{8}{2} = 4$

Second example:

What is the quotient of 10 times 5 – 2 and 3?

$\ 10 \cdot \frac{(5 – 2)}{3} = 10$
Now let’s get to the simple word problems.
With these kind of tasks it is very important to carefully read what you have and what you have to get. Be sure to read the task a couple of times before writing anything down.
Second thing you do is to write information you have one by one, and be sure to mark them appropriately.
Third and last is to simply solve your math problem.

Example 1.
Martha loves chestnuts. During the fall she was walking down the street every day to collect them. On the first day she collected 5 chestnuts. On the second day she collected 3, but on her way home lost 2 of them. Her father came and gave her exactly the amount of chestnuts she already had. How many does she have now?

1. day – collected 5
2. day – collected 3, lost 2

Before the day three she had $5 + 3 – 2 = 6$. This means that her father gave her 6 more, which means that she has $6+6=12$ chestnuts.

How would you write this problem in a single mathematical expression?

number of chestnuts = $\ (5 + 3 – 2) \cdot 2 = 12$

Example 2.
Pier was a famous painter and had a display of his paintings. He had only one week to prepare 8 paintings. By Tuesday he had already prepared 2 paintings, but he knew that Wednesday and Thursday are the days when he could not paint. How many paintings minimum must he need to paint on Friday, Saturday and Sunday?

Solution:
One week has 7 days. If we know that he can’t work on Wednesday and Thursday, that means that he has only 5 days to prepare paintings. If he has finished two paintings by Tuesday, this means that:

8 – 2 = 6 – he is left with 6 paintings more to prepare
7 – 4 = 3 – and he has 3 days to do it
$\frac{6}{3}=2$
This means that he has to paint two paintings a day minimum.
How would you write this problem in one expression? Multi-step word problems

Multi-step word problems usually include few parentheses.
You solve them step by step. First step is to write down all information you can extract from the task. This will help you to easily see what to do and to realize how to do it.

Example 3.
Ella and Martin are going to buy some ice cream for their families and themselves. One ice cream costs 2 dollars. Ella will buy 5 ice creams, and Martin will buy 4 ice creams. How much more money did Ella spend than Martin?

1. Ice cream = 2 dollars
2. Ella bought 5 ice creams -> she spent $\ 5 \cdot 2$ dollars = 10 dollars
3. Martin bought 4 ice creams -> he spent $\ 4 \cdot 2$ dollars = 8 dollars
4. the difference in the amount of money that Ella spent and the amount of money that Martin spent-> 10 dollars – 8 dollars = 2 dollars

The formula: $\ 5 \cdot 2 – 4 \cdot 2 = 10 – 8 = 2$
Or if you already know that all ice creams cost 2 dollars this problem can also be written as: $\ 2 \cdot ( 5 – 4) = 2$

Next thing that may occur in your problem solving are inequalities. If you are not sure about what it is or how it is solved you can look it up in inequalities lesson.

Example 4.
Maria and Peter are competing in gathering postcards. Maria has 2 postcards more than Peter. If Maria is getting two postcards a week and Peter is getting three postcards a week, will Maria still have more postcards than Peter in three weeks?

Let’s mark Peter’s postcards with a ‘P’, and Maria’s with an ‘M’.
M = P + 2 (Maria has two more postcards than Peter)

After three weeks they will have this much postcards:
$\ M = P + 2 + 3 \cdot 2 = P + 8$ (she receives two postcards for three weeks)
$\ P = P + 3 \cdot 3 = P+ 9$ ( P is the start value, and he receives three postcards for three weeks)
$\ P + 8 < P + 9$ this means that Peter will have one postcard more.

In an algebraic expression:
‘O’ will represent our missing inequality sign.

Maria O Peter

The start values

$\ P + 2 > P$

After three weeks

$\ P + 2 + 3 \cdot 2 O P + 3 \cdot 3$
$\ P + 8 O P + 9$
$\ P < P + 1$
$\ M < P$

Cross topic arithmetic

Word problems can also be given by ratios, rational numbers and percentages. They are solved in the same way you’d solve any other word problem, you just have to be careful of what you take percentage of and what you add it with. It isn’t hard but can be messy.

To remind yourself how to calculate with percentages, read lesson about percents.

Example 5.
You found two cubes. One weighs 4 pounds and second one weighs 9 pounds. What is the ratio between their weights?

This is one of the fundamental tasks. You simply write it in a fraction and shorten the fraction if possible.

$\frac{(weight1)}{(weight2)} = \frac{4}{9}$

The ratio of their weights is 4 : 9.

Example 6.
If Anna bought 25 chocolate bars and Michael came and ate 10% of them, how many does she have left?

The starting value is 25. This is how many chocolate bars Anna had in the beginning.

Now we want to know how many of them Michael ate: 10% from 25 we calculate as

$\ 25 = \frac{10}{100} \cdot 25 = 2.5$

Michael ate 2.5 bars. Since Anna had 25 chocolate bars, now she has $\ 25 – 2.5 = 22.5$ chocolate bars.

Example 7. (slightly more complicated task)
Martha had a big bowl of strawberries. This bowl contained 50 strawberries. Firstly she ate 14% of them. After one hour she came back and ate 10% of the remaining strawberries. If she had to leave 50% from the whole bowl, how many more strawberries is she allowed to eat?

The starting value is 50 strawberries.

If she ate 14% of them, she ate: $\frac{14}{100} \cdot 50 = 7$ which means she had $\ 50 – 7 = 43$ strawberries left.

When she came back she ate 10% more, but now the value you take the 10% of is 43 because she already ate the first 14% of strawberries.

This means that she ate 10% from 43:

$\frac{10}{100} \cdot 43 = 4.3$.

Now she has $\ 43 – 4.3 = 38.7$ strawberries left.

Now we have to see how many of them she can’t eat. If she can’t eat 50% of the starting value, she has to leave 25 strawberries. ($\frac{50}{100} \cdot 50=25$)

She has 38.7 and has to leave 25 -> she can eat $\ 38.7 – 25 = 13.7$ strawberries.

Number patterns

Number pattern is a property of a continuous sequence of numbers which helps us anticipate how a certain sequence will behave after a lot of steps.

Here we’ll be observing a few simple patterns that you may encounter.

First are the repeating numerals.

These patterns are easy to recognize and continue the pattern.

For example:
The pattern with only one or more repeating numerals: Task 1. Continue the sequence with two more numerals: 8 7 6 5 8 7 6 8
If we examine this sequence we’ll see that the numbers 8, 7, 6 are repeating. This means that the next two numerals in the given sequence will be 7 and 6.

Second are patterns that involve addition.

This sequence is made by simply adding one fixed number onto another. How will you recognize this pattern? You’ll simply subtract two adjacent numbers, the left one from the right one, and do this few times and see if you always get the same number. In case you do get the same number, this will be the number you’ll always add onto your start value. This number is called their common difference.

You are familiar with the following sequence:

$\ 1, 2, 3, 4, 5, 6, 7…$

This is a sequence that is made by the start value 1 and with constantly adding number 1.
The start value is always the first number in your sequence, and the number you’re constantly adding is
$\ 2 – 1 = 1; 3 – 2 =1; 4 – 3 = 1$ and so on, this means that the common difference is 1.

Of course, the common difference can be any real number. For example examine the following sequence and find the missing element.

$\ 1, \frac{3}{2}, 2, … , 3, \frac{7}{2}, 4, …$

The first thing to do is to determine which kind of sequence this is. Because we only learned one type, which are the repeating decimals, which this clearly isn’t, this has to do something with addition.
Again, we’ll find the difference of two adjacent numbers.

$\frac{3}{2} – 1 = \frac{1}{2}, 2 – \frac{3}{2} = \frac{1}{2}, \frac{7}{2} – 3 = \frac{1}{2}, 4 – \frac{7}{2} = \frac{1}{2} …$

This means that this sequence is made by adding 1/2 on our start value 1.

This means that the missing value is $\ 2 + \frac{1}{2} = \frac{5}{2}$

We can check it by subtracting $\frac{5}{2}$ from 3, if we get $\frac{1}{2}$ the number we’ve guessed is correct.

$\ 3 – \frac{5}{2} = \frac{1}{2}$
But things don’t have to be necessarily so easy. Sequences can be mixed up:

$\ 1, 4, 2, 5, 3, 6, 4, 7, 5, 8, 6…$

This sequence can be broken into two separate sequences:

$\ 1, 4, 2, 5, 3, 6, 4, 7, 5, 8, 6…$

First sequence 1, 2 ,3, 4, 5, 6… where our start value is 1, and the common difference is 1, and second sequence 4, 5, 6, 7, 8 where our start value is 4 but the common difference is also 1.
The common difference can be any real number, so it can be a negative number, too. For example, if your start value equals 4, and the common difference is -1, the sequence would be:

$\ 4, 3, 2, 1, 0, -1, -2…$

One of the most famous sequences of this sort is the Fibonacci sequence.

It looks like this:

$\ 1, 1, 2, 3, 5, 8, 13, 21…$

These elements have no common difference. But if you go and try to find it:
$\ 21 – 13 = 8; 13 – 8 = 5…$ See the pattern? Every following number in Fibonacci sequence is given as the sum of its 2 predecessors. This pattern can be seen in many examples in nature: the petals in the flowers, proportions of human bodies, growth of any living thing and so on.

The last thing we’re going to mention is the sequence in which the elements are bounded by some sort of multiplication.

For example:

$2, 4, 8, 16, 32, 48, 96…$

This sequence is made by the start value 2 that is constantly multiplied with 2. If we want to know which is the number we’re constantly multiplying with, we’d divide a number in a sequence with his predecessor. If that number is the same in every element, we got our common factor.

Example 8. Find the missing number in a sequence.

$\ 5, 15, 75, x, 675$

First, we’ll find the common factor:
$\ 15 : 5 = 3, 75 : 15 = 3$. This means that in the given sequence our common factor will be 3.
This leads us to the fact that our element is: $\ 75 \cdot 3 = 225$.
Also: $\ 675 : 225 = 3$

Of course, problems containing sequences, can get a lot more complicated than this. You can also have powering or roots included or mix of any mathematical operations.
The best thing to learn how to recognize these patterns is to construct your own. Give it a try and find some new sequences.