Basic Algebra/Working with Numbers/Adding Rational Numbers

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Vocabulary

Numerator
Denominator
Irreducible

Lesson

It is easy to add fractions when the denominators are equal. For example, adding 310 and 210 is very simple, just add the numerators and you have the numerator of the resulting fraction:

  • 310+210=510=12

Notice the simplification: five parts out of ten is the half of the parts. Unfortunately, it is not always so simple. Sometimes we need to add fractions that have different denominators. Before we can add them, we must alter the fractions so that their denominators are the same. We can do this by multiplying each fraction by the number one which doesn't change the value of the fraction). However, the form of the number one will itself be represented as a fraction whose denominator and numerator are equal, and under our control. For example, all of these fractions are equal to one:

  • 22=33=107107=1

Knowing this, we can change the denominators of the fractions so that the denominators of both are the same. For example:

  • 13+12=1×23×2+1×32×3=26+36=56

In this case we changed both fractions so that they each had a denominator of 6.

More complicated fractions

In these cases, we can guess which multiplication to do, but sometimes, it is not that easy. For example, adding 123456 and 234120.


The simplest general method is to multiply the numerator and denominator of the first fraction by the denominator of the second fraction and vice-versa. The resulting denominators will both be the product of the two original denominators. In this case :

  • 123456+234120=123×𝟏𝟐𝟎456×𝟏𝟐𝟎+234×πŸ’πŸ“πŸ”120×πŸ’πŸ“πŸ”=1476054720+10670454720=14760+10670454720=12146454720

We obtain generally big numbers which is not optimal because the fraction can most of the time be written with smaller numbers.


The second is more subtle. Instead of multiplying by the actual denominators, we multiply by the smallest possible number for each side so that we obtain the same denominator. For example:

  • 16+14=1×26×2+1×34×3=212+312=512

We only multiplied by 2 in the first fraction and by 3 in the second fraction. The resulting fraction, 512 is optimal, which we call irreducible.

Note that 2 is the half of 4=2×2 and 3 the half of 6=3×2. We did not multiply by the given denominators, we avoided to multiply by the factor 2. Let's take the previous example and find the factors composing the numbers...

123=πŸ‘×41 and 456=2×228=2×2×114=2×2×2×57=2×2×2×πŸ‘×19
234=2×117=2×3×39=𝟐×πŸ‘×3×13 and 120=2×2×3×10=2×𝟐×πŸ‘×2×5

We can see that we can simplify 123456 by 3 which gives 412×2×2×19 and simplify 234120 by 2 Γ— 3 which gives 392×2×5. Remember that multiplying by the same number the numerator and the denominator does not change the value. The same is true when dividing by the same number.

Now comes a question : which is the smallest integer that contains the factors 2×2×2×19 and the factors 2×2×5. It is the number that has just all these factors in correct number: 2×2×2×5×19=760.

To attain this number, we must multiply in the first fraction by 5 and in the second by 2 Γ— 19. So, finally we have:

  • 123456+234120=412×2×2×19+392×2×5=41×πŸ“760+39×𝟐×πŸπŸ—760=205760+1482760=1687760

This fraction is simpler as the first obtained 12146454720.

Both fractions are equal: 1687760=12146454720

But the factor between the two fractions is 72!

Practice Games

put links here to games that reinforce these skills

Practice Problems

Use / as the fraction line and put spaces between the wholes and fractions! <quiz display=simple points="1/1"> { |type="{}"} 12+14={ 3/4_5 }

{ |type="{}"} 13+16={ 3/6|1/2_5 }

{ |type="{}"} 14+16={ 5/12_5 }

{ |type="{}"} 18+14={ 3/8_5 }

{ |type="{}"} 18+12={ 5/8_5 }

{ |type="{}"} 23+19={ 7/9_5 }

{ |type="{}"} 24+12={ 4/4|2/2|1/1|1_5 }

{ |type="{}"} 88+12={ 12/8|3/2|1 4/8|1 1/2_5 } </quiz> Template:BookCat Template:Simple Page Navigation