Prove that square root of 5 is irrational
To prove that square root of 5 is irrational, we will use a proof by contradiction. What is a proof by contradiction?.
Suppose we want to prove that a math statement is true. Simply put, we assume that the math statement is false and then show that this will lead to a contradiction.
If it leads to a contradiction, then the statement must be true
To show that
√5
is an irrational number, we will assume that it is rational
Then, we need to find a contradiction when we make this assumption
If we are going to assume that
√5
is rational, then we need to understand what it means for a number to be
rational
Basically, if square root of 5 is rational, it can be written as the ratio of two numbers as shown below:
Square both sides of the equation above
Multiply both sides by y
^{2}
5 × y
^{2} =
x^{2}
/
y^{2}
× y
^{2}
We get 5 × y
^{2} = x
^{2}
{
Another important
concept before we finish our proof: Prime factorization
Key question: is the number of prime factors for a number raised to the second power an even or odd number?
For example, 6
^{2}, 12
^{2}, and 15
^{2}
6
^{2} = 6 × 6 = 2 × 3 × 2 × 3 (4 prime factors, so even number)
12
^{2} = 12 × 12 = 4 × 3 × 4 × 3 = 2 × 2 × 3 × 2 × 2 × 3 (6 prime factors, so even number)
15
^{2} = 15 × 15 = 3 × 5 × 3 × 5 = (4 prime factors, so even number)
There is a solid pattern here to conclude that any number squared will have an even number of prime factors
In order words, x
^{2} has an even number of prime factors
}
Let's finish the proof then!
5 × y
^{2} = x
^{2}
Since 5 × y
^{2} is equal to x
^{2}, 5 × y
^{2} and x
^{2} must have the
same number of prime factors
We just showed that
x
^{2} has an even number of prime factors
y
^{2} has also an even number of prime factors
5 × y
^{2} will then have an odd number of prime factors.
The number 5 counts as 1 prime factor, so 1 + an even number of prime factors is an odd number of prime factors
5 × y
^{2} is the same number as x
^{2}. However, 5 × y
^{2} gives an odd number of prime factor while x
^{2} gives an even number of prime factors
This is a contradiction since a number cannot have an odd number of prime factors and an even number of prime factors at the same time
The assumption that square root of 5 is rational is wrong. Therefore, square of 5 is irrational

Mar 19, 18 05:53 PM
Triangle midsegment theorem proof using coordinate geometry and algebra
Read More
New math lessons
Your email is safe with us. We will only use it to inform you about new math lessons.