To prove that 5 is an irrational number, we first need to understand what an irrational number is. In mathematics, an irrational number is any number that cannot be expressed as a ratio of two integers, meaning that the decimal expansion of an irrational number is non-terminating and non-repeating.
Now, let’s suppose that 5 is a rational number, then we can express it as a ratio of two integers, let us say p and q.
Therefore, 5 = p/q, where p and q are co-prime integers.
Now, let’s simplify the above expression by multiplying both sides by q, we get:
5q = p
This means that p is a multiple of 5, since 5q is a multiple of 5.
Therefore, we can express p as p = 5k, where k is an integer.
Substituting the above value of p in the equation 5 = p/q, we get:
5 = 5k/q
Simplifying the above expression, we get:
1 = k/q
This means that k is equal to q, and since p and q are co-prime integers, k cannot be equal to q.
This contradiction proves that our assumption that 5 is a rational number is false, and hence, 5 is an irrational number.
We have proven that 5 is irrational by assuming that it was rational and arriving at a contradiction. Therefore, we can conclude that 5 cannot be expressed as a ratio of two integers, which makes it an irrational number.
Table of Contents
How to prove difference between irrational and rational number is irrational?
To prove that the difference between an irrational number and a rational number is irrational, we need to use the principles of proof by contradiction.
Let’s assume that the difference between an irrational number (say, a) and a rational number (say, b) is rational. So, there exists another rational number (say, c) such that a – b = c.
We know that a is irrational and b is rational, which means that a cannot be expressed as a rational number (i.e., a cannot be expressed as a quotient of two integers). Conversely, b can be expressed as a quotient of two integers, say p and q, where q is not equal to zero.
Now, we can substitute b with p/q in the equation a – b = c, which gives us a – p/q = c.
We can then move a to the other side of the equation to get a = c + p/q.
Since c and p/q are both rational numbers, their sum is also a rational number. Therefore, a can be expressed as a sum of two rational numbers, which contradicts our assumption that a is irrational.
So, we have arrived at a contradiction, proving that the difference between an irrational number and a rational number is indeed irrational.
The proof by contradiction shows that if a is irrational and b is rational, then a – b is irrational. This property is fundamental in real analysis, and it helps us understand the nature of irrational numbers and their relationships with rational numbers.
