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.
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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.