Proving the Irrationality of ( sqrt{5} ) – A Step-by-Step Approach
We often encounter the concept of irrational numbers in mathematics, and one of the prime examples is ( sqrt{5} ). How can we prove that ( sqrt{5} ) is irrational? This article will guide you through a detailed, rigorous proof using the method of contradiction. By the end, you'll understand the core logic and the mathematical reasoning behind this proof, essential for anyone interested in advanced mathematics and mathematical proofs.
Introduction to the Proof by Contradiction
A proof by contradiction is a valuable method in mathematics, where we assume the opposite of what we want to prove and show that this assumption leads to a logical contradiction. In our case, we will assume that ( sqrt{5} ) is rational and demonstrate why this assumption cannot be true.
Step-by-Step Proof
Assumption
Let us begin by assuming that ( sqrt{5} ) is rational. If ( sqrt{5} ) is rational, it can be expressed as a fraction of two integers in simplest form, ( frac{p}{q} ), where ( p ) and ( q ) are coprime (i.e., they have no common factors other than 1) and ( q eq 0 ).
Squaring Both Sides
Starting with the assumption that ( sqrt{5} frac{p}{q} ), we square both sides of the equation:
[ (sqrt{5})^2 left(frac{p}{q}right)^2 ]
This simplifies to:
[ 5 frac{p^2}{q^2} ]
To find an equation in terms of ( p^2 ) and ( q^2 ), we multiply both sides by ( q^2 ) :
[ p^2 5q^2 ]
Analyzing ( p^2 )
From the equation ( p^2 5q^2 ), it is evident that ( p^2 ) is a multiple of 5. Since 5 is a prime number, this implies that ( p^2 ) must also be a multiple of 5. As a result, ( p ) itself must be a multiple of 5. Let's express ( p ) as:
[ p 5k ]
where ( k ) is an integer.
Substituting Back
Next, we substitute ( p 5k ) back into the equation ( p^2 5q^2 ):
[ (5k)^2 5q^2 ]
Expanding the left side, we get:
[ 25k^2 5q^2 ]
Dividing both sides by 5:
[ 5k^2 q^2 ]
Analyzing ( q^2 )
From the equation ( 5k^2 q^2 ), it is clear that ( q^2 ) must also be a multiple of 5. As before, since 5 is a prime number, ( q^2 ) must be divisible by 5, implying that ( q ) itself must be a multiple of 5.
Contradiction
At this point, we have established that both ( p ) and ( q ) are multiples of 5. This means that ( p ) and ( q ) share a common factor of 5, contradicting our initial assumption that ( p ) and ( q ) are coprime (no common factors other than 1).
Conclusion
Since our assumption that ( sqrt{5} ) is rational leads to a contradiction, we conclude that ( sqrt{5} ) must be irrational.
Related Keywords
irrational numbers, proof by contradiction, prime numbers
Conclusion
The proof that ( sqrt{5} ) is irrational is a beautiful example of how mathematical reasoning can reveal the true nature of numbers. By understanding and applying the method of contradiction, we have established that ( sqrt{5} ) cannot be expressed as a simple fraction, solidifying its status as an irrational number.