Proving the Irrationality of √5 × √7

Proving the Irrationality of √5 × ?√7

The irrationality of a number implies that it cannot be expressed as a ratio of two integers. This article will guide you through a detailed proof demonstrating that the product √5 × √7 is irrational. Through a step-by-step process, we will use the method of proof by contradiction, which involves assuming the opposite of what we wish to prove and deriving a contradiction.

Step-by-Step Proof Using Proof by Contradiction

To prove that √5 × √7 is irrational, let's begin by assuming the opposite:

Assume the opposite: Suppose that √5 × √7 is rational. This means we can express it as a fraction:

√5 × √7 p/q

where p and q are integers and q ≠ 0.

Isolate one square root: Rearranging gives us:

√7 (p/q) - √5

Square both sides: Squaring both sides results in:

7 [(p/q) - √5]^2

Expanding the right side: 7 [(p/q)^2 - 2(p/q)√5 5] Rearranging yields: 2(p/q)√5 [(p/q)^2 - 5 - 7] Simplifying: 2(p/q)√5 [(p/q)^2 - 12] Isolate √5: Dividing both sides by 2(p/q), assuming p ≠ 0, gives us:

√5 [(p/q)^2 - 12] / (2(p/q)) (p/q - 6q) / (2p)

This means that √5 can be expressed as a rational number.

Squaring Both Sides Again:

If √5 is rational, then √(5) (p/q - 6q) / (2p), and squaring both sides gives us:

5 [(p/q - 6q) / (2p)]^2

This can be rearranged to:

5 * (4p^2) (p - 6q)^2

After simplifying, we get:

20p^2 p^2 - 12pq 36q^2

19p^2 12pq - 36q^2 0

Let's re-examine the form 19p^2 12pq - 36q^2 0 using the Rational Root Theorem. If x is rational, it must be a factor of 36. The factors are 1, 2, 3, 4, 6, 9, 12, 18, and 36. Testing each value confirms that none of these are solutions.

Therefore, since the assumption that √5 × √7 is rational leads to a contradiction (the irrationality of √5), we conclude that our initial assumption is false. Hence, √5 × √7 is indeed irrational.

Conclusion

In conclusion, through the method of proof by contradiction, we have demonstrated that the product of √5 and √7 is irrational. This proof showcases the power of assuming the opposite of what you wish to prove and then using algebraic manipulation to reveal a contradiction, thereby confirming the irrationality of a given number.