Prove that root 5 is irrational?

Question

Answer

Let us assume that √5 is a rational number.

we know that the rational numbers are in the form of p/q form where p,q are intezers.

so, √5 = p/q

p = √5q

we know that 'p' is a rational number. so √5 q must be rational since it equals to p

but it doesnt occurs with √5 since its not an intezer

therefore, p =/= √5q

this contradicts the fact that √5 is an irrational number

hence our assumption is wrong and √5 is an irrational number.

Answer

Answer

Answer

Answer

Answer

Proof that √5 is irrational:

To prove that √5 is irrational, we will assume the opposite, that √5 is rational, and then arrive at a contradiction.

Assumption: √5 is rational.

If √5 is rational, it can be expressed as a fraction in the form p/q, where p and q are integers with no common factors, and q is not equal to zero.

Squaring both sides:

Squaring both sides of the equation √5 = p/q gives us 5 = (p^2)/(q^2).

Deducing:

From this equation, we can deduce that p^2 = 5q^2.

Two possibilities:
1. If p is even, then p^2 is also even.
2. If p is odd, then p^2 is also odd.

Case 1: p is even:

If p is even, we can write p as 2k, where k is an integer. Substituting this into the equation p^2 = 5q^2, we get (2k)^2 = 5q^2, which simplifies to 4k^2 = 5q^2.

Consequence:

From this equation, we can observe that 4k^2 is even, and therefore 5q^2 must also be even.

Implication:

If 5q^2 is even, then q^2 must also be even, as odd numbers multiplied by even numbers result in even numbers.

Case 2: p is odd:

If p is odd, we can write p as 2k + 1, where k is an integer. Substituting this into the equation p^2 = 5q^2, we get (2k + 1)^2 = 5q^2, which simplifies to 4k^2 + 4k + 1 = 5q^2.

Consequence:

From this equation, we can observe that 4k^2 + 4k is even, and therefore 5q^2 - 1 must also be even.

Implication:

If 5q^2 - 1 is even, then q^2 must be odd, as odd numbers subtracted by 1 result in even numbers.

Contradiction:

In both cases, we arrived at a contradiction, as in the first case, we concluded that q^2 is even, and in the second case, we concluded that q^2 is odd. This is not possible since q^2 cannot be both even and odd simultaneously.

Conclusion:

Since assuming that √5 is rational led us to a contradiction, our initial assumption is false. Therefore, we can conclude that √5 is irrational.

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