Prove that root 5 is Irrational – √5 is an irrational number

Question: Which of the following statements is the most logical way to prove that root 5 is irrational?

  1. Root 5 is an integer.
  2. Root 5 is a rational number.
  3. Root 5 is an irrational number.
  4. Root 5 can be expressed as a fraction of two integers, where the denominator is not equal to 0.

Answer: √5 is an irrational number.

Solution:

  • A rational number is a number that can be expressed as a fraction of two integers, where the denominator is not equal to 0. An irrational number is a number that cannot be expressed as a fraction of two integers.
  • To prove that root 5 is irrational, we can use a proof by contradiction. We start by assuming that root 5 is rational. We can then square both sides of this equation to get a^2 = 5b^2.
  • Since 5 is prime, this means that 5 must divide a^2. But 5 also divides 5b^2. Therefore, 5 must divide both a and b. This contradicts our assumption that a and b are integers with no common factors other than 1.

So, we have √5 = p/q.

Now, let’s square both sides of the equation to eliminate the square root:

(√5)^2 = (p/q)^2 5 = p^2/q^2

Now, rearrange the equation:

5q^2 = p^2

From this equation, we can see that p^2 is a multiple of 5. This implies that p must also be a multiple of 5 since squaring an integer will result in a multiple of 5. So, we can write p as p = 5k, where k is an integer.

Substituting p = 5k back into our equation:

5q^2 = (5k)^2 5q^2 = 25k^2 q^2 = 5k^2

Now, we see that q^2 is also a multiple of 5, which means q must also be a multiple of 5.

However, this contradicts our initial assumption that p and q have no common factors other than 1 because both p and q are divisible by 5. This contradiction shows that our assumption that √5 is a rational number must be false.

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