Prove that √3 is irrational.
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Step-by-Step Solution
Step 1: Assume the opposite
To prove that 3 is irrational, we use the method of contradiction. We start by assuming the opposite, that 3 is a rational number. A rational number can always be expressed as a fraction ba, where a and b are integers, b=0, and a and b have no common factors other than 1 (i.e., they are coprime).
Step 2: Square both sides
Now, we square both sides of the equation 3=ba to eliminate the square root. This gives us 3=b2a2. Rearranging the terms, we get 3b2=a2. This equation implies that a2 is a multiple of 3, which means a itself must be a multiple of 3.
Step 3: Substitute 'a' as a multiple of 3
Since a is a multiple of 3, we can write a=3c for some integer c. Substituting this into the equation 3b2=a2, we get 3b2=(3c)2, which simplifies to 3b2=9c2. Dividing both sides by 3, we find b2=3c2. This implies that b2 is a multiple of 3, and therefore b must also be a multiple of 3.
Step 4: Reach a contradiction
From the previous steps, we concluded that both a and b are multiples of 3. This means that a and b have a common factor of 3. However, our initial assumption was that a and b are coprime, meaning they have no common factors other than 1. This creates a contradiction.
Step 5: Conclude irrationality
Since our assumption that 3 is rational leads to a contradiction (that a and b have a common factor of 3, despite being assumed coprime), our initial assumption must be false. Therefore, 3 cannot be rational, which means it must be irrational.