Prove that 3 is irrational.
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Step-by-Step Solution
Step 1: Assume 3 is rational
To prove that 3 is irrational, we will use the method of proof by contradiction. We start by assuming the opposite, that 3 is a rational number. If 3 is rational, it can be expressed as a fraction a/b, where a and b are integers, b=0, and a and b have no common factors other than 1 (i.e., they are in their simplest form or coprime).
Step 2: Square both sides
To eliminate the square root, we square both sides of the equation. This operation maintains the equality if both sides are positive. This gives us an expression relating a2 and b2.
Step 3: Rearrange the equation
Multiply both sides by b2. This equation shows that a2 is a multiple of 3. This implies that a2 is divisible by 3.
Step 4: Conclude that 'a' is divisible by 3
According to a property of prime numbers, if a prime number p divides a2, then p must also divide a. Here, 3 is a prime number, and since 3b2=a2, it means 3 divides a2, so 3 must divide a.
Step 5: Express 'a' as a multiple of 3
Since a is divisible by 3, we can write a as 3c, where c is some integer. This substitution will allow us to find a similar relationship for b.
Step 6: Substitute 'a' back into the equation
Substitute the expression for a (i.e., 3c) back into our rearranged equation 3b2=a2. Squaring 3c gives 9c2.
Step 7: Simplify and conclude that 'b' is divisible by 3
Divide both sides of 3b2=9c2 by 3 to get b2=3c2. This shows that b2 is a multiple of 3, which implies that b2 is divisible by 3. Therefore, similar to the case of a, 3 must also divide b.
Step 8: Reach a contradiction
We have shown that both a and b are divisible by 3. This means that 3 is a common factor of a and b. However, our initial assumption stated that a and b have no common factors other than 1. This contradiction arises from our initial assumption that 3 is rational.
Step 9: Conclude that 3 is irrational
Since our assumption led to a contradiction, our initial assumption must be false. Therefore, 3 cannot be rational, which means it must be an irrational number.