Prove that the square root of 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. 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 and rearrange
Next, we square both sides of the equation 3=ba to eliminate the square root. This gives us 3=b2a2. Multiplying both sides by b2 results in the equation 3b2=a2. This equation implies that a2 is a multiple of 3.
Step 3: Deduce a is a multiple of 3
According to a fundamental property of numbers, if a prime number (like 3) divides the square of an integer (a2), then it must also divide the integer itself (a). Therefore, a must be a multiple of 3. We can express this by writing a=3c, where c is some integer.
Step 4: Substitute a=3c and simplify
Now, we substitute a=3c back into our equation 3b2=a2. This gives us 3b2=(3c)2, which simplifies to 3b2=9c2. Dividing both sides by 3, we get b2=3c2. This new equation implies that b2 is also a multiple of 3.
Step 5: Deduce b is a multiple of 3
Similar to the reasoning for a, since 3 divides b2, it must also divide b. Therefore, b is a multiple of 3.
Step 6: Reach a contradiction
From our deductions, we found 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 7: Conclusion
Since our initial assumption led to a contradiction, the assumption must be false. Therefore, 3 cannot be a rational number, which means it must be an irrational number.