Prove that the square root of 5 is irrational.
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Step-by-Step Solution
Step 1: Assume 5 is rational
We begin by assuming, for the sake of contradiction, that 5 is a rational number. By definition, a rational number can 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., the fraction is in its simplest form).
Step 2: Square both sides
To eliminate the square root, we square both sides of the equation. This gives us 5=b2a2. We can then rearrange this equation to 5b2=a2.
Step 3: Analyze divisibility
The equation a2=5b2 implies that a2 is a multiple of 5, and thus a2 is divisible by 5. A fundamental property of numbers states that if the square of an integer is divisible by a prime number, then the integer itself must also be divisible by that prime number. Since 5 is a prime number, it follows that a must be divisible by 5.
Step 4: Express a in terms of 5k
Since a is divisible by 5, we can express a as 5k for some integer k. This substitution will allow us to explore the implications for b.
Step 5: Substitute a back into the equation
Now, we substitute a=5k back into our equation 5b2=a2. This yields 5b2=(5k)2. We will simplify this expression to find information about b.
Step 6: Simplify and analyze divisibility of b
Simplifying (5k)2 gives 25k2, so the equation becomes 5b2=25k2. Dividing both sides by 5, we get b2=5k2. This means b2 is divisible by 5. By the same property used earlier, if b2 is divisible by 5, then b must also be divisible by 5.
Step 7: Reach a contradiction
From our analysis, we found that both a and b are divisible by 5. This means that a and b share a common factor of 5. However, our initial assumption was that the fraction ba was in its simplest form, meaning a and b have no common factors other than 1. This creates a contradiction.
Step 8: Conclude 5 is irrational
Since our initial assumption (that 5 is rational) led to a contradiction, this assumption must be false. Therefore, 5 cannot be expressed as a fraction of two integers and is thus an irrational number.