Prove that 5 is irrational and find whether 3/13 is a terminating or non-terminating repeating decimal.
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Step-by-Step Solution
Step 1: Proof by Contradiction for 5
To prove that 5 is irrational, we use the method of proof by contradiction. We start by assuming the opposite, that 5 is a rational number. If 5 is rational, it can be expressed as a fraction ba, where a and b are coprime integers (meaning they have no common factors other than 1) and b=0.
Step 2: Square both sides and rearrange
Squaring both sides of the equation 5=ba gives 5=b2a2. Multiplying both sides by b2 yields 5b2=a2. This equation implies that a2 is a multiple of 5, which means a must also be a multiple of 5.
Step 3: Substitute a=5c and simplify
Since a is a multiple of 5, we can write a=5c for some integer c. Substituting this into the equation 5b2=a2, we get 5b2=(5c)2, which simplifies to 5b2=25c2. Dividing both sides by 5 gives b2=5c2. This implies that b2 is a multiple of 5, and therefore b must also be a multiple of 5.
Step 4: Conclusion for 5
We have shown that both a and b are multiples of 5. This means that 5 is a common factor of a and b. However, our initial assumption was that a and b are coprime integers, meaning they have no common factors other than 1. This is a contradiction. Therefore, our initial assumption that 5 is rational must be false. Hence, 5 is irrational.
Step 5: Analyze the denominator for 3/13
To determine if a rational number qp (in simplest form) has a terminating or non-terminating repeating decimal expansion, we examine the prime factors of its denominator q. If the prime factorization of q contains only 2s and/or 5s, the decimal expansion is terminating. Otherwise, it is non-terminating repeating.
Step 6: Conclusion for 3/13
The prime factorization of the denominator, 13, is simply 13. Since 13 is not of the form 2m×5n (where m and n are non-negative integers), the decimal expansion of 133 will be non-terminating and repeating.