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KiRa [710]
3 years ago
7

Consider the following theorem and proof.Theorem: The number âš2 is not rational number.Proof: Let's suppose âš2 is a rational n

umber.Then we can write âš2 = a/b where a, b are whole numbers, b not zero.We additionally assume that a/b is simplified to lowest terms, i.e., a and b have no common factors. Both of the numbers a and b cannot be even.From the equality âš2 = a/b it follows that 2 = a2/b2 .So, a2 = 2 · b2. Showing that the square of a is an even number since it is two times something.From this we know that a itself is also an even number.In symbols, a = 2k where k is a natural number. If we substitute a = 2k into the original equation 2 = a2/b2, this is what we get:2=(2k)2/b22=4k2/b22*b2=4k2b2=2k2Showing that b2 is even because it is 2 times a number.This implies that b itself is also even.Therefore the supposition made in the first line cannot be true and we can conclude that the number âš2 is not rational number.thereforeThe statement "Both of the numbers a and b cannot be even." is justified by the fact that The assumption that âš2 is a rational number. The assumption that a/b is simplified lowest terms. The assumption that âš2 is an irrational number. The fact that b divides a evenly. The fact that a and b are whole numbers.
Mathematics
1 answer:
ololo11 [35]3 years ago
7 0

Answer:

The statement "Both of the numbers a and b cannot be even." is justified by the fact that a/b is simplified lowest terms

Step-by-step explanation:

We need to  show that the √2 is an irrational number.

And from the given steps of proof stated in the question, we need to find the assumption that justifies the fact : " Both of the number a and b cannot be even".

First take the given options :

Option a : √2 is a rational number

√2 being an rational or irrational has no relation of a and b to be even or odd. So, this option is rejected.

Option B : a/b is simplified lowest terms

This shows that a and b are not even because if a and b are even then a/b can be simplified in other lowest term.

Option c : √2 is a irrational number

Similarly, By using the inverse part of Option A, option c is also rejected.

Option d : The fact that b divides a evenly

This only shows that the a is even. This does not give any idea about b is even or not. So option D is also rejected.

Option E : The fact that a and b are whole numbers

This fact does not imply that the a and b are even or odd. So option E is also rejected.

Hence, The statement "Both of the numbers a and b cannot be even." is justified by the fact that a/b is simplified lowest terms

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