Respuesta :
Answer:
By contradiction it can be proved that:
Average of four real numbers will be greater than or equal to at least one of the four real numbers.
Step-by-step explanation:
Method of contradiction means we assume opposite to the facts to be proved and then we contradict our assumption.
As a result, we prove the fact.
Here, let the four number be:
[tex]p, q, r, s[/tex]
The average will be Sum of all numbers divided by count of numbers.
[tex]\dfrac{p+ q+ r+ s}4[/tex]
Now, let us assume the opposite that the average is less than all the numbers.
i.e.
[tex]\dfrac{p+ q+ r+ s}4 <p[/tex]
[tex]\dfrac{p+ q+ r+ s}4 <q\\\dfrac{p+ q+ r+ s}4 <r\\\dfrac{p+ q+ r+ s}4 <s[/tex]
Now, let us add all of them:
[tex]\dfrac{p+ q+ r+ s}4 +\dfrac{p+ q+ r+ s}4 +\dfrac{p+ q+ r+ s}4 +\dfrac{p+ q+ r+ s}4 < p+q+r+s\\\Rightarrow \dfrac{4(p+q+r+s)}4<p+q+r+s\\\Rightarrow p+q+r+s<p+q+r+s[/tex]
Which can never be possible hence, our assumption is contradicted.
our assumption is wrong.
Therefore, by contradiction it is proved that:
Average of four real numbers will be greater than or equal to at least one of the four real numbers.
An average of four real numbers will be greater than or equal to at least one of the four real numbers.
What is average?
An average is simply defined as the mean of the given set of numbers. The mean is said to be an arithmetic mean. It is the ratio of the sum of the observation to the total number of observations.
Prove the following using proof by contradiction.
The average of four real numbers is greater than or equal to at least one of the numbers.
Method of a contradiction means we assume opposite to the fact to be proved and then we contradict our assumption.
Let the four numbers be a, b, c, and d.
The average will be
[tex]\rm \dfrac{a+b+c+d}{4}[/tex]
Now let us assume the opposite that the average is less than all the numbers.
[tex]\rm \dfrac{a+b+c+d}{4} < a\\\\\rm \dfrac{a+b+c+d}{4} < b\\\\\rm \dfrac{a+b+c+d}{4} < c\\\\\rm \dfrac{a+b+c+d}{4} < d[/tex]
Let us add all of them.
[tex]\rm \rightarrow \dfrac{a+b+c+d}{4} +\dfrac{a+b+c+d}{4} +\dfrac{a+b+c+d}{4}+\dfrac{a+b+c+d}{4} < a+b+c+d\\\\\rightarrow \dfrac{4(a+b+c+d)}{4} < a+b+c+d\\\\\rightarrow a+b+c+d < a+b+c+d[/tex]
That can never be possible hence, our assumption is contradicted.
Our assumption is wrong.
Therefore, by contradiction, it is proved that;
An average of four real numbers will be greater than or equal to at least one of the four real numbers.
More about the average link is given below.
https://brainly.com/question/897199
