Show that For all real numbers $a, b$: If $0 < a < b$, then $\frac{a+b}{3} \in [a, b]$ is false by providing counter example.

Question

We are given few exercise questions that include statements that are false and we are required to prove it by providing counter example for each. Given below is statement is such.

Statement: For all real numbers $a, b$: If $0 < a < b$, then $\frac{a+b}{3} \in [a, b]$

My Issue: The issue is I have taken hours but I can't find any example and I tend to think that the statement might be true. It makes sense as well because a and b are positive when they are added and divided by $3$ we are basically dividing interval $[a, b]$ into $3$ intervals each with length equal to the $\text{answer}=\frac{(a+b)}{3}$ and hence it will lie in this interval of $[a,b]$ as well.

Who is wrong? My professor or I? Kindly guide.

Answer 1

When doing this question, I started with following test cases:

1. $a,b$ both positive and far apart relative to $0.$
2. $a,b$ both positive and close to each other, relative to $0.$

An example of Case $1$ is $a=1, b= 100$. But this fails to be a counter-example.

An example of Case $2$ is $a=100, b= 101$. Then, $(a+b)/3 \approx 67 \not\in [100,101].$ So this is a counter-example.

If none of these were counter-examples, you would have to keep coming up with qualitatively different test cases until you find one that is a counter-example (since the question tells you there is one).

Answer 2

One way to think about the term $\frac{a+b}3$ is to rewrite it as $\frac{a+b+0}3$. In this form, it is clear that it is the average of $a, b$, and $0$. Finding a counterexample for $\frac{a+b}3\in [a,b]$ is to find $a,b$ which are close to each other and far from $0$.

Take $a=7, b=8$. Then $\frac{a+b}3=\frac{15}3=5\not\in[7,8]$.

Your intuition is almost correct, and the statement would hold if the denominator in the fraction was $2$ (since now we'd be taking the average of just $a$ and $b$, which is naturally somewhere between the two). It's just that division by larger numbers now pushes $\frac{a+b}n$ towards 0 and outside $[a,b]$.