write a convincing argument that if a and b are real numbers,?

Question

a

Answer 1

This is just the average. The average of two numbers is always between the numbers. Use the Triangle Inequality theorem.
But I got to say, Enough already!
It's one thing to help a kid struggling with basic college math to understand how to do what they need to do, so that they can get past the basic math requirements and into their field of study. And it's one thing to help a person working in advanced math to get a handle on something, once or twice.
But you've asked quite a few of questions all in a row, about the kind of things one would do early on in a course on number theory. What's the deal?
Are you a professor looking for fresh approaches to present this material to the students? Or are you a math major just starting out and finding oneself in over one's head?
If the former, nice going, but please use a little disclosure, because otherwise you're using our work without credit.
If the latter, suggest you do some serious thinking about your approach. Things are going to get a lot tougher than this and we won't always be around to help you.

Answer 2

(i) a (ii) a<(a+b)/2
Statement : (i) ///////> (ii)

Let "d" be the real number that equals the half of the sum of two real numbers a and b (not necessarily integers)

(iii) d = (a+b) / 2

To convince ourselves that the statement is true we pictured (iv) : a three sided statement, as follows
(iv) a+a < a+b < b+b

statement (iv) comes from our hypothesis (i)

Thus, dividing each three sides of (iv) by two, we get

(v) a < d < b

Since it is basicly clear that (v) == (ii), we just conclude our demonstration by repeating the now prooved statement :
(i) ///////> (ii)

If more explanations are necessary
- The symbol '///////>' is a conditional proposition
- we can show that d>a by writing equations as follows
d = (a+b)/2 = (a+a+b-a)/2
d = ((a+a)/2) + ((b-a)/2)
d = a + ((b-a)/2)
a + ((b-a)/2) > a
d > a

- we can also show that d d = a+((b-a)/2)
(b-a)/2 < b-a
a+ (b-a)/2 < a+b-a
d < b

- Note that the same equations apply if we invert a and b (b < a for instance)

Answer 3

If a and b are real numbers then the average of a and b would be (a+b)/2. Averages of real number always fall between the ranges (inclusive) of the highest and lowest numbers being averaged, in this case the average will be between or equal to a and b. Since we know that a
a<(a+b)/2
2a a something we know to be true

(a+b)/2 a+b<2b multiply both sides by 2
a with something we know to be true

Just make sure you give me credit when you turn this into your teacher.

Answer 4

if
a then
a+b therefore
a+b<2b
therefore
(a+b)/2
Now
if a then
a+a therefore
2a therefore
a<(b+a)/2
therefore
a<(a+b)/2

Since

(a+b)/2
and

a<(a+b)/2

Thus

a<(a+b)/2