Hi Guys,

I've been stuck with the proof of this particular statement, specifically the comments at the end of the statement. As follows:

 

Prove that If a < b are real numbers then a < a + b/2 < b. (How do you know that 2 > 0^#? What exactly is 2?)

 

My attempt, as follows:

 

since a < b, then 0 < b - a (by definition.)

 

*assuming that we know what b/2 is, and that b/2 < 0 such that -(b/2) > 0

 

then, we know that: b - a - b/2 = b - (a + b/2) > 0 ----> b > a + b/2. NB:[-a - b/2 = -(a + b/2) can be justified]

 

the a< a + b/2 ---> don't know how to go about proving this part!

 

*Can this assumption be justified.

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^#I know that 1 > 0 (I've proven this in an exercise. Using the Trichotomy Law and that for any a in R, we have a² > 0.)

 

Using that fact, we have 1 + 1 = 2 > 0.

 

is this correct ?

 

 

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Your help is most appreciated .

Prove that If a < b are real numbers then a < a + b/2 < b

The statement as you put it is not true and should therefore resist being proven. Take for example a=-3 and b=-2. You'd get -3 < -3 + -2/2 = -4, which is obviously not true. Did you possibly mean

prove that if a < b are real numbers then a < (a + b)/2 < b ?

The statement as you put it is not true and should therefore resist being proven. Take for example a=-3 and b=-2. You'd get -3 < -3 + -2/2 = -4, which is obviously not true. Did you possibly mean

prove that if a < b are real numbers then a < (a + b)/2 < b ?

 

Hi,

Thank you very much for your quick reply. I see your point, but I took the question directly out of the textbook. Maybe the author forgot to put in the parentheses.

 

How would one go about proving the above statement, as you stated. I will start working on it as well, before looking at your reply.

 

Much appreciated .

2a < a+b < 2b is easier to prove.

2a < a+b < 2b is easier to prove.

 

I'll go about doing this, here is my attempt at the solution:

 

Let a; b > 0 then a + b > 0.

 

Now, 2 > 0 and a > 0, then 2a > 0.

 

write 2a = a + a, hence 2a = a + a > 0, since a < b

 

we find that a + a < a + b.

 

Also, 2b = b + b > 0, where a < b.

we have a + b < b + b.

 

And so 2a < a + b < 2b.

 

 

That was for the case a, b > 0. Now if you want to consider for any real number, I will use the following result (Which I have proven earlier.)

 

Result: If a < b, then a + c < b + c.

 

to show that 2a < a + b, using the above result, set c = a.

 

And, similarly to show that a + b < b + b = 2b, using the above result, set c = b.

 

Is this correct?

 

 

Also what does the author mean by:

 

What exactly is 2 anyway?

Should I say it's the multiplicative of 2^-1

 

 

Once again your help is appreciated .

 

Kind Regards.

That was for the case a, b > 0. Now if you want to consider for any real number, I will use the following result (Which I have proven earlier.)

 

Result: If a < b, then a + c < b + c.

 

to show that 2a < a + b, using the above result, set c = a.

 

And, similarly to show that a + b < b + b = 2b, using the above result, set c = b.

 

Is this correct?

That's what I had in mind.

That's what I had in mind.

 

Thanks for the guidance!

 

Really appreciate it!

This and similar propositions are easy to prove once one has a good definition of real numbers - see www.realnumbers.me.uk for example