Cyclic Means
The mean of the two numbers a and b is c .
The mean of the two numbers b and c is a .
The mean of the two numbers c and a is b .
Is it possible for all the statements above to be fulfilled at the same time for distinct numbers a , b and c ?
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3 solutions
But you haven't show that it cannot be true if they are not equal. You only show that it can be true if they are equal.
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No.Read properly.I stated 3 requirements of the question(regarding a,b,c) and concluded,in order to satisfy all three that they should be equal.I deduced it.If there were more possibilities,then i would get them at the end of the deduction,but i got only 1,that is,all three equal.
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So you have showed that it should be equal, but not must be equal. Please write you entire thoughts. If you only get 1, it does not mean other people cannot get another. Prove that no one else can.
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@William Nathanael Supriadi – The method of Deduction in mathematics is a very interesting phenomenon in logic.If,through deduction,we were to get more possibilities,then deduction would surely provide it.But if only 1(or even 0) possibility existed,deduction method would end up in some single result(or contradictory result,respectively).That is how it works.So I am not saying that others can't get another result,but my deduction(in fact,any deduction) is Universal .
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@Anandmay Patel – I don't understand any of your explanation about deduction and therefore I still couldn't accept it. Can you show me how the method of deduction works? Or at least provide a link to a wiki about it?
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@William Nathanael Supriadi – Hey!Quite a long time you took to respond anyway...Take this^ question for example.Here are 2 possibilities.1.)Yes,there exists such three numbers.2.)No,there does not exist such 3 numbers.Now,by some method,I deduced my result by stating 3 conditions and applying logic about their simultaneous truth-holding,that is,the conditions hold true altogether at the same time.Right?Now,suppose that some other person proved some other situation that there exists such 3 numbers,then why my legal deduction went wrong?This,in real world never happens,because we have a contradiction-free system of logic.So,my deduction is legal and will always hold true,and there is a very powerful reason of not having some other possibility other than that of mine(otherwise there will be contradiction everywhere).Of course,there may be other methods through which this question can be solved,that is,other ways of deducting the same universal answer.By now i have realized that this is going very much big so i stop here.Got it?Logical hmm?
If I have still not answered your actual question(if you think),then please ask it again and more precisely,otherwise.
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@Anandmay Patel – Sorry for not being able to reply fast. I'm quite busy preparing for my final test. I'm giving you an example for this then. Suppose I claim that abs(x) = x, and I prove it by means of putting x=1,2,3, and so on. And since that is my deduction, and it is correct (in my opinion), then my claim is correct, right? No. Some other people would plug in x=-1 and show me that I'm wrong. Why can it be done? Obviously because I am not proving anything yet. Showing that it works for some values doesn't prove that it works for all values, or that it works only for the values we work out with. You need to show why they can't be different too. By the way, this conversation is getting interesting :)
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@William Nathanael Supriadi – Hi!Yes indeed this is becoming interesting:)You are right in your argument.According to me,this type of argument must have arisen in the early days of the discovery of the math subject.If we prove some result to be true for some numbers,then not necessarily its true for others.Correct.But,to tackle this,a new branch of mathematics was formed,Algebra.Here we introduce the concept of generalization.Variables STAND for all numbers.Here,in this question^,we used variables a,b,c to describe the question and prove the answer.We proved it in the language of variables implies that we proved for all numbers that the variable standed for,that is,all numbers!That is why this 'generalized' result is trivially true for all numbers.You got what I am trying to say?I will be happy hearing from you:)
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@Anandmay Patel – Now, back to the point: how are you so sure that the three statements in the problem cannot be fulfilled with a, b, and c not all being equal?
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@William Nathanael Supriadi – Okay.See my 3 statements.The first one tells that a,c,b form an AP.But I think that this language will not be interesting.So forget APs.Consider the first statement equivalent to the statement:c is between a and b.(And of course,it is also equivalent to Chung Kevin's first statement.).The second statement says that a is between b and c.The third statement says that b is between c and a.This pattern is impossible in the non trivial way,as you can observe.Trivially,it is possible if all the three numbers are equal.
Now,without loss of generality,assume that my conclusion is not always true.So lets assume(without loss of generality) that b and c are equal BUT a is different.Now,take a look at the second statement.According to it,a is between b and c.But b and c are equal!And there is no number between them,obviously.So in order to satisfy statement 2,here,you need 'a' to be equal.Likewise can be thought of for all other possibilities.
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@Anandmay Patel – Now that's the formal proof I was expecting! The argument is both logical and acceptable. Therefore, case closed! Now, can some admin convert the last comment into a solution?
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@William Nathanael Supriadi – Great!I have added this in the solution.Thanks!This discussion was really fruitful:)
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@Anandmay Patel – You're welcome!
Let's prove this by contradiction. Without loss of generality, assume that a < b . As c is the mean of a and b , then c must either lie strictly between a and b , or is equal to both a and b . Since our assumption stated that a = b , c must lie strictly between a and b ( a < b < c ). By similar argument on second line in the question, we can also conclude that a strictly lies between b and c . However, we have already shown that a is less than both b and c . We have a contradiction here, and therefore, our original assumption was wrong. All three variables must be equal.
While this assumption is being made in the comments the question does not state that a b c cannot all be equal. It states ever without the exclusion of a=b which would mean c=a c=b. While this is the only case in which this is true the question does not exclude this therefore. The answer should be yes
From the statements we can deduce that
a
+
b
=
2
c
∧
a
+
c
=
2
b
⇒
3
c
−
b
=
2
b
⇒
b
=
c
.
Hence, the answer is
No
.
The first statement tells us that a , c , b form an AP.
The second statement tells that b , a , c form an AP.
The third statement tells that c , b , a form an AP.
(CLEARLY) These three statements can be simultaneously true iff a = b = c .
There is a more Formal Proof of this:
The first statement tells that a,c,b form an AP.But I think that this language will not be interesting.So forget APs.Consider the first statement equivalent to the statement:c is between a and b.(And of course,it is also equivalent to Chung Kevin's first statement.).The second statement says that a is between b and c.The third statement says that b is between c and a.This pattern is impossible in the non trivial way,as you can observe.Trivially,it is possible if all the three numbers are equal.
Now,without loss of generality,assume that my conclusion is not always true(That is,my conclusion is not true for all such triplets).So lets assume(without loss of generality) that b and c are equal BUT a is different.Now,take a look at the second statement.According to it,a is between b and c.But b and c are equal!And there is no number between them,obviously.So in order to satisfy statement 2,here,you need 'a' to be equal.Likewise can be thought of for all other possibilities