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Posts posted by triclino


    Since you have already demonstrated that you can correctly perform the arithmetic of fractions, I see no reason why you can't work out the answer to this for yourself.


    So are they the same?



    When you have done that you might like to consider this.



    Most people just use the formulae you quoted in your post#1 without thinking about the detail or the rules.


    The rules are:


    1. The % must all be calculated on the same base.

    2. The % must be mutually exclusive. This is why I made such a fuss about the difference between and and or in my post#7

    To show this try calulating the following %.


    [math]\left( {100} \right)\left( {\frac{{200}}{{500}}} \right) = a\% [/math]


    [math]\left( {100} \right)\left( {\frac{{260}}{{500}}} \right) = b\% [/math]


    [math]\left( {100} \right)\left( {\frac{{300}}{{500}}} \right) = c\% [/math]


    [math]\left( {100} \right)\left( {\frac{{360}}{{500}}} \right) = d\% [/math]



    They are all reckoned on the same base in accordance with rule 1.


    I have already discussed rule 2 in post#7, read it again.


    now calculate

    [math]\left( {b - a} \right) + \left( {d - c} \right)[/math]



    What do you notice?





    To put it extremely simply, you are mixing 2 pools of people into one. 20% of stomach patients are from a different pool than the 30% of liver patients.


    If person A has 10 apples and person B has 10 bananas, and they are each given an additional 5 of their fruit respectively, you would say there is was an 100% increase in fruit, whereas there was only a 50% increase in their fruit.


    This is because the added change was 10 out of 20 fruit, and not 10/10.


    Why dont' you read the whole thread ??

  2. if you add:

    [math]\frac{Larger_1-Smaller_1}{Smaller_1}*100 +\frac{Lalger_2-Smaller_2}{Smaller_2}*100[/math]

    Are you going to get:

    [math] = \left( {100} \right)\left( {\frac{{{\rm{Larger1 + Larger2 - Smaller1 - Smaller2}}}}{{{\rm{Smaller1 + Smaller2}}}}} \right)[/math] ????

  3. ok,let's make it easier:


    suppose a No goes from 300 to 360,then its %change is:


    [math]\frac{360-300}{300}*100[/math] =20%


    And suppose another No goes from 200 to 260,then its % change is:


    [math]\frac{260-200}{200}*100[/math] = 30%


    Now total % change is :


    [math]\frac{620-500}{500}*100[/math] = 24%


    Since now we have pure Nos and not different populations shouldn't the total % change be 50%

  4. It looks ok to me - why do you think it should be 50%?

    20% change in stomach patients and 30% change in liver patients gives a total 50% and not 24%

    A 20% change in one population plus a 30% change in another population cannot be added to give a 50% change in a total population

    give me a mathematical or logical reason why not

  5. A % change for stomach patients within a year from 300 to 360 is:


    [math] \frac{360-300}{300}*100= [/math] 20%


    A % change for liver patients within a year from 200 to 260 is:


    [math]\frac{260-200}{200}*100 =[/math] 30%


    Hence Total % change in stomach and liver patients within a year is:


    [math] \frac {620-500}{500}*100=[/math] 24%


    But it should be 50% shouldn"t it ??

  6. In proving that :[latex]x\leq y\wedge 0\leq z\Longrightarrow xz\leq yz [/latex] we have the following proof:


    Let ,:[latex]x\leq y\wedge 0\leq z[/latex] and

    let ,[latex]\neg(xz\leq yz) [/latex].......................................................1

    But from (1) and using the trichotomy law we have :yz<xz.And using the fact [latex]0\leq z[/latex] we have for 0= z, y0<x0 => 0<0 , a contradiction since ~(0<0)


    Hence [latex] xz\leq yz[/latex]


    No - if you checked my first response I said that epsilon was very small. Sorry I had thought you would know this terminology; epsilon is used to denote a very small addition of arbitrary smallness.


    What I am saying in my posts is that x is the tiniest bit bigger than a and that a is equal to t




    In the whole of mathematics can you show me a single definition defining the exprssion " small epsilon"?

    In the definition of limits we say for all epsilon biger tha 0 ,there exists a delta biger than 0 such that: e.t.c e.t.c

    99% of the problems in analysis stert with the expession . Let ε>0, or given ε>0

    Yes you can use small or big epsilon,or delta .

    But if you do not clearly show their relatin to other variables of the problem or between them it is catastrophic.

    In the limits for example ,you have to clearly show the relatio9n between epsilon and delta, whether the epsilon and deltas are small or big,or whatever

    In your solution what will happen if for your " small epsion" we have a small : b-a??

    Besides when you say : " for a small epsilon ",one may say ,ok But how small .

    Mathematics is relations between variables and certainly expressions or words of : small,big ,tiny e.t.c do not show relations

  8. "By the way in the OP it is r and not ε" - exactly and for the inquality to work epsilon is smaller than r (ie now matter hw small r is there is a smaller real that we designate epsilon)


    the inequality becomes

    0<|a+ε-a|<r =>


    Suppose r=3b-a

    And since you clame the epsilon that you introduce in the problem to be ε<r then we can say that ε can be equal to 2b-a<3b-a=r

    Hence if we put


    Does this x satisfy also the relation a<x<b ?

    Furthermore when we introduce a new variable into a problem we have to quantify that variable to make our formula a well formed formula. And since you introduce ε how do you quantify it?

  9. This reasoning is actually correct, though if you were to present the proof (in a paper, or for an assignment, etc.), then you'd want to flesh it out a bit, and some of the wording is a bit odd.






    But my substitution in my 2nd proof isnt it the same with the substitution in my 1st proof??


    BECAUSE when i substitute a=b in the inequality a+c<b+c (1) of my 2nd proof isnt the SAME like i combine a=b And a<b??


    a+c<b+c is the result of adding c to both sides of the inequality a<b


    Like b+c<a+c was the result of adding c to both sides of the inewuality b<a


    a logical inequality (a fault), surely you can't believe equal numbers can in fact be unequal? Maybe on the quantum level this can be true, but not with an a/b test.


    Can you perhaps refer me to any mathematical or logical book that I can find the definition of the phrase "logical inequality" ??


    No equal Nos are not unequal that is why we end up with a contradiction

  11. Situation 1: A = B

    Situation 2: A > B


    Given this difference, the answer found in Situation 1 cannot be used as a substitute in Situation 2. A cannot be equal to B and also greater than B at the same time.






    Is then the following proof without using contrapositive but only using contradiction, correct ??


    Let, a+c<b+c...............................................................................................................................................................................(1)


    Let, ~(a<b).......................................................................................................................................................................................(2)


    Then by using (2) and and the law of trichotomy we have: a=b or b<a


    For a=b => a+c<a+c ,by substituting into (1)


    For b<a => b+c<a+c => a+c<a+c,by using the transitive property of inequalities and (1)


    Hence a+c<a+c ,contradiction and


    Thus ~~(a<b) => a<b

  12. Yes i understand ,but to convince me, you have to produce a law of logic or mathematics that will oppose my substitution


    On the other hand Suppes's subtitution law is more convincing than your justification


    Also that substitution law includes all cases ,even my case otherwise it wold be aspecific law applying only in certain cases.


    But i do not detect that in the definition of the law.

  13. First, about your wording: instead of "Let ~(a<b)" and "Let (a+c)<(b+c)", it's more correct to say "Assume ~(a<b) and (a+c)<(b+c)."


    Second, your proof involves the two cases a=b and b<a. Substitution of the first into the second isn't a valid step, since for all real numbers a and b, a cannot simultaneously be equal to b and greater than b. That is to say, these are entirely separate cases and should not be mixed as you've done here.


    Third, what axioms/results are you allowed to use? If you're allowed to make use of the fact that (b<a) => (b+c)<(a+c), for instance, then assuming you can also make use of additive inverses and the associativity of addition, you can do a pretty short direct proof of this cancellation law without having to resort to an indirect proof of the contrapositive. I'd be a bit more clear and detailed here, but this seems like it might be a homework question, so I don't want to give too much away.




    This is incorrect. As detailed before, the entire substitution is invalid, but even in the context of the OP's reasoning (the apparent point was to show the logical conflict between (a+c)=(b+c) and (b<a) => (b+c)<(a+c)), the equality shouldn't be carried over.

    First i suggest you write to all those authors that they use the word "let" instead of suppose in their proofs to correct their style of writing

    Second,on page104 in Suppes's book ,Introduction to Logic i find the following rule concerning substitution:


    "if S is an open formula,from S and t_1 =t_2,or from S and t_2=t_1 we may deriveT,provided that T results from S by replacing 0ne or more occurrences of t_1 in S by t_2".


    So in our case if we put:


    S = b+c<a+c


    t_1= a+c


    t_2 = b+c


    then the derivable formula T IS



  14. in proving the cancellation law in inequalities in real Nos we have the following proof:



    Let : ~(a<b)

    Let : a+c<b+c


    Since ~(a<b) and using the law of trichotomy we have : a=b or b<a


    for a=b => a+c=b+c.....................................................................(1)


    for b<a => b+c<a+c......................................................................(2)


    and substituting (1) into (2) we have : a+c<a+c ,contradiction.


    Hence ~(a+c<b+c)


    Thus by contrapositive we have: a+c<b+c => a<b


    is that proof correct??

  15. I did define it.



    No, that is not how I will define force, and that's not how Newton defined it either. Newton gave F=ma as an equation of motion, not a definition of force. He defined force as a quantifiable "push or pull," and he gave an example of a force: gravitation. The Lorentz force is another example. F=ma is not, nor has it ever been, a definition of force. Acceleration is absurdly simple to define: the time derivative of velocity.


    Before we continue, I think we deserve to know what point you're trying to make. Because you're coming across as an obnoxious troll who just wants to argue for the sake of arguing. It's clear you didn't come hear because you were confused and wanted clarification, you came here to be annoying.

    iN an axiomatic development we have the udefined or primitive concepts.


    For example in Modern Geometry the primitive or undefined concepts are: point ,line,on,between,congruent.


    So in an axiomatic development of Newtonian Physics i was trying to establish what are the primitive concepts .


    Is mass one of the undefined concepts??


    In your definition of force what is a pull or push??


    Notiice in any definition if we do not have some udefined concepts the definition is worthless

  16. Are you just trying to be obnoxious? In classical physics, mass is the constant of proportionality relating how much an object accelerates to how much net force was applied to it. On Earth, the easiest way to determine the mass of something would be to place it on a scale (measure its weight), then divide by the acceleration due to gravity at Earth's surface which is ~9.8 m/s2. If you don't have the luxury of being able to measure something's weight (if you're in orbit around a planet, in deep space, etc.), then I suppose the next easiest way to determine mass is to attach the object to the end of a spring with known spring constant, then set it into simple harmonic motion. If you measure its period to good accuracy, then its mass is [math]m=T^2 k /4 \pi^2[/math].

    I did not ask : How we measure mass ,but how we define mass


    Howevetr if we define mass as the constant of proportionality connecting force and accelaration,then i will ask you what is force or accelaration.


    Then you will say : force is the product of mass accelaration .That will lead us into a never ending cycle,because then i will ask you what is mass

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