triclino

Given 0<a<π/2 , b>0 find a c>0 such that : for all ,x : if 0<x<π/2 and |x-a|<c ,then |((sqrt sinx) +1)^2-((sqrt sin a) +1)^2|<b

Mathematics certainly is not the science that you can make stories out of nothing

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] ????

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%

20% change in stomach patients and 30% change in liver patients gives a total 50% and not 24% give me a mathematical or logical reason why not

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 ??

I have a problem with the following inequality : I can go as far: But i have problem with 0=ΑΒ > A=0 ή Β=0 part.

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]

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

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 x=a+ε=a+(2b-a)=2b 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?

If you put t=a and x=a+ε ,then the inequality becomes: |a+ε-a|<ε => ε<ε, a contradiction By the way in the OP it is r and not ε

Given the reals a<b find a, t such that : For all r>0 there exist an , x such that : 0<|x-t|<r and a<x<b

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

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