# triclino

Senior Members

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1. ## the definition of mass

What is matter What is a number of something What is greatness of something
2. ## the definition of mass

What is the quantity of matter What is momentum
3. ## the definition of mass

What is the definition of mass??
4. ## gravitational force

The strongest recorded
5. ## gravitational force

What is the biggest gravitational force in the Universe??
6. ## intermediate value theorem

We know that $a\leq c\leq b$ ,since c is the Sup of S. Hence $b-c\geq 0\Longrightarrow \frac{b-c}{2}\geq 0$ And in the case where b-c/2 >0 we can choose ΄0<δ <b-c/2 . But in the case where b-c/2 =0 ,if we choose δ<b-c/2 , then δ will be negative. But we want δ positive. Where in my previous posts there as the slightest indication of that claim
7. ## intermediate value theorem

Even better ,that fixes the value of δ stronger. By us??
8. ## intermediate value theorem

The value of δ is defined by the price of ε. When we say that for every ε>0 there exists a δ>0 such that .......e.t.c....e.t.c we mean that for every value of ε>0 there corresponds a particular volue of δ. In our case for ε=u-f©>0 there corresponds a particular $\delta >0$. We do not how much is that δ. However we can investigate with the help of trichotomy law what happens if : $\delta\leq b-c\vee \delta>b-c$. On the other hand there is no theorem,definition ,axiom to support the expression : choose $\delta >b-c$
9. ## intermediate value theorem

If we assume : $\delta< b-c$ we have to consider what happens if $\delta\geq b-c$ Since from trichotomy law we have: $\delta< b-c \vee \delta\geq b-c$
10. ## intermediate value theorem

if a=1 then f(a)=1 and b =2 then f(b) =1/2. Hence u must be between : 1 AND 1/2 . Go and read the theorem in wiki ,before you come into wrong conclusions
11. ## intermediate value theorem

For the following proof of the intermediate value theorem ,which i found in wikipedia: Proof: Let S be the set of all x in [a, b] such that f(x) ≤ u. Then S is non-empty since a is an element of S, and S is bounded above by b. Hence, by completeness, the supremum c = sup S exists. That is, c is the lowest number that is greater than or equal to every member of S. We claim that f© = u. Suppose first that f© > u, then f© − u > 0. Since f is continuous, there is a δ > 0 such that | f(x) − f© | < ε whenever | x − c | < δ. Pick ε = f© − u, then | f(x) − f© | < f© − u. But then, f(x) > f© − (f© − u) = u whenever | x − c | < δ (that is, f(x) > u for x in (c − δ, c + δ)). This requires that c − δ be an upper bound for S (since no point in the interval (c − δ, c] for which f > u, can be contained in S, and c was defined as the least upper bound for S), an upper bound less than c. The contradiction nullifies this paragraph's opening assumption. Suppose instead that f© < u. Again, by continuity, there is a δ > 0 such that | f(x) − f© | < u − f© whenever | x − c | < δ. Then f(x) < f© + (u − f©) = u for x in (c − δ, c + δ). Since x=c + δ/2 is contained in (c − δ, c + δ), it also satisfies f(x) < u, so it must be contained in S. However, it also exceeds the least upper bound c of S. The contradiction nullifies this paragraph's opening assumption, as well. We deduce that f© = u as stated. I have the folowing questions: 1)How does the author come to the conclusion that ,c-δ is an upper bound for S ,in this part of the proof: "This requires that c − δ be an upper bound for S (since no point in the interval (c − δ, c] for which f > u, can be contained in S, and c was defined as the least upper bound for S), an upper bound less than c. The contradiction nullifies this paragraph's opening assumption" 2) How does the author conclude that ,x = c+δ/2 is contained in S ,since he has not proved that $c+\frac{\delta}{2}\leq b$ in the following part of the proof: "Suppose instead that f© < u. Again, by continuity, there is a δ > 0 such that | f(x) − f© | < u − f© whenever | x − c | < δ. Then f(x) < f© + (u − f©) = u for x in (c − δ, c + δ). Since x=c + δ/2 is contained in (c − δ, c + δ), it also satisfies f(x) < u, so it must be contained in S" 3) Nowhere in the proof proves that : a<c<b
12. ## prove -0 =0

You converted a two line proof into a long,long story: We have : 0x =(0+0)x=0x+0x,by distributive property,and the axiom : x+0=x But : ox +0 =0x (by x+0=x again). Hence : ox =0 ,by cancellation law
13. ## prove -0 =0

It is a completely different business in proving the uniqueness of the zero ,from the proof : -0 = 0 How x0 = x0+x0 imply x0 =0?? Also this is the real zero as kavlas said,so we have the axioms of the reals.And one of them is : x+0 =x
14. ## prove -0 =0

I have seen a lot of mathematical proofs ,but a correct one with a wrong justification?? x*0=0 ,is a theoerem not a definition,it can be proved
15. ## Mastering Symbolic Logic

derive S&N from: 1. (V<->N)&S 2. (GvS)->(V&N) 3. S..............................1,Simplification 4. GvS...........................3,Addition 5. V&N............................2,4,M.Ponens 6. N...............................5,Simplification 7. S&N...........................3,6,Conjuction derive ~Z from: 1. G 2. (G&D)->~Z 3. G->~(Zv~M) 4. (Zv~M)vD 5. ~(Zv~M)....................................1,3,M.Ponens 6. D...............................................4,5,Disnjuctive Syllogism 7. G&D............................................1,6,Conjuction 8. ~Z.............................................2,7,M.Ponens derive SvE: 1. (D<->Z)->E 2. Z&D 3. Z->(Xv(D<->Z)) 4. X->S 5. Z....................................................2,Simplification 6. [Xv(D<=>Z)].....................................3,5,M.Ponens 7. (X=>S)&[(D<=>Z)=>E].......................1,4,CONJUCTION 8. SvE..................................................6,7 ,Constructive dilemma derive GvN: 1. ~U 2. CvB 3. (HvB)->(I&G) 4. C->U 5. ~C...............................................1,4,M.Tollens 6. B..................................................2,5,Disjunctive syllogism 7. HvB................................................6,Conjunction 8. I&G.................................................3,7,M.Ponens 9. G.....................................................8,Simplification 10. GvN................................................9,Conjuction derive K: 1. (XvP)&R 2. P->Q 3. X->Z 4. (ZvQ)->(K&J) 5.XvP.................................................1,Simplification 6. (X =>Z)&(P =>Q)...............................2,3,Conjuction 7. ZvQ.................................................5,6, Constructive dilemma 8. K&J..................................................4,7,M.Ponens 9. K......................................................8,Simplification derive ZvU: 1. WvA 2. V&G 3. ~N->Z 4. W->~N 5. (VvH)->(A->U) 6. W =>Z...............................................4,3,Hypothetical syllogism 7. V.......................................................2,Simplification 8. VvH....................................................7,Addition 9. A => U................................................5,8,M.Ponens 10. (W=>Z)&(A=>U)..................................6,9,Conjunction 11. ZvU...................................................1,10, Constructive dilemma derive I: 1. Tv~P 2. ~T&(Q->H) 3. (A->T)&(X->I) 4. (~T&~P)->(AvX) 5. i WILL let you do this last one
16. ## Divergent sequence

Will the above change the very fact that you do not know how to prove that the sequence does not converge ,by using the negation of the definition of convergence ??. No ,definitely no And the funny thing is that,as usually, i get red flags from the forum,approving your pompous ignorance and rudeness But i do not care .You know why?? Because that strengthens my belief : That nobody in this forum can give a proof the way that the OP asked
17. ## Proof with rigorous definition of limit (exponential function)

We have that: $lim_{x\to 0}\frac{e^x-1}{x} = 1$ if 0<x<1,then we have:$1\leq\frac{e^x-1}{x}\leq\frac{1}{1-x}$....................................................................1 if -1<x<0 ,then we have: $\frac{1}{1-x}\leq\frac{e^x-1}{x}\leq1$..................................................................2 And we observe that : $\lim_{x\to 0}\frac{1}{1-x} = 1$ ,hence $lim_{x\to 0}\frac{e^x-1}{x} = 1$ The proof of the inequalities (1) and (2) is based on the fact that : $\forall x(x\in R\Longrightarrow e^x\geq 1+x)$
18. ## Divergent sequence

He asked for the 'ε' approach ,did he not??? What is the matter,apart from not knowing how to solve the problem the way that the OP asked you have problems with the language??
19. ## Divergent sequence

This is the wrong approach definitely. The sequence has two sub sequences each having a limit. But the sequence itself has no limit. So the uniqueness of limit approach will not work. The proof of non existence of the limit for this particular sequence using the 'ε'definition is very difficult to prove indeed. I wonder if anyone in this forum can do it. But let us put things in their correct prospective . First of all the correct definition for convergence is the following: $x_{n}$ converges in R if and only if................. $\exists a (a\in R\wedge\forall\epsilon(\epsilon>0\Longrightarrow\exists k(k\in N\wedge\forall n(n\geq k\Longrightarrow |x_{n}-a|<\epsilon)))$ Meaning : There exists an aεR ,such for all ε>0 there exists a kεN such that for all $n\geq k$ $|x_{n}-a|<\epsilon$ Now the negation of the above statement ( which b.t.w is also very difficult to prove) is the following: $\forall a(a\in R\Longrightarrow\exists\epsilon(\epsilon>0\wedge\forall k(k\in N\Longrightarrow\exists n(n\geq k\wedge |x_{n}-a|\geq\epsilon))))$ Meaning: Given any real No a we must be able to find an ε>0 such that for any Natural No k we have to find a natural No n ,$n\geq k$ such that :$|x_{n}-a|\geq\epsilon$. At a first glance this is very difficult to assimilate ,let alone to ,to prove . However this is analysis . Where do we start ?? As we usually do with nearly all limits we start from the end of the statements (this is a trial and error process) and work our way opposite to proof that we have to write later. I.e in our case let us start with $|x_{n}-a|\geq\epsilon$ inequality. We observe that for n even the inequality becomes : $|1-a|\geq\epsilon$, and with n odd the inequality becomes : $|3-a|\geq\epsilon$. Well,half of the work is done. Can you carry on from here ...................................???
20. ## Limits

Obviously the -1 limit is of the :$lim_{n\to\infty}x-\sqrt{x^2+2x}$ ,which you will get,if you multiply by the : $\frac{x+\sqrt{x^2+2x}}{x+\sqrt{x^2+2x}}$ fraction,then divide numerator and denominator by x and then take the limit For the other limit do exactly the same and the result will be :$\frac{a-b}{2}$
21. ## open sets

Prove that (0,2) is open in $(R^2,d)$ where d is the discrete metric: .....................................................d(x,y) = 0 ,if x=y........................................................... ......................................................d(x,y) =1 ,if $x\neq y$..................................... Obviously we have to prove that: for all xe(0,2) ,there exists ε>0 such that B(x,ε) is a subset of (0,2) ,or for all xε(0,2) ,there exists ε>ο such that yεB(x,y) iplies $0<y_{1}<2$,where x =$(x_{1},0)$ and y=$(y_{1},y_{2})$ AM I correct??
22. ## inequalities,2 solutions

To clarify the case even further ,perhaps ,i must mention the logics behind solution No 1. in solution No1 it was proved that: if x>0 ,then $2+x^2 <0$. Now thru the law of contrapositive we can say: $\neg(2+x^2)<0\Longrightarrow\neg (x>0)$.........................................1 But we know that: $2+x^2\geq 0$..............................................................................................................2 Also from the law of trichotomy we have: $2+x^2\geq 0 \Longrightarrow\neg(2+x^2<0)$.......................................................3 From (1) and (3) and using hypothetical syllogism we can conclude: $2+x^2\geq 0\Longrightarrow\neg (x>0)$........................................................4 From (2) and (4) and using M ponens we have: $\neg(x>0)$...........................................................................................5 And since again by trichotomy law we have: x>0 or x<0 ( x=0 is not allowed by the nature of the inequality) and we have proved $\neg(x>0)$ we can conclude : x<0. As you may very well see we have no need in examining the case x<0
23. ## inequalities,2 solutions

Which law of logic was violated??
24. ## inequalities,2 solutions

Suppose we were given the following inequality :$\frac{x-2}{x}>x+1$ ,and the following two solutions were suggested: Solution No 1: for x>0 the inequality becomes (after multiplying across):$x-2> x^2 +x$ ,or $2+x^2<0$,but since $2+x^2\geq 0$ we have a contradiction ,hence the inequality is satisfied for all x<0 Solution No 2:$\frac{x-2}{x}>x+1\Longleftrightarrow\frac{x-2}{x}-(x+1)>0$ . And working on the left hand side of the equation we have : $\frac{-2-x^2}{x}>0$ ,or $\frac{2+x^2}{x}<0$,and since $2+x^2\geq 0$ we conclude that x<0 . Hence the inequality is satisfied for all ,x<0 Wich of the above solutions is right and which is wrong.Or are they both right,or are they both wrong??
25. ## sequence convergence

Can we consider that answer as a proof??
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