triclino

THANK YOU. Would you consider the following as rigorous: 1) (-a)0 =0..............................................by using the theorem:[math]\forall x(x0=0)[/math] 2) b+(-b)=0.............................................by using the axiom:[math]\forall x(x+(-x)=0)[/math] 3)(-a)[b+(-b)] =0....................................by substituting (2) into (1) 4) (-a)[b+(-b)] = (-a)b +(-a)(-b)...............by using the axiom:[math]\forall x\forall y\forall z[x(y+z)=xy+xz][/math] 5) (-a)b+(-a)(-b)=0..................................by substituting (4) into (3) 6) 0b =0.....................................................by using the theorem:[math]\forall x(0x=0)[/math] 7) a+(-a)=0..............................................by using the axiom:[math]\forall x(x+(-x)=0)[/math] 8) [a+(-a)]b = 0........................................by substituting (7) into (6) 9) [a+(-a)]b = ab+(-a)b............................by using the axiom:[math]\forall x\forall y\forall z[(x+y)z = xz+yz][/math] 10) ab+(-a)b =0........................................by substituting (9) into (8) 11) ab+(-a)b = (-a)b+ab............................by using the axiom:[math]\forall x\forall y(x+y=y+x)[/math] 12) (-a)b+ab =0.........................................by substituting (11) into (10) 13) (-a)b+(-a)(-b)=(-a)b+ab......................by substituting (12) into (5) 14) (-a)(-b) = ab........................................by using the cancellation law:[math]\forall x\forall y\forall z( x+y=x+z\Longrightarrow y=z)[/math]

can somebody give the definition of rigorous proof?? How for example one would write a rigorous proof for: (-x)(-y) = xy for all x,y. thank you

Since the limit of [math] \frac{lnx}{\frac{1}{sinx}}[/math] is indeterminate as x goes to zero ,we use L'hospital"s rule and: [math]\ lim_{x\to 0}\sin x \ln x[/math] = [math]\lim_{x\to 0}-\frac{\frac{1}{x}}{\frac{cosx}{sinx^2}}[/math] = [math]\lim_{x\to 0}-\frac{sinx^2}{xcosx}[/math] = [math]\lim_{x\to 0}-\frac{sinx}{x}\frac{sinx}{cosx}[/math] = -(1).[math]\frac{0}{1}[/math] = 0 Now since [math] lnx^{sinx}[/math] goes to 0 , [math] x^{sinx}[/math] goes to 1 as x goes to zero Note that [math] \lim_{x\to 0} {xlnx}\neq 0[/math] ,but 0.[math]-\infty[/math] which is indeterminate

Somebody asked me if i knew what the Rule of conditional proof is and how is used in mathematics and particularly in analysis. can anyone help with respect to that??

Some body sujested the following proof which i do not understand: [math](x,y)\in f\Longrightarrow (x,y)=(1,a)\vee(x,y)=(2,b)[/math][math]\Longrightarrow (x,y)=(1,a)\vee (x,y)=(2,b)\vee (x,y)=(1,b)\vee (x,y)=(2,a)[/math] [math]\Longrightarrow (x,y)\in A\times B[/math] I do not know how we go from the 1st implication to the 2nd

Thanks. but it is the details of the proof that i got confused. For example in proving that :[math] f\subset A\times B[/math], i started: Let [math](x,y)\in f\Longrightarrow (x,y) = (1,a)\vee (x,y)=(2,b)[/math],then??

Given that the definition of a function f, from A to B (sets) is: 1) f is a subset of AxB............................................................................... [math]f\subset AxB[/math] ,and 2)for all xεΧ ,there exists a unique yεΥ SUCH that (x,y)εf .......................................... [math]\forall x [x\in X\Longrightarrow\exists! y(y\in Y\wedge (x,y)\in f)][/math] prove that the following is a function from A to B,where A = {1,2} B = { a,b } f = { (1,a),(2,b) }

First of all i never quoted anybody and write underneath ga ga ra ba dum,thus showing disrespect for his/her post. Ga ga ra ba dum was intended to ease the tension and cause laughter. Sense of humor in any human activity,i think is absolutely necessary

Ga ga ra ba dum . Ga ga ra ba dum

Such a town does not exist, at least on earth. There is no difference between "for all" and "for any","ANY mathematician- logician" can tell you that

SO if somebody tells you that any bus goes to L.A YOU gonna be waiting for bus No 9, for example. I am pretty sure you gonna be the last one to arrive in L.A

In wikipedia the proof that i could get is the following one ; "Let V be the vector space of all functions of a real variable t. Then the functions [math]e^t[/math] and [math]e^{2t}[/math] in V are linearly independent. Proof Suppose a and b are two real numbers such that [math]ae^t + be^{2t}[/math] = 0 for all values of t. We need to show that a = 0 and b = 0. In order to do this, we divide through by [math]e^t[/math] (which is never zero) and subtract to obtain [math]be^t[/math] = −a. In other words, the function [math]be^t[/math] must be independent of t, which only occurs when b = 0. It follows that a is also zero." Which very much implies that the definition of inderendence of functions is ; for all a,b,t : [math]ae^t + be^{2t}=0\Longrightarrow a=b=0[/math] But again in this definition if we put a=1,b=-1 ,t=0 ,then the definition gives a false unswer,because [math] ae^t +be^{2t}=0[/math],but[math] a\neq 0[/math] and [math] b\neq 0[/math]. And since in logic : true implying false is ,false ,the definition brakes down for these values |

Your definition of linear independence would make the functions: f(x) = x and g(x) =2x linearly independent,because for a=1 ,b=1 there exists,x=2 such that ,[math]ax +2bx\neq 0[/math] But are these two functions linearly independent??

I am sorry ,but i fail to see the difference between "for all x" and "for any x". Besides taking the negation of the above formula we get: there exists a,b ,x such that [math]ae^x +be^{2x}=0[/math] and [math]a\neq 0[/math] and [math]b\neq 0[/math],a formula true for a=1,b=-1 and x=0. Hence the formula for linear independence is not true ,thus the functions are linearly dependent

The problem with this definition is the following: If we define the functions to be linearly independent,iff ,for all a,b,x [math] ae^x+be^{2x}=0\Longrightarrow a=b=0[/math],then in this definition if we put: a=1,b=-1,x=0 we have [math] ae^x + be^{2x}=0[/math],but [math]a\neq 0 [/math]and [math]b\neq 0[/math]. Which shows that the definition is not applicable. Any suggestions??

Thanks . But in a system where the primitive symbols are: +,=, and the constants are : 0 ,and the only axioms are : 1) a+ (b+c) = (a+b) + c ,for all a,b,c 2) a+b = b +a ,for all a,b 3) a +0 =a ,for all a the substitution : x= a+b+c ,would be allowed??

What theorem or axiom allow us to do the substitutions : x= a + b ,or y= a + b + c, or x= 2y + z^2 e.t.c , e.t.c some times in a mathematical proof ??

IN LINE (3) of the formal proof, the assumption is not to start a contradiction or otherwise called RAA,But to start a conditional proof. DO you know the difference between ,the law of conditional proof and that of RAA (= Reductio ad Absurdum ) A book that explains quite well the difference between the two and gives examples is: THEORY AND PROBLEMS OF LOGIC ,in Schaum's OUTLINE SERIES. I repeat in a formal proof there are not any hidden formulas. everything is explicitly mentioned Your assumption is an imaginary one . IF you think that you can write a formal proof with the assumption [math]y\neq z[/math] then do it .

O.K ,thanks tomorrow i will produce the 1st of the proofs,because now is very late in the island of Corfu and i must get some sleep Merged post follows: Consecutive posts merged Here is the direct formal proof for the uniqueness of zero. But before we start the proof let: (A) denote an axiom and (L) a law of logic. 1) for all ,x : x + 0 = x......................................................................(A) and in quantifier form : [math]\forall x ( x + 0 =x)[/math] 2)there exists a ,y ,for all , x : x + y = x........................................from (1) and using Existential Introduction (L) and in quantifier form: [math]\exists y\forall x ( x + y =x)[/math] 3)for all ,x : x + y = x and for all , x : x + z = x.........................................an assumption to start the conditional proof (L) and in quantifier form : [math]\forall x ( x + y = x)[/math] and [math]\forall x ( x + z = x)[/math] The above assumption tell us the following: Let us assume that y and z are any two identity elements for which the identity axiom holds 4) for all ,x : x + y =x...................................................................from (3) and using addition elimination (L).Another name for addition elimination is Conjunction elimination. and in quantifier form : [math]\forall x ( x + y = x)[/math] 5) z + y = z................................................................................from (4) and using Universal Elimination where we put x=z (L) 6) for all ,x : x + z = x...................................................................from (3) and using again addition elimination (L) And in quantifier form : [math]\forall x ( x + z = x)[/math] 7) y + z = y ................................................................................from (6) and using again Universal Elimination where we put x= y (L) 8) for all,a for all , b : a + b = b + a....................................................(A) and in quantifier form : [math]\forall a\forall b ( a + b = b + a )[/math] 9) z + y = y + z............................................................................from (8) and using Universal Elimination where we put a = z and b = y (L) 10) y + z = z ...............................................................................by substituting (9) into (5) (L) 11) y = z ....................................................................................by substituting (10) into (7) (L) 12) [math]\forall x ( x + y = x )[/math] and [math]\forall x ( x + z = x )[/math][math]\Longrightarrow y= z[/math].......................................from (3) to (11) and using the law of conditional proof (L) 13) [math]\forall y\forall z [ (\forall x ( x + y = x)\wedge\forall x ( x + z = x ))\Longrightarrow y = z][/math]..........................................................from (12) and using Universal Introduction for y and z (L) This last statement tell us that: for any two identity elements in the set for which the identity axiom applies ,then these two elements are identical SO far we have proved the following: [math]\exists y\forall x ( x + y = x )[/math] and [math]\forall y\forall z[(\forall x ( x + y = x )\wedge\forall x ( x + z = x))\Longrightarrow y= z ][/math] ,which is the definition of uniqueness of zero As you very well know to have a proof by contradiction ,at least in one of the steps of the formal proof we must have the appearance of a contradictory statement of the form : Q AND NOT Q. Certainly we do not have that in the above proof . I may add also that a similar formal proof appears in ANGELO'S MARGARIS book ,named : FIRST ORDER MATHEMATICAL LOGIC ,page 121. In the same book on page 109 ,an excellent definition of the uniqueness appears as well

When we are in doubt for the nature of a proof ,whether it is contradictory or not ,one must use "the top gun" of logic-mathematics scientific investigation,the formal proof or otherwise called "the stepwise proof" In a formal proof each step of the proof is justified and the laws of logic and axioms,theorem or the definitions involved in the proof ,are explicitly mentioned. And very clearly one ,with no doubts at all , can decide whether the proof is contradictory or not. Hence ,if the forum allows ,I am prepared to produce such a proof in which the non contradictory nature of the uniqueness proof will triumphantly emerge. In contrast, I am also prepared to produce a complete formal proof using contradiction

me no understand ( ANGELO MARGARIS ,first order mathematical logic,page109) By the way,what is the difference between a syntactical and a semantical proof?? Why, do you have a secondarystream logic ?? Merged post follows: Consecutive posts merged Mention one book that labels the uniqness proof as a contradictory one Thats a clever way avoiding the whole issue

A direct proof for the uniqness of zero consists of two parts: .....................................PART A..................................... In part A we prove the existence of zero: From the axiom : for all x ( x + 0 =x) ,we simply say ,there exists a y for all x such that x + y = x . So far we have proved the existence of an identity element .......................................PART B........................................... In part B we prove the uniqness of y WE SAY lets bring in ANY other element z and see what happens : For those two elements we have the same axiom holding i.e for all x : x + y = x........................................................1 for all x : x + z = x...........................................................2 Now in (1) we put x=z and in (2) we put x= y and the result is; z + y = z and y +z = y and since z + y = y + z ,we have y=z THERE IS NOTHING CONTRADICTORY SO FAR. TO SAY THAT: ASSUMING THAT THERE DOES NOT exists one element ,but there exist two is WRONG. HOW do you know ,that if there does not exist one bus in a town there exist two and not a million In a contradictory proof you have to negate both parts A and B. SO you can say: O.K , LETS assume that: The statement,that there exists a unique y such that: for all x ( x + y =x ) it is not true what happens ?? 1) for all y there exists an x such that ( x + y = x).This the negation of PART A ......................................OR............................................. 2) there exist a y and there exist a z such that : for all x ( x + y =x) and for all x ( x + z = x) BUT ALSO [math] y\neq z[/math] .This is the negation of PART B. And in examining both cases we must end up with a contradiction

Mr Moderator ,thank you for your wellcoming to the forum and your valuable advice but, a proof that cannot be found in books we cannot call homework. The forum that Bignose suggested ( MHF) ,have two proofs ,one for the uniqness of zero and one for the uniqness of the inverse ,which are not proofs BY CONTRADICTION. The book that Bignose suggested it is difficult to get. SO FOR ONE MORE TIME I BEG THE WHOLE FORUM IF ANYONE KNOWS THE PROOF BY CONTRADICTION,PLEASE WRITE IT HERE. Note: in my post #7 i pointed out the difference between a direct and indirect proof ( = contradiction),which i suspect many people mix up,for that particular proof ,that is

The proof written in the MHF that you refered me to, IS NOT A PROOF BY CONTRADICTION

Producing a particular proof YOU DO NOT HELP ME with my work because the contradictory proof it never crossed my mind,untill you mentioned it, YOU SIMPLY add valuable knowledge to the forum