triclino

Bignose ,i looked uppon many books of linear and abstract algebra and none of them gives a proof by contradiction. So please be so kind ,since you suggested and i pressume you know,as to produce,if you wish, the proof by contradiction

CAN somebody,please write the definition for the linear independence of the following functions?? [math] e^x,e^{2x}[/math]

Please!!! my intentions are not to throw anything at anybody's face. If you read my posts again in one of them i said i could not do the proof by using contradiction and i asked for more details. Yet i cannot do the proof by contradiction,and if anybody else can,please do so

How can you spoon feed me ,if the rigorous proofs of the above exist in books as you claim. All you had to do from the very beginning was to refer me to the particular book(s) ,instead of suggesting a method which you cannot follow to the very end

So we want to prove that: There exists a unique y for all x such that: x+y =x And now using contradiction we must negate the above statement AND the question now is what is the negation of the above statement?? In quantifier form the above statement is: [math]\exists y!\forall x[ x+y=x][/math]...... ([math]\exists !y[/math] means there exists a unique y) , which by definition is equivalent to: [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 in words is: there exists a, y for all x such that :x+y =x AND FOR all y,z :if for all x ( x+y=x) and for all x(x+z=x) ,then y=z Now to negate the above is not an easy task. That is why i asked Bignose to give more details when he suggested the proof by contradiction. However if anybody insists in a proof by contradiction ,i think, she/he must show more details

I am sorry, i should have pointed out that rigorous proof means ,a proof in which every line is justified by the appropriate axiom. The proof by contradiction that you suggest ,i do not know how to do , so please if you wish show more details

I am not a student and this is not homework . if you noticed this a rigorous proof and not simply a proof .next step to rigorous proof is a formal proof. I am simply interested to see how other people apart from my self are approaching the problem

given a set with the symbols : " + " for addition " - " for inverse the constants : 1 0 AND the axioms: for all a,b,c : a+(b+c) = (a+b) + c for all a : a+0 = a for all a : a +(-a) =0 for all a,b : a+b = b+a Give a rigorous proof of the following: 1) 0 is unique 2) the inverse of a ,-a is unique 3) for all x,y : -( x+y) = -x +(-y)

HERE you have : 1=1 OR -1 = 1 ,which is a true statement. The statement : 1 = 1 AND -1 = 1 IS a wrong statement Other statements that are correct are: 1=1 or 3>9 -3 = 7 or ln 1 = 0 [math]\sqrt{x^2} = |x|[/math] or 1<0 1=1 = 2 or 2 + 3 =5 But the statements: -3 =7 or 7<0 1=1 and 1<0 are wrong

obviously when you root both sides the result is not :i/1 =1/i ,because : i/1 =1/i <=====> [math] i^2 = 1\Longleftrightarrow -1 = 1[/math] Now the statement, -1 =1 in any line in any proof can result in any conclusion right or wrong For example can result to false statements ,like: 5=7 ,1>4 , [math] x^2<0[/math] ln(-2) = 0 e,t,c ,e,t,c in the following way: -1 =1 [math]\Longrightarrow [(-1 =1 )[/math]or [math](x^2<0)]\Longleftrightarrow[(-1\neq 1)\Longrightarrow (x^2<0)][/math] and since [math] -1\neq 1[/math] we conclude that: [math] x^2<0[/math]