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

I didn't insist on anything, I was trying to give a hint, and I still am not going to provide a detailed proof because I don't like spoon feeding anybody. If you really are driven to find someone else to do the work for you, these proofs can be found in the literature (and no, I'm not going to tell you which book).

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

Proofs by contradiction usually work well for the first two -- that is, assume that there isn't a unique zero and show that assuming that that leads to conclusions that break the axioms, and therefore there is a unique zero. Same with stating with a non-unique inverse.

 

Whether it is homework for a class or not, the rule remains the same. Members can give clue or hints (like the above), but we still don't just do all the work for someone else. If you want to post your proof or a few different ones in the interest of discussing them, that's fine. But, the forum justifiably so will always err on the side of caution. And the cautious approach here is that this looks like homework, so I really hope that no one just posts answers.

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

There's a (written or unwritten, I am not sure) policy on sfn that others should not do your homework for you. Try it yourself. If you get stuck or your attempts fail, then ask for help and present what you did/tried so far.

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)

I am a bit confused by this. Here is the equation:

 

[math]

a=b[/math]

[math]a^{2}=b^{2}[/math]

[math]a^{2}-b^{2}=0[/math]

[math](a+b)(a-b)=0[/math]

[math]a=b,-b[/math]

let [math]a=1[/math]

[math]1=1,-1

[/math]

 

I understand that this cannot be true but why does it work algebraically? To my understanding, if [math] a=b [/math], than [math] a^{2}=b^{2}[/math] but if [math] a^{2}=b^{2} [/math], than [math] a\neq b [/math] What am I not understanding?

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

The versin of the proof that I've always seen was 1=2, but I'm sure it could be rigged up for 2 + 2 = 5

 

1=1

-1=-1

-1/1 = 1/-1

root both sides

i/1 = 1/i

add 3/2i to both sides

3/2i + 1/i = i/1 + 3/2i

multiply both sides by i

3/2 + 1 = -1 + 3/2

2.5 = .5

 

Ok. so it didn't equil 1=2, but its the principle of the thing...

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]