rigorous proof

[evagelos]

I was asked to write a rigorous proof for the following theorem:

0x = 0 ,for all x.

Is the following rigorous proof correct??

1) 0x = 0x+0.............................................. .......by using the axiom:for all ,a : a+0=a

2) x+(-x) = 0................................................. ...by using the axiom: for all ,a: a+(-a) = 0

3) 0x = 0x +(x+(-x))..........................................by substituting (2) into (1)

4) 0x+(x+(-x)) = (0x+x)+(-x)...............................by using the axiom:for all a,b,c:a+(b+c)=(a+b)+c

5) 0x = (0x+x)+(-x)............................................by substituting (4) into (3)

6) 0x+x = x+0x.............................................. ....by using the axiom:for all a,b:a+b=b+a

7) 0x = (x+0x)+(-x)............................................by substituting (6) into (5)

8) 1x = x................................................. .........by using the axiom:for all,a:1a = a

9) 0x = (1x+0x)+(-x)..........................................by substituting (8) into (7)

10) 1x+0x = (1+0)x............................................ by using the axiom: for all a,b,c:(a+b)c= ac+bc

11) 0x = (1+0)x+(-x)..........................................by substituting (10) into (9)

12) 1+0 = 1................................................. ....by using the axiom:for all,a:a+0=a

13) 0x = 1x+(-x)...............................................b y substituting (12) into (11)

14) 1x = x................................................. .......by using the axiom:for all,a:1a = a

15) 0x = x+(-x)................................................ by substituting (14) into (13)

16) x+(-x) = 0................................................. .by using the axiom:for all,a:a+(-a) = 0

17) 0x = 0................................................. ......by substituting (16) into (15)

Thanx ,any help will be wellcomed

[EnumaElish]

I see redundant steps. How about:

Let n be an integer.
0x = n
0x - x = n - x
(0 - 1)x = n - x
-x = n - x
x - x = n
What can you say about n?

[evagelos]

I see redundant steps. How about:

Let n be an integer.
0x = n
0x - x = n - x
(0 - 1)x = n - x
-x = n - x
x - x = n
What can you say about n?
Thanks ,a good way to find how much is 0x.

B,t.w, x is a real No
There is an even shorter proof :

0x =0 <===> 0x +x =0+x <===> x(0+1) = x <===> x=x.

But in a rigorous proof we must show the axioms involved .

Where are the redundant steps??

Thanks again

[EnumaElish]

You are correct, I should have written "n is real."

As for redundancy, I think you can start with 6; because 0x + x = x + 0x as a postulate. I am not saying you are wrong, but I do not see why you cannot start with 6.

[statdad]

"But in a rigorous proof we must show the axioms involved."

No, in a rigorous proof you must make sure details are explicit. Stating the axioms is not a requirement.

Is there a reason for this discussion?

[evagelos]

No, in a rigorous proof you must make sure details are explicit. Stating the axioms is not a requirement.



yes i agree after that statement there is no reason for further discussion

[EnumaElish]

evagelos, I think statdad is referencing to widely accepted standards of proof in general math, statistics, and related fields.

If your instructor/professor has explicitly asked you to state each axiom, then those specific instructions take precedence over general ones.

[statdad]

EnumaElish is correct; my response was a little terse, my apologies, but I simply did not want to get involved in another long "gotcha" post involving what is and what is not a "rigorous proof".

Never email in haste - never post on a forum in haste: words I need to live by.