I was asked to write a rigorous proof for the following theorem:(adsbygoogle = window.adsbygoogle || []).push({});

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)...............................................by 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

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# Rigorous proof

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