Simple proof?

x*0=x*(0+0)=

x*0+x*0.

The first equality is because 0 is an additive identity;

the second from the distributive law.

Now add the negative of x*0 to both sides:

0=x*0+(-)x*0=

(x*0+x*0)+(-)x*0=

x*0+(x*0+(-)x*0)

=x*0 +0=

x*0.

The first equality is from the definition of a negative;

the second from the previous result; the third from

associativity; the fourth from the definition of a negative;

the fifth from the fact that 0 is an additive identity.

this is a simple question but has a deeper implication. the reason why (usually) X*0=0 is that 0 is the additive identity element for Z, Q , R, and C. N however does not have an additive identity simply because 0 is not a member of N.

therefore, although X*0=0 for our usual Z, Q, R, and C operations, it might not be always so.

x has an additive inverse, -x.

0 = (-x) + x

= (-x) + x*1 [ x = x*1 ]

= (-x) + x*(1+0) [ 1 = 1 + 0 ]

= (-x) + (x*1 + x*0) [ distributive law ]

= (-x) + (x + x*0) [ x*1 = x ]

= ((-x) + x) + x*0 [ Associative rule for addition ]

= 0 + x*0 [ -x + x = 0 ]

= x*0 [ 0+z = z ]

Well, if you were to see this question,

x.0 = x.(1-1) = x.1 - x.1 = x - x = 0

(Unless you were to ask another stupid question to prove x.1=x)

So that's why kids nowadays are taught:

x.0=0.

any number times 0 is going to be 0 simple as that you cant multiple something by nothing.

Simple proof? Oh yes indeed.

Without much fancy ‘footwork’:

0=1/infiniry or x/infinity =0 iff x is finite (negative infinity<x<infinity (or x is finite) )

Have fun

I'm sure this isn't proper, but:

x*2=x+x

X*1=x

X*0=

(nothing = 0)

let 0=a-b

then x.0 = x(a-b)

=xa - xb

but if a-b=0 then a=b

xa-xb = xa-xa = xb-xb = 0

so x.0 = 0