Show that 0 is an absorbing element? (First Year Uni, Pure Maths)

1) Show that 0 is absorbing element for x, i.e. that

for all a in set Z : 0a = 0,

using the basic properties of + and x in set Z.


And I'll sneak in a second question here...

2) Denote by -1 the opposite of 1. Prove that (-1)a = -1 for any a in set Z.


I'm really unsure about how to answer these after several attempts. Will appreciate any help, even if it's not a solution -- tips or recommended material for this stuff is fine.

This seems so pointless lol

0 is 0, nothing more.

That's coming at it from the wrong angle. You need to show that 0a=0a+0a and then use cancellation to show that 0a=0; you're not showing that "if 0a=0 then (0+0)a=0".

Thanks for the comments. After reading through decided to give it another shot... though I still feel kinda uncomfortable with my answer. Any more help would be useful.

An attempt:

Question
1) Show that 0 is an absorbing element for x, i.e. that

for all a in set Z : 0a = 0,

using the basic properties of + and x in set Z.

Solution attempt

0a + 0a = 0(a + a)

Using distributive law we can say

for all a, 0 in set Z : 0(a + a) = 0a + 0a.

If, for all a in set Z : 0a = 0, is true then

a + a = a and 0a + 0a = 0.

Addition is commutative in set Z, so

a + a = a + a.

Using the neutral element property, we can say

a + a = a + a = a for all a in set Z

Therefore, 0 satisfies the value for a.



In summary,

0 = a.

So,

for all a, 0 in set Z : 0(a + a) = 0a + 0a,

can be rewritten as,

for all a in set Z : 0a = 0.

Nope, you've done this the wrong way round. You've said "if a=0 then a satisfies a logical consequence of a being equal to zero". Essentially, you've said "assume A is true; A=>B and B is true, therefore A is true", which is bad logic. Another example of doing this would be to say: assume 1=2. Then 0*1=0*2, i.e. 0=0, which is true, and therefore 1=2. But this is obviously nonsense.

However, you can use the fact that 0=0+0, which you know is true without making any assumptions beyond the axioms, and work from there. Try multiplying both sides by a and seeing what happens.

You were definitely right. My post was confusing.

Solution attempt 3

0a + 0a = 0a.

Additive Inverse...

0a + (-0a) = 0.

Therefore,

(0a + 0a) - 0a = 0a - 0a,

becomes 0a = 0.

Yup. But all of your "therefore" and "becomes" waffle is a bit unnecessary. You could just set it out like:

$0+0=0$
$\Rightarrow (0+0)a=0a+0a=0a$
$\Rightarrow 0a+0a+(-0a) = 0a+(-0a)$
$\Rightarrow 0=0a$

Nice and compact