# If a is a real number, then a*0=0. 1. a*0+0=a*0 (additive identity postulate) 2. a*0=a*(0+0) (substitution principle) 3. a*(0+0)=a*0+a*0 (distributive postulate) 4. a*0+0=a*0+a*0 I'm lost here, wanna say its the transitive 5. 0+a*0=a*0+a*0 (commutative postulate of addition) 6. 0=a*0 (cancellation property of addition) 7.a*0=0 (symmetric postulate) So I'm not sure what to put down for the 4th step. The theorem and proof were given and I had to list the postulates for each step.

Theorem: If a is a real number, then $a\cdot 0=0$.

1. $a\cdot 0+0=a\cdot 0$ (additive identity postulate)
2. $a\cdot 0=a\cdot \left(0+0\right)$ (substitution principle)
3. $a\cdot \left(0+0\right)=a\cdot 0+a\cdot 0$ (distributive postulate)
4. $a\cdot 0+0=a\cdot 0+a\cdot 0$ I'm lost here, wanna say its the transitive
5. $0+a\cdot 0=a\cdot 0+a\cdot 0$ (commutative postulate of addition)
6. $0=a\cdot 0$ (cancellation property of addition)
7. $a\cdot 0=0$ (symmetric postulate)

So I'm not sure what to put down for the 4th step. The theorem and proof were given and I had to list the postulates for each step.
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bgr0v
Just put steps 1,2 and 3 together and you obtain step 4.
$a\cdot 0+0=a\cdot 0=a\cdot \left(0+0\right)=a\cdot 0+a\cdot 0.$

Awainaideannagi
Step 4 (not 5) is the one you are confused on. And it is indeed the transitive property of equality.
But step 2 actually skips a step and is not at all the associattive postulate. You should first say 0=0+0 (additive identity) then (2) follows by substitution.