Theorem: 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 transitive5. 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.

Question

Theorem: 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&#039;m lost here, wanna say its the transitive5.   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&#039;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.

bgr0v · Accepted Answer

Just put steps 1,2 and 3 together and you obtain step 4.  a  ⋅  0  +  0  =  a  ⋅  0  =  a  ⋅  (  0  +  0  )  =  a  ⋅  0  +  a  ⋅  0.

Awainaideannagi · Accepted Answer

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.

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