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.

PoentWeptgj

PoentWeptgj

Answered question

2022-07-18

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 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.

Answer & Explanation

bgr0v

bgr0v

Beginner2022-07-19Added 13 answers

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

Awainaideannagi

Beginner2022-07-20Added 5 answers

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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