PoentWeptgj

2022-07-18

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

1. $a\cdot 0+0=a\cdot 0$ (additive identity postulate)

2. $a\cdot 0=a\cdot (0+0)$ (substitution principle)

3. $a\cdot (0+0)=a\cdot 0+a\cdot 0$ (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.

1. $a\cdot 0+0=a\cdot 0$ (additive identity postulate)

2. $a\cdot 0=a\cdot (0+0)$ (substitution principle)

3. $a\cdot (0+0)=a\cdot 0+a\cdot 0$ (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.

bgr0v

Beginner2022-07-19Added 13 answers

Just put steps 1,2 and 3 together and you obtain step 4.

$a\cdot 0+0=a\cdot 0=a\cdot (0+0)=a\cdot 0+a\cdot 0.$

$a\cdot 0+0=a\cdot 0=a\cdot (0+0)=a\cdot 0+a\cdot 0.$

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.

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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