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 2022-07-18 Answered
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.
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (2)

bgr0v
Answered 2022-07-19 Author has 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.

We have step-by-step solutions for your answer!

Awainaideannagi
Answered 2022-07-20 Author has 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.

We have step-by-step solutions for your answer!

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

asked 2022-07-06
I read that the following can be proved using Bertrand's postulate (there's always a prime between n and 2 n): N N , there exists an even integer k > 0 for which there are at least N prime pairs p, p + k.
But I have no idea how to prove it. Any help would be much appreciated.
Many thanks.
asked 2022-07-15
Is there a form of non-Euclidean geometry in which perpendicular lines never cross, or cross twice or something?
asked 2022-08-17
Bertrand's postulate says:
For every n > 1 there is always at least one prime p such that n < p < 2 n.
Is the following statement:
For every n > 3 there is always at least one prime p such that F n < p < F n + 1 ( F n is n-th Fibonacci number).
also valid?
If it is invalid, is there a finite or infinite number of ns such that there is no prime between F n and F n + 1 ?

This question is inspiblack by another question. I feel intuitively that it may be interesting, but don't have enough number theory background to tackle it.
asked 2022-05-09
Bertrand's postulate

claim :

for all 5 < n N there is at least two prime numbers different in the opening interval ( n , 2 n ).Use reinforcement of Bertrand's postulate in order to prove that :
for all 10 < n N Exists that has at least two Prime factors in the factorization of n ! which appear with a power of 1.
Example : n = 11 the primes 7,11 are such prime's meet the conditions.
But , for n = 10 it dosen't meet the conditions because only 7 appears with power 1 in the factorization of 10!.

Hint : Consider two cases, n even or n odd

Attempt:
we need to use the claim in order to Implement the solution

Case (1): n even
if n is even we can rewrite n = 2 t
if we use the claim we can get ( 2 t , 4 t )

Case (2): n odd
if n is odd we can rewrite n = 2 t + 1
if we use the claim we can get ( 2 t + 1 , 2 ( 2 t + 1 ) )
asked 2022-05-07
How can I prove the following two questions:

Prove using Peano's Postulates for the Natural Numbers that if a and b are two natural numbers such that a + b = a, then b must be 0?

Prove using Peano's Postulates for the Natural Numbers that if a and b are natural numbers then: a + b = 0 if and only if a = 0 and b = 0?

I understand the basic postulates, but not sure how to apply them to these specific questions. The questions seem so basic and obvious, but when it comes to applying the postulates I am lost.
asked 2022-06-25
Lorenz, Every single line through a point within an angle will meet at least one side of the angle.
I know I have to Show that the parallel postulate 5 implies lorenz, and then lorenz implies parallel postulate 5.
Assume postulate 5 . So we are given AB and a point C not on AB. Choose B on AB draw CD to construct angle ECD= angle BDC.
I just don't get what Lorenz postulate means. Thats where I am getting stuck.
asked 2022-05-09
I came up with this question- how would you show 4 not equal to 6 (or m not equal to m+n ( n not 0)), using only Peano's Postulates?

I can see a number of things go wrong- for instance the Principle of Mathematical Induction seems to fail. Also possibly 0 seems to be in the image of the successor function in that case.

New questions