In Peano arithmetic addition is usually defined with the following two postulates: ( 1

auto23652im 2022-07-07 Answered
In Peano arithmetic addition is usually defined with the following two postulates:
( 1 a ) : p + 0 = p
( 2 a ) : p + S ( q ) = S ( p + q )
Lets say I put the successor term of the second postulate on the left? Namely:
( 1 b ) : p + 0 = p
( 2 b ) : S ( p ) + q = S ( p + q )
Are these two definitions of addition equivalent?
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)

amanhantmk
Answered 2022-07-08 Author has 17 answers
Logically perhaps. But note that it is clear that the first set (a) is consistent: the right-side addend is either 0 or a successor, so at most one clause applies in every situation.

This is not the case for the second set (b), where there are additions to which both clauses apply and you will have to decide which one you will use. For example, consider S ( t + 0 ). The first clause says that this should be reduced to S ( t ). The second clause says that it should be reduced to S ( t + 0 ). Now it is true if we kept on reducing, we would eventually get to the same place, S ( t ), regardless of which steps we took. But to be certain that your definition makes sense, you would need an induction proof that shows that there is no reduction for any initial expression where the result depends on the particular steps taken. (This property is called “confluence”; without it we can't really say that an expression has a well-defined ‘value’.)

With (a) there are never any choices so no such proof is needed. At most one rule applies to every expression, you apply the one rule until there is no longer a + sign and you have to stop.

You should probably change (2b) to say 0+p=p for this reason.

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

Esmeralda Lane
Answered 2022-07-09 Author has 7 answers
Well, they are certainly not logically equivalent: without making any further assumptions/axioms, you cannot derive the one from the other.
Now, if you include the axiom of induction, then you can derive the second definition from the first… but I am not sure if you can derive the first from the second even if you have the inductive axiom.
I do know that if you have the axiom of induction, you can derive the following pair of statements from the first definition:
0 + p = p
s ( p ) + q = s ( p + q )
and from this pair, you can derive the first pair using induction. So, within the context of the inductive axiom, those two pairs are equivalent.

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 2022-06-29
I am beginning real analysis and got stuck on the first page (Peano Postulates). It reads as follows, at least in my textbook.

Axiom 1.2.1 (Peano Postulates). There exists a set N with an element 1 N and a function s : N N that satisfies the following three properties.
a. There is no n N such that s ( n ) = 1.
b. The function s is injective.
c. Let G N be a set. Suppose that 1 G, and that g G s ( g ) G. Then G = N .

Definition 1.2.2. The set of natural numbers, denoted N , is the set the existence of which is given in the Peano Postulates.

My question is: From my understanding of the postulates, we could construct an infinite set which satisfies the three properties. For example, the odd numbers { 1 , 3 , 5 , 7 , }, or the powers of 5 { 1 , 5 , 25 , 625 }, could be constructed (with a different s ( n ), of course, since s ( n ) is not defined in the postulates anyway). How do these properties uniquely identify the set of the natural numbers?
asked 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.
asked 2022-05-29
Hi guys I just need to know if my answer is right.

The question is

1) Euclids 4 t h postulate is "That all right angles are equal to one another". Why is this not obvious?

My answer:

When I read this question I am like it is obvious, so I got kind of confused. But I took a crack at it anyways.

If you read the postulate. This is not obvious because you dont know if the right angle is a right angle. What if a triangle was drawn differently but with one line perpendicular to another. You need to know if the line is perpendicular or not, and that is why it is not obvious.

Could someone tell me if that is right or should I add in more info.
asked 2022-05-09
Suppose I have this boolean expression:
W'XYZ + WX'YZ + WXY'Z + WXYZ' + WXYZ
How would I go about simplifying this without using a K-map? Using K-map, the simplified form is XYZ + WXY + WXZ + WYZ. I read about the redundancy theorem somewhere, would rather not use that as well.
asked 2022-08-16
Two questions came to mind when I was reading the proof for Bertrand's Postulate (there's always a prime between n and 2 n):
(1) Can we change the proof somehow to show that: x > x 0 , there exists a prime p [ x , a x ], for some a ( 1 , 2 )?
(2) Suppose the (1) is true, what is the smallest value of x 0 ?
I'm not sure how to prove either of them, any input would be greatly appreciated! And correct me if any of the above statement is wrong. Thank you!
asked 2022-07-01
In Introduction to Metamathematics, Kleene introduces a formal system where the first three postulates in the group for propositional calculus are:
1 a . A ( B A ) 1 b . ( A B ) ( ( A ( B C ) ) ( A C ) ) 2. A , A B B
As far as I understand, 1 a and 2 are typical for Hilbert-style deductive systems, but 1 b is not. A more traditional choice, serving pretty much the same purpose (e.g. proving A A to start with) would have been:
( A ( B C ) ) ( ( A B ) ( A C ) )
What is the rationale for the unique choice made for 1 b in Introduction to Metamathematics?
asked 2022-05-28
I don't find Postulate 13 of Spivak's Calculus trivial, nor can I understand why it's true.
Postulate 13: Every non-empty set of real numbers that is bounded above has a least upper bound (sup).
Why is this postulate true? Any proof/intuition behind it?

Edit: Let me pose a few questions.
1. Suppose I decide to devise a pathological function R R which is bounded above but has no supremum. Why will I fail to find such f? If you simply take it as an axiom, there's no guarantee I won't be successful.
2. Suppose S is an arbitrary non-empty set of real numbers that is bounded above. Does there exist an algorithm to determine s u p ( S )?

New questions