Mathematical Reasoning and Proof

This assignment focuses on mathematical reasoning concepts such as induction and strict order relations. It delves into proving statements using mathematical induction and analyzing the properties of a strict order relation defined on the successor set of natural numbers. The assignment includes step-by-step solutions and explanations for various proofs, demonstrating how to apply these concepts in practice.

Running head: MATHEMATICAL REASONING
Mathematical Reasoning
Student
Institutional Affiliation

2
MATHEMATICAL REASONING
QUESTION 1
With regards to the usual arithmetic operations in conjuction with the property of induction on
nutural numbers, addition by reccursion from successor fuction is defined as
n+ 0=n ∀ n ∈ N
n+ ( k +1 )=(n+k )+1
Our task is to prove that n+ k=k + n by inducction on n
It clearly follows that 0+k=k=k +0 since the set of natural numbers is the closure of the empty
set under succesor.
NB: The empty set is 0 that is 0=ϕwhich has no predecessor
Now suppose n> 0be the successor of ksuch that n=k +1
Then it follows that
1+n=1+(k + 1)
¿( 1+ k )+1 by definition of addition property
¿ n+1 …………….*
{Remember 1+n=S ( 0 ) +n=S ( 0+ n )=S ( n ) }
Now with assuption that n+ k=k + n for some k in the set of natural numbers (k ∈ N ), it can
clearly be deduced that
n+ ( k +1 ) = ( n+ k ) +1 by definition of additive operation
( k +n ) +1with regards to the assumption above
¿ k + ( n+1 ) by definition of +¿ ………………**
Hence the proof

3
MATHEMATICAL REASONING
Clarification:
From * the statement holds for S ( n )
That is 1+n=n+1 but 1=S ( 0 ) → n+S ( 0 ) =S ( 0+n ) =S ( n )
With the assumption that n+k=k + n ,we’ve also seen that the statement holds for S( n+ k ) i.e.
n+ ( k +1 )=n+ S ( k )=S ( n+k ) since k +1=S (k ) by definition
QUESTION 2
A relation < on the succesor set S(x ) is a strict order if
a) Irreflexive: S ( n ) < S( n) does not hold for any S ( n ) a predicessor set of S (x )
b) Asymmetric: if S ( k ) < S(n) then S ( n ) <S(k) does not hold for any
S ( k ) , S ( n ) predecessors of S ( x)
c) Transitive: S ( m ) < S (k ) and S ( k ) <S(n) implies S ( m ) <S ( n )for any S(m), S(k), and
S(n) predecessors of S(x)
QUESTION 3
The irreflexive property follows clearly by defition of the succesor function on the empty set,
that is
0=∅
taking the same trend with each natural number as the set of all its predicessors, it is clear that
S ( n ) < S(n) does not hold for any S ( n ) as a predicessor set
For transitivity:
If S ( m ) , S(k) and S ( n ) as predecessors of S ( x ) with k =m+1 and n=k +1

4
MATHEMATICAL REASONING
For the first case, S ( m ) <S (k )
Since k =m+1 , it implies S ( m ) < S ( k ) =S ( m+1 ) =S ( 1+m ) =1+ S(m) hence the prove
For the second case, S ( k ) <S(n)
Since n=k +1 , it implies S ( k ) < S ( n ) =S ( k +1 ) =S ( 1+k ) =1+S (k ) hence the prove
The two cases clearly implies that S ( m ) <S (n) that is
S ( m )< S ( n )=S ( k +1 ) =S (1+ k )=1+ S ( k ) =1+ S ( m+1 ) =1+S ( 1+m )=1+ ( 1+S ( m ) ) =1=1+S(m)
hence the prove for transitivity
Combining irreflexive and transitive properties above it canclearly be deduced that if S ( k ) < S(n)
then S ( n ) < S( k) by the above definition of the relationship between n and k (assymetric
property), tha is to say since k < n holds, then n< k and k =n cannot hold