Exploring Natural Numbers and Mathematical Proofs in Divisibility

This homework assignment explores mathematical reasoning techniques used to determine the divisibility of numbers by 3, 9, and 11. Initially, it examines a trick involving reversing digits of a number, where specific conditions ensure divisibility by either 3 or 9 other than when using the digit '3'. The proof involves analyzing base 10 representations and their algebraic manipulations to confirm that particular numbers are divisible by these bases. Extending this concept, the assignment asks for similar tricks applicable for testing divisibility by 11 in base 10. It introduces a proposition stating that any positive natural number is divisible by 11 if the difference between sums of its digits at odd and even positions is also divisible by 11, supported by algebraic expression and proof. The document serves as an academic exercise in understanding mathematical proofs, divisibility rules, and higher-level arithmetic concepts within the realm of foundations of mathematics.

Running head: MATHEMATICAL REASONING
Mathematical Reasoning
Name
Institutional Affiliation

2
MATHEMATICAL REASONING
Question 1
Suppose n=¿. Show that 3∨n ⟺3∨ ( dk + dk−1+ …+d1+d0 ) .For which natural number ¿ 1
other than 3 does this trick work?
Proof:
We first need to show that indeed the trick works for n=¿
Suppose that 3∨n, where n=¿, and expressing ¿
With this notation ¿ and taking base 10 into consideration ¿, 10i can be expressed as ¿ that is
10i=¿. Hence
dk x 10k+ dk−1 x 10k−1+ dk−2 x 10k−2+ dk−3 x 10k−3+ …+d3 x 103 +d2 x 102 +d1 x 101 +d0=¿
dk x ¿
¿ ¿
{¿ ¿
{¿ ¿
Since 3∨¿. Therefore
3∨n−{¿ ¿. {HINT: if 3|n and 3|k, then 3|n-k}
But n={¿ ¿
Implying3∨{¿ ¿

3
MATHEMATICAL REASONING
That is 3∨ {dk +dk−1 +dk−2 +dk−3 +… d3 +d2 +d1 +d0 }
Conversely, suppose that3∨ {dk +dk−1 +dk−2 +dk−3 +… d3 +d2 +d1 +d0 }. Then 3∨{¿ ¿. That is 3∨¿n
HINT: if 3|q and 3|k, then 3|q+k
It can easily be seen the same trick above works when 3 is replaced by 9. That is
Since 9∨¿. Therefore
9∨n−{¿ ¿.
But n={¿ ¿
Implying9∨{¿ ¿
That is 9∨ {dk +dk −1 +dk −2 +dk −3 + … d3 + d2 + d1 +d0 }
Conversely, suppose that9∨ {dk +dk −1 +dk −2 +dk −3 +… d3 + d2 + d1 +d0 }. Then 9∨{¿ ¿. That is 9∨n
Thus the same trick works for 9
Question 2

4
MATHEMATICAL REASONING
Given a base b, for which integers c is it true that whenever n=¿ ¿ We have
c∨n ⟺ c∨dk +dk−1 +…+d1 +d0?
Soln.
The integers c are 3∧9 as clearly proved in question 1.
That is 3∨n ⟺3∨dk +dk −1 +…+ d1+d0 and
9∨n ⟺ 9∨dk+ dk−1+ …+d1 +d0whenever n=¿ ¿
Question 3
Devise a similar trick for testing for divisibility by 11 in base 10. Type equation here .
As a proposition, we know that any positive natural number n is divisible by 11 iff the difference
of the sums of digits in the odd and even positions results into a number that is divisible by 11
…. *
Proof:
For a more simplified formulae, we will suppose the number n=d4 d3 d2 d1 d0
Note that 10000=9999+1
1 000=1001−1
1 00=99+1

5
MATHEMATICAL REASONING
1 0=11−1
All the above numbers 9999, 1001, 99, and 11 are divisible by 11 as proven by proposition *
above
Thus we can express n=d4 d3 d2 d1 d0 in the form
n=11. k +d4 −d3+ d2+ d1+ d0(devised formulae)
Proof:
Suppose that 1 1∨d4 d3 d2 d1 d0 , and since 11∨11. k, we have 11∨¿, that is
11∨ ( d4−d3+ d2+ d1 + d0 )
Conversely, suppose that 11∨ ( d4−d3+ d2+ d1 + d0 )
Since 11∨11. kit implies 11∨11. k + ( d4−d3 + d2+d1 + d0 ) that is , 11∨n

6
MATHEMATICAL REASONING

7
MATHEMATICAL REASONING