Why does "carrying" work in addition? I've been asked this many times from my most inquisitive students. Grab a pencil and paper and follow me...
 
We'll work on a simple example. If you want to try a more complex example, just post your question to this post.
 
On your piece of paper write this,
 
  35
+47 
 
If you already know how to add this, go ahead and complete the problem. Your answer should be 82, and you should have carried a 1 over the tens place. So why does this “carrying” method work?!

Do you agree that
35 = 30 + 5, and that
47 = 40 + 7? Yes, right?

Do you agree that 5 + 7 = 12? Of course, right?!

So we can write the following,
35 + 47
= 30 + 5 + 40 + 7
= 30 + 40 + 5 + 7
= 30 + 40 + 12        (here we are substituting 5 + 7 for 12)
= 30 + 40 + 10 + 2  (by substituting 12 for 10 + 2)

Now we can add in our head WITHOUT “carrying”.
  30
+40
+10
+  2
  82
By “carrying” a 1, you are essentially splitting the 12 into a 10 and 2, and that’s why carrying works!

How many Z+ ≤ 1000 have distinct digits and are even?

Solution.

Single digit numbers:        _____

                                    {2, 4, 6, 8} = 4 digits

Double digit numbers:       _____   _____

                                    4 digits  {0, 2, 4, 6, 8}

                                                5-1=4 digits

                                                exclude the digit used in the first position

Triple digit numbers:        _____   _____   _____

                                    4 digits 4 digits 5-2=3 digits

                                                           exclude the 2 digits already used in previous positions

The only four digit number we have is 1000, but it does not meet our conditions.

So we have, 4 + (4 x 4) + (4 x 4 x 3) = 68 distinct even digits ≤ 1000.

Don't you feel like this sometimes when you write proofs?! Ha, writing proofs can seem nutty <:-)

Math Help Forum

Check out this great forum. It contains threads discussing actual textbook problems. It is the best online source I have found.

www.mathhelpforum.com

1.4-6: Use a direct proof to show that the product of 2 odd integers is odd.

Section 1.4 Problem 6.

Proof. Assume that x and y are odd integers. By definition of odd, there are integers a and b such that x = 2a + 1 and y = 2b + 1. By substitution

xy = (2a + 1) (2b + 1)

= 4ab + 2a + 2b + 1

= 2(2ab + a + b) + 1.

Since 2, a, and b are integers and integers have closure with respect to multiplication and addition, then 2ab + a + b is an integer. By definition of odd, it follows that the product of xand y is odd. ☐

1.4-18a: Prove that if n is an integer and 3n + 2 is even, then n is even using a proof by contraposition.

Section 1.4 Problem 18a.

Proof. Let n be an integer. Assume that n is odd. By definition of odd, there exists an integer a such that n = 2a + 1. By substitution,

3n + 2 = 3(2a + 1) + 2

= 6a + 3 + 2

= 6a + 5

= 2(3a + 2) + 1.

Since 3, a, and 2 are integers and integers have closure with respect to multiplication, then 3a + 2 is an integer. Therefore by the definition of odd, 3n + 2 is odd. ☐

1.4-26: Prove that if n is a positive integer, then n is even if and only if 7n + 4 is even.

Proof. Let n be a positive integer. Assume that n is even. By definition of even, this means that there exists an integer a such that n = 2a. By substitution

7n + 4 = 7(2a) + 4

= 14a + 4

= 2(7a + 2).

Since 7, a, and 2 are integers and integers have closure under addition and multiplication, then 7a + 2 is an integer. Therefore by the definition of even, 7n + 4 is even.

Now assume that n is odd. By the definition of odd, this means there is an integer b such that n = 2b + 1. By substitution

7n + 4 = 7(2b + 1) + 4

= 14b + 7 + 4

= 14b + 11

= 2(7b + 5) + 1.

Since 7, b, and 5 are integers and integers have closure under addition and multiplication, then 7b + 5 is an integer. Therefore by the definition of odd, it follows that 7n + 4 is odd.

Since n is odd implies that 7n + 4 is also odd, then clearly 7n + 4 is even implies n is even must be equivalently true.

Furthermore, since n is even implies 7n + 4 is even, and conversely 7n + 4 is even implies n is even are true, then by the definition of a biconditional statement it follows that the original proposition n is even if and only if 7n + 4 is even must also be true. ☐