Download 1.3.1 notes

DAY 7 – MATH 3
SEPTEMBER 15/16, 2016
LESSON 3 – ALGEBRAIC REASONING
AND PROOF (1.3.1)
• Lesson objectives:
• Use algebraic notation – letters, expressions, equations, and
inequalities – to represent general patterns and relationships
among variables
• Use algebraic transformations of expressions, equations,
and inequalities to establish general propositions about
quantitative relationships
• Look at the the yellow box on page 52
•
•
•
•
•
Pick an integer between 0 and 20
Add 5 to your number
Multiply the result by 6
Divide that result by 3
Subtract 9 from that result
INVESTIGATION 1
• Algebra and Number Magic – Problem #1
• Come up with an algebraic expression
• Review order of operations
•
•
•
•
A. write the algebraic expressions for each
B. write a final, condensed expression, then simplify
C. how did I find out your starting number?
D. how would that change affect how I found out
your starting number?
SUMMARIZING #1
Part A and B:
6(n + 5)
- 9 becomes 2n +1
3
Part C: To recover the number, subtract 1 from the
ending number, then divide by two
Written algebraically (x is the ending #): n = x -1
2
Part D: if the third step is “divide by 2” the expression
would be 6(n + 5)
2
- 9 which becomes 3n + 6
(not as easy to do in one’s head)
INVESTIGATION 1
• Problem #2 – new number trick
• A. find the algebraic expression, and how to
“decode”, or find the starting number
• B. why does “decoding” work?
SUMMARIZING #2
Part A: This procedure can be represented by
5(2n + 3)- 7 which becomes 10n +8
• The decoding rule will be to subtract 8 from the end
number and divide that intermediate result by 10
Part B: To find the start number, solve 10n +8 for “n”
• Written algebraically where x is the ending number:
x -8
n=
10
INVESTIGATION 1
• Problem #3 – your own number trick
• Problem #4 – Explaining number patterns
•
•
•
•
A. complete the table
B. calculate ALL the differences, then find a pattern
C. see the algebraic pattern, then try to simplify it
D. come up with an if/then statement….
SUMMARIZING #4
Part A: fill in the table
n
…
5
6
7
8
…
n
n+1
Sn
…
25
36
49
64
…
n2
(n+1)2
Part B: the differences between successive square
integers form the sequence of positive odd numbers
2
2
S
S
=
(n
+1)
n
Part C: n+1
n
= (n 2 + 2n +1) - n 2
= 2n +1
#4 CONTINUED
Part D: Some examples…
• If you calculate differences between successive
terms in the sequence of square numbers, then the
result will be the sequence of odd numbers
• If one square number is subtracted from the next,
then the result will be an odd number
•
2
2
(n
+1)
n
If you calculate
for any whole number n,
then the result will always be the odd number 2n +1