STRATEGY FOR MATHEMATICAL PROOF
A Coggle Diagram about (If we use
, If we use DIRECT PROOF
and If 5x - 7 is even, then x is odd
), The proof consists of two steps:
, HOW TO CHOOSE? If a direct proof appears problematic, the next most natural strategy to try is contrapositive proof.
, METHOD 4 : MATHEMATICAL INDUCTION
, the most straightforward approach to prove a conditional statement
The first strategy you should try when attempting to prove any statement is to use the direct proof method as it is straightforward.
By using this method, we assume p (n is even) is true, therefore q (7n + 4 is even) is also true
, If n is an even integer then 7n + 4 is an even integer SOLUTION:
Suppose n is even,n = 2a
7n + 4 = 7(2a) + 4
= 14a + 4
= 2 (7a + 2)
= 2 b, b = 7a + 2Therefore, 7n + 4 is an even integer.
, It is an alternative to direct proof & sometimes it is much easier than direct
, METHOD 2 : CONTRAPOSITIVE PROOF
, METHOD 1 : DIRECT PROOF
, Can be referenced as the sequential effect of falling DOMINOES.
, Used to prove all the statements are true when we have a set of statement S1,S2,S3,Sn...
, METHOD 3 : CONTRADICTION PROOF
, HOW DOES IT WORK?
By assuming the statement we want to prove is FALSE, then show that this assumption leads to nonsense.
, Can be used to prove any kind of statement
, (*) For n > 1, 2 + 2^2 + 2^3 + 2^4 + ... + 2^n = 2^n+1 – 2
Let n = 1. Then:2 + 2^2 + 2^3 + 2^4 + ... + 2^n = 2^1 = 2
2^n+1 – 2 = 2^1+1 – 2 = 2^2 –2 = 4 – 2 = 2So (*) works for n = 1.Assume, for n = k, that (*) holds; that is, that2 + 2^2 + 2^3 + 2^4 + ... + 2^k = 2^k+1 – 2Let n = k + 1.2 + 2^2 + 2^3 + 2^4 + ... + 2^k + 2^k+1
= [2 + 2^2 + 2^3 + 2^4 + ... + 2^k] + 2^k+1 = [2^k+1 – 2] + 2^k+1
= 2×2^k+1 – 2
= 2^1×2^k+1 – 2
= 2^k+1+1 – 2
= 2^(k+1)+1 – 2Then (*) works for n = k + 1.
, If a,b ∈ Z, then a^2- 4b ≠ 2.
Suppose this proposition is false,
a,b ∈ Z for which a^2-4b=2.From this equation, we get a^2= 4b+2= 2(2b+1), so a^2 is even.
a^2 is even, it follows that a is even, so a=2c for some integer c.Now plug a=2c back into the boxed equation to get (2c)^2-4b=2,
so 4c^2-4b= 2. Dividing by 2, we get 2c^2-2b=1.Therefore 1= 2(c^2 -b), and because c^2 -b∈ Z, it follows that 1 is even.We know 1 is not even, so
we were wrong to assume
the proposition was false.
Thus the proposition is true.