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Help Jehoram Isiah N. Navarro! Anyone can contribute an answer, even nontutors.
Please help me rearrange the statements.
The topic we're facing is about Summation Identity and how do we prove it by using Induction. Unfortunately, we don't know proof by Induction because our teacher has yet to discuss it.
How do you prove an identity?
What is Summation Identity?
What is Proof by Induction?
Then, we assume this identity is true for some n = m (the fifth one in the list) and rewrite the entire equation using n instead of m (the last option in the list) and we proceed to find out if the n = m + 1 case is true or not (the sixth one in the list).
This is followed by long computation (the third one in the list) and a conclusion that the n = m + 1 case is also true (the second one in the list).
By the principle of induction, the entire statement is true for all n (first one in the list).
The idea why we prove the first case and the n = m + 1 case is that n and m can take any number from 1 onwards. Suppose the first statement is true. Since the first statement is true (m = 1), the second statement must be true. Since the second statement is true (m = 2), the third statement must be true. Since the third statement is true (m = 3), the fourth statement must be true. And so on.
We can even do a proof by induction on simple formulae such as a^2  b^2 = (a + b) (a  b).
Mathematical induction can take on many different forms, but the general method is like this.
1. Look at the statement provided.
2. Prove it for n = 1 (or the lowest value possible, like n = 3 if n is given to be more than 2). Term numbers are always integers.
3. Now we have established that the starting case is true.
4. We assume that the mth statement is true for some real integers m.
4. We must prove that the case is true for the “m + 1”th statement as well for the identity to hold.
5. We replace n by m + 1 and reduce it to a equation which contains lots of m + 1.
6. Step 5 establishes that the for every true mth statement, the (m + 1)th statement must be true.
7. So every statement in the list must be true, since having the starting first statement being true means the second statement must be true. Since having the second statement is true, the third statement must also be true. And so on. It’s like dominoes. One tile falls and the rest will follow suit one by one.
To prove that n2  1 = (n + 1) (n  1) is true for all positive integers n.
Step 1: Prove that the lowest possible case (here, n = 1 since it is a positive integer) is true
LHS: 1^2  1 = 0
RHS: (1 + 1) (1  1) = 0
So the base case n = 1 is true.
Step 2: Assume the case n = m is true for some real integer m
Step 3: We work out the case n = m + 1
For n = m + 1,
LHS
= n^2  1
= (m + 1)^2  1
= m^2 + 2m + 1  1
= m^2 + 2m
= m (m + 2)
= (m + 2) (m)
We rewrite this as
= [ (m + 1) + 1] [ (m + 1)  1]
which is simply none other than (n + 1) (n  1). This version in n need not be written.
So the case n = m + 1 is also true.
Step 4: Conclusion
By the principle of mathematical induction, the statement n^2  1 = (n + 1) (n  1) is true for all positive integers n.
You're a lifesaver (╥﹏╥)
Also, I'm glad that you included a simple version of your clarification regarding the topic. It helped me a lot in comprehending it. ✧◝(⁰▿⁰)◜✧