Prove or disprove (discrete math)Prove or disprove the following statement. The difference of the square...

Luz Stokes

Luz Stokes

Answered

2022-07-16

Prove or disprove (discrete math)
Prove or disprove the following statement. The difference of the square of any two consecutive integers is odd
This is working step: let m , m + 1 be 2 consective integers:
( m + 1 ) 2 m 2
m 2 + 1 + 2 m m 2
1 + 2 m
If m is odd then 2 m = even, if m is even then 2 m = even, then adding 1 will make it odd.
Can you please advise me if my working is the right step and could I answer like this?

Answer & Explanation

esbalatzaj

esbalatzaj

Expert

2022-07-17Added 15 answers

Step 1
Your proof looks correct. You might want to make it more clear that you are saying ( m + 1 ) 2 m 2 = ( m 2 + 2 m + 1 ) m 2 = 2 m + 1when you do your arithmetic.
Step 2
You also don't need to consider the cases where m is even and odd separately: since 2m is a multiple of 2, it must be even, and so you can conclude that 2 m + 1 is not evenly divisible by 2, so it is odd.
Raynor2i

Raynor2i

Expert

2022-07-18Added 6 answers

Step 1
What does it mean for an integer ℓ to be odd exactly? It means that we may express ℓ in the form = 2 n + 1, where n Z (note that n can be any integer).
Now consider your question:
The difference of the square of any two consecutive integers is odd.
Thus, assume (as you did in your first attempt) that m is an arbitrary integer in Z; then, we are dealing with the difference (1) ( m + 1 ) 2 m 2 .
Step 2
Now expand (1) as Strants did in his answer (except with a slight modification):
( m + 1 ) 2 m 2 = ( m 2 + 2 m + 1 ) m 2 = 2 m + 1 = ,
where Z . We know Z because adding two integers, namely 2m and 1, yields an integer, namely ℓ. Here's the important part: notice what form ℓ takes. We have that = 2 m + 1 , where m is an arbitrary integer in Z. Thus, by definition, we can see that ℓ is an odd integer.
Maybe this will clear anything up you did not quite get before.

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