Algebra Question: 2r - 2s - 1 = 2(r-s-1) + 1 (?)

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In summary, the conversation is about a discrete math problem that requires proving that the difference of an even integer and an odd integer is an odd integer. Through the use of equations and a step-by-step solution attempt, it is shown that the difference can be expressed as 2(r-s-1)+1, with the extra one coming from adding 0 in the form of -1+1. The simplicity of writing the difference as 2(r-s)-1 is also discussed, but it is noted that odd numbers are typically written as 2(some letter or quantity)+1.
  • #1
Animuo
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Homework Statement


Well, the problem is a discrete math problem to prove that the difference of an even integer and an odd integer is an odd integer.

Homework Equations


If you let m = an even integer, and n = an odd integer then m = 2r for some integer r, while n = 2s + 1 for some integer s.

The Attempt at a Solution


Then
m - n = 2r - (2s + 1) = 2r - 2s - 1

at the next step, the answer in the book shows
2(r-s-1)+1, from there I can follow the rest of the steps to show that an integer is odd but I'm confused as to where this extra one is coming from. I know it's just a basic factoring question, but I'm a little rusty, maybe somebody could help me.. thanks.
 
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  • #2
Animuo said:

Homework Statement


Well, the problem is a discrete math problem to prove that the difference of an even integer and an odd integer is an odd integer.

Homework Equations


If you let m = an even integer, and n = an odd integer then m = 2r for some integer r, while n = 2s + 1 for some integer s.

The Attempt at a Solution


Then
m - n = 2r - (2s + 1) = 2r - 2s - 1

at the next step, the answer in the book shows
2(r-s-1)+1, from there I can follow the rest of the steps to show that an integer is odd but I'm confused as to where this extra one is coming from. I know it's just a basic factoring question, but I'm a little rusty, maybe somebody could help me.. thanks.

2r - 2s - 1 = 2r - 2s - 1 - 1 + 1 = 2r - 2s - 2 + 1 = 2(r - s - 1) + 1

In the second step above, I added 0, in the form of -1 + 1. You can always add 0 to an expression to get an identically equal expression.
 
  • #3
animuo said:

Homework Statement


well, the problem is a discrete math problem to prove that the difference of an even integer and an odd integer is an odd integer.

Homework Equations


if you let m = an even integer, and n = an odd integer then m = 2r for some integer r, while n = 2s + 1 for some integer s.

The Attempt at a Solution


then
m - n = 2r - (2s + 1) = 2r - 2s - 1

at the next step, the answer in the book shows
2(r-s-1)+1, from there i can follow the rest of the steps to show that an integer is odd but I'm confused as to where this extra one is coming from. I know it's just a basic factoring question, but I'm a little rusty, maybe somebody could help me.. Thanks.

-1 = -2 + 1

rgv
 
  • #4
Aha, can't believe I missed such an easy solution, guess it's good to humble one down, thanks for the help mates... I'm new to this forum is there anything like +rep or likes that I can give?
 
  • #5
What am I missing? Why do you have to put it as 2(r-s-1)+1 to show it is odd? Why isn't it acceptable (and slightly simpler) just to put it as 2(r-s)-1 ?
 
  • #6
Animuo said:
Aha, can't believe I missed such an easy solution, guess it's good to humble one down, thanks for the help mates... I'm new to this forum is there anything like +rep or likes that I can give?

Unfortunately no, you can't give rep; but I know some would have a whole lot more than the rest of us!

oay said:
What am I missing? Why do you have to put it as 2(r-s-1)+1 to show it is odd? Why isn't it acceptable (and slightly simpler) just to put it as 2(r-s)-1 ?

It's probably because odd numbers are usually written in the form 2(some letter or quantity) + 1 instead of 2(some letter or quantity) - 1.
 
  • #7
Bohrok said:
It's probably because odd numbers are usually written in the form 2(some letter or quantity) + 1 instead of 2(some letter or quantity) - 1.
Yep, that's what I assumed. Seems a bit daft to me.
 

FAQ: Algebra Question: 2r - 2s - 1 = 2(r-s-1) + 1 (?)

What is the purpose of using parentheses in this equation?

The parentheses help to group certain terms together and indicate the order of operations to be followed when solving the equation. In this case, the terms within the parentheses should be simplified first before combining with the rest of the equation.

How do I solve for the variables in this equation?

To solve for the variables, you can use the distributive property to remove the parentheses and then combine like terms. In this case, you would distribute the 2 to each term within the parentheses, simplify, and then combine like terms on each side of the equation.

Can I solve this equation without using parentheses?

Yes, you can solve this equation without using parentheses by using the distributive property to remove the parentheses and then combining like terms. However, using parentheses can make the equation easier to read and follow, especially in more complex equations.

How do I know if I have solved the equation correctly?

You can check your solution by plugging the values you obtained for the variables back into the original equation. If the equation is true, then your solution is correct. You can also use a calculator to verify your solution.

Can this equation have multiple solutions?

Yes, this equation can have multiple solutions. In this case, the variables r and s can take on different values as long as the equation is satisfied. Some equations may have only one solution, while others may have an infinite number of solutions.

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