B Doing proofs: Setting an expression as a variable

[opus]

In a book I'm reading, I'm told to prove that: If m and n are odd, then (m)(n) is odd.

The proof goes as such:

Let m=(2a+1) and n=(2b+1)

Then,
mn= (2a+1)(2b+1) = 4ab+2a+2b+1 = 2(2ab+a+b)+1 = 2t+1 ; where t= 2ab+a+b

Two questions:

When we take an expression, and assign it to a single variable, what rules must we abide by? Can we do this for any situation? My reasoning tells me the step is correct because the definition for an odd number is 2t+1, with t being any natural number. However, how can we be sure that the expression, for which we defined as t, will give a natural number? What's to say that expression can't give a negative number, which would make the solution to the proof false?

My second question is in regards to the definition of a prime as it relates to this proof. To my understanding, there are two ways to define a prime number.
1) 2m-1, where m is any integer
2) 2n+1, where n is any natural number

Would both of these definitions work for this proof?

 

[andrewkirk]

However, how can we be sure that the expression, for which we defined as t, will give a natural number?

It is part of the axioms of arithmetic on the natural numbers that the addition of two natural numbers gives a natural number, and the multiplication of two natural numbers does so too. Since a and b are assumed to be natural numbers and the expression 2ab+a+b = ((((2*a)*b)+a)+b) is obtained by doing two multiplications and two additions, starting with natural numbers, it follows that the value of the expression is a natural number.

My second question is in regards to the definition of a prime as it relates to this proof

I think you mean 'odd number' rather than 'prime', no? Which one works depends on what definition of natural numbers you are using. Some definitions include 0 while others start at 1. For the former we could not use 2m-1 as that does not give a natural number when m=0. I think it is more common to include 0 in the natural numbers (the Peano axiomatisation, linked above, includes 0), in which case we need to use 2m+1 to denote an odd number.

 

[fresh_42]

In a book I'm reading, I'm told to prove that: If m and n are odd, then (m)(n) is odd.

The proof goes as such:

Let m=(2a+1) and n=(2b+1)

Then,
mn= (2a+1)(2b+1) = 4ab+2a+2b+1 = 2(2ab+a+b)+1 = 2t+1 ; where t= 2ab+a+b

Two questions:

When we take an expression, and assign it to a single variable, what rules must we abide by? Can we do this for any situation? My reasoning tells me the step is correct because the definition for an odd number is 2t+1, with t being any natural number. However, how can we be sure that the expression, for which we defined as t, will give a natural number?

Very good observation. It has to be shown. However, it's quite obvious for $a$ and $b$ are. But formally, it has to be shown.

What's to say that expression can't give a negative number, which would make the solution to the proof false?

No. Odd and even extend to negative integers as well. All that counts is, whether the remainder of a division by two is one (odd) or zero (even). Thus zero is even.

My second question is in regards to the definition of a prime as it relates to this proof. To my understanding, there are two ways to define a prime number.
1) 2m-1, where m is any integer
2) 2n+1, where n is any natural number

Those are not prime. They are odd. And as said above, the sign is irrelevant.

Would both of these definitions work for this proof?

Basically yes. But only if the author had been carefully enough and spoke of integers instead of natural numbers, in which case the trouble starts: is zero a natural number, or do we start with one? If we write odd numbers $2m-1$, then $m\ge 1$, if we write $2n+1$ then $n\ge 0$, but does this collide with what we said about zero being a natural number or not? Forget about $\mathbb{N}$ here, $\mathbb{Z}$ is the relevant set for number, group, ring and in part module theory.

 

[opus]

Ok that clears things up for me. And yes, I meant to say "odd" not "prime".
In the book, the author distinguishes the difference between 2n+1 and 2n-1, saying that we can write an odd integer as 2n+1 if we allow n to be a natural number. I just wasn't sure if they were saying different things to accomplish the same end, and if that had any effect on the proof.
Thank you!