Prove that if m and n are integers and mn is even, then m is even or n is even.

Cariglinom5

Cariglinom5

Answered question

2021-11-06

Prove that if m and n are integers and mn is even, then m is even or n is even.

Answer & Explanation

menerkupvd

menerkupvd

Beginner2021-11-07Added 12 answers

Assume m=2k+1,n=2l,k.lZ
Then mn=(2k+1)(2l)=2(2kl+l)=2r, where r=2kl+lZ
Then mn is even.
Similarly when m=2k, n=2l+1, k.lZ
Assume m=2k+1, n=2l+1, k.lZ
Then mn = (2k+1)(2l+1)=4kl+2(k+l)+1=2(2kl+k+l)+1=2r+1 is odd, where r=2kl+lZ
Its
nick1337

nick1337

Expert2023-05-10Added 777 answers

Step 1:
Firstly, we can use proof by contradiction.
Assume that m is an odd integer and n is an odd integer. By definition, an odd integer can be expressed as 2k+1, where k is an integer.
Therefore, we can write m as 2a+1 and n as 2b+1, where a and b are integers.
Now, let's calculate the product mn:
mn=(2a+1)(2b+1)=4ab+2a+2b+1
Step 2:
Observe that mn can be expressed in the form 2(2ab+a+b)+1. This means mn is odd since it has the form 2k+1 for some integer k.
However, our initial assumption is that mn is even. This contradicts our assumption that both m and n are odd integers.
Therefore, our assumption that both m and n are odd integers must be false. At least one of them must be even.
Hence, we have proven that if m and n are integers and mn is even, then m is even or n is even.
madeleinejames20

madeleinejames20

Skilled2023-05-10Added 165 answers

We aim to prove the statement: 'If m and n are integers and mn is even, then m is even or n is even.'
Let's assume the negation of the statement, which is: 'If m and n are integers and mn is even, then m is odd and n is odd.'
Suppose m and n are odd integers. By definition, an odd integer can be expressed as 2k+1, where k is an integer.
So, we can write m as 2a+1 and n as 2b+1, where a and b are integers.
Now, let's calculate the product mn:
mn=(2a+1)(2b+1)=4ab+2a+2b+1
Notice that mn can be expressed in the form 2(2ab+a+b)+1. This means mn is odd since it has the form 2k+1 for some integer k.
However, we assumed that mn is even, which contradicts our assumption that both m and n are odd integers.
Therefore, our assumption that both m and n are odd integers must be false. At least one of them must be even.
Hence, we have proven that if m and n are integers and mn is even, then m is even or n is even.
Eliza Beth13

Eliza Beth13

Skilled2023-05-10Added 130 answers

To prove the statement 'If m and n are integers and mn is even, then m is even or n is even,' we can use proof by contrapositive.
Proof:
We will prove the contrapositive statement, which is: 'If m and n are integers and m is odd and n is odd, then mn is odd.'
Let's assume that m and n are odd integers, i.e., m=2k+1 and n=2l+1, where k and l are integers.
Substituting these values into the expression mn, we have:
mn=(2k+1)(2l+1)
Expanding the expression, we get:
mn=4kl+2k+2l+1
Rearranging the terms, we have:
mn=2(2kl+k+l)+1
We can see that mn can be expressed in the form 2p+1, where p=2kl+k+l. Since p is an integer, mn is odd.
Therefore, we have proved the contrapositive statement. By the principle of contrapositive, the original statement is also true.
Hence, if m and n are integers and mn is even, then m is even or n is even.

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