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

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menerkupvd · Accepted Answer

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

nick1337 · Accepted Answer

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&#039;s calculate the product mn:mn=(2a+1)(2b+1)=4ab+2a+2b+1Step 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 · Accepted Answer

We aim to prove the statement: &#039;If m and n are integers and mn is even, then m is even or n is even.&#039;Let&#039;s assume the negation of the statement, which is: &#039;If m and n are integers and mn is even, then m is odd and n is odd.&#039;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&#039;s calculate the product mn:mn=(2a+1)(2b+1)=4ab+2a+2b+1Notice 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 · Accepted Answer

To prove the statement &#039;If m and n are integers and mn is even, then m is even or n is even,&#039; we can use proof by contrapositive.Proof:We will prove the contrapositive statement, which is: &#039;If m and n are integers and m is odd and n is odd, then mn is odd.&#039;Let&#039;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+1Rearranging the terms, we have:mn=2(2kl+k+l)+1We 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.

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

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