Nathalie Fields

Answered

2022-07-17

Direct Proof Discrete Math.

Show that if n is an odd integer, then ${n}^{2}$ is odd.

Proof : Assume that n is an odd integer. This implies that there is some integer k such that $n=2k+1$. Then ${n}^{2}=(2k+1{)}^{2}=4{k}^{2}+4k+1=2(2{k}^{2}+2k)+1$. Thus, ${n}^{2}$ is odd.

Why does the solution assume n to be $2k+1$?

How do you know ${n}^{2}$ is odd based on $2(2{k}^{2}+2k)+1$?

I don't see how $2k+1$ for n and 2(2k2+2k)+1 for ${n}^{2}$ means an odd integer. Some clarification would be helpful.

Show that if n is an odd integer, then ${n}^{2}$ is odd.

Proof : Assume that n is an odd integer. This implies that there is some integer k such that $n=2k+1$. Then ${n}^{2}=(2k+1{)}^{2}=4{k}^{2}+4k+1=2(2{k}^{2}+2k)+1$. Thus, ${n}^{2}$ is odd.

Why does the solution assume n to be $2k+1$?

How do you know ${n}^{2}$ is odd based on $2(2{k}^{2}+2k)+1$?

I don't see how $2k+1$ for n and 2(2k2+2k)+1 for ${n}^{2}$ means an odd integer. Some clarification would be helpful.

Answer & Explanation

iljovskint

Expert

2022-07-18Added 18 answers

Step 1

An integer m is odd if and only if it can be written as the sum of an even integer and 1, if and only if there exists an integer q such that $m=2q+1.$.

Step 2

In the proof, assumption that n is odd implies the existence of an integer k such that $n=2k+1$ (that is the only if part of the lemma). As for ${n}^{2}=2(2{k}^{2}+2k)+1$, letting $q=2{k}^{2}+2k$ we have ${n}^{2}=2q+1,$, from which it follows that n2 is odd (that is the if part of the lemma).

An integer m is odd if and only if it can be written as the sum of an even integer and 1, if and only if there exists an integer q such that $m=2q+1.$.

Step 2

In the proof, assumption that n is odd implies the existence of an integer k such that $n=2k+1$ (that is the only if part of the lemma). As for ${n}^{2}=2(2{k}^{2}+2k)+1$, letting $q=2{k}^{2}+2k$ we have ${n}^{2}=2q+1,$, from which it follows that n2 is odd (that is the if part of the lemma).

enmobladatn

Expert

2022-07-19Added 6 answers

Step 1

By definition:

integer n is even iff there exists an integer k such that $n=2k$

integer n is odd iff there exists an integer k such that $n=2k+1$

Step 2

So in the first step we use the definition to go from n is odd to $n=2k+1$, and in the other step we again use the definition to go from ${n}^{2}=2(2{k}^{2}+2k)+1$ to ${n}^{2}$ is odd.

By definition:

integer n is even iff there exists an integer k such that $n=2k$

integer n is odd iff there exists an integer k such that $n=2k+1$

Step 2

So in the first step we use the definition to go from n is odd to $n=2k+1$, and in the other step we again use the definition to go from ${n}^{2}=2(2{k}^{2}+2k)+1$ to ${n}^{2}$ is odd.

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