# True or false statements odd integers

One is false one is true write the negation for the false one and prove both of them

a) for all integers a, there exists an integer b so that a+b=odd

b) there exists an integer b such that for all integers a, a+b=odd

I said (a) is false and (b) is true. I wrote the negation for (a) then I split it in two cases. Case 1: b is even. Let b=2k and let a=0 then a+b=2k which is even. Case 2: b is odd. Let b=2k+1 and let a=-1 then a+b=2k

Then for (b) I set up two cases again. Case 1: a is even. A=2k and b=1 then a+b=2k+1 which is odd Case 2: a is odd, let a=2k+1 and b=0 then a+b=2k+1 which is odd

Input in whether I'm right or not would be great thanks

## 2 Answers

I’m afraid that you got them exactly backwards. Let’s take a look.

(a) For each integer $a$ there is an integer $b$ such that $a+b$ is odd.

Think of this in terms of a game. I give you an integer $a$, and you win if you can find an integer $b$ such that $a+b$ is odd; if you cannot find such a $b$, I win. The assertion (a) says that you can always win, no matter how I choose my $a$.

Suppose I give you the integer $101$; can you find an integer to add to it to make an odd integer? Sure: it’s already odd, so just add $0$. Suppose instead that I give you the integer $100$; can you find an integer to add to it to make an odd integer? $0$ won’t work, but $1$ will: $100+1=101$, which is certainly odd. Was there anything very special here about $101$ and $100$? No: all I used was the fact that $101$ is odd, so that adding $0$ was bound to give me an odd total, and the fact that $100$ is even, so that adding $1$ would give me an odd total. If I give you any odd integer $a$, you can use $b=0$ and be sure that $a+b=a+0=a$ will be odd. And if I give you any even integer $a$, you can use $b=1$ and be sure that $a+b=a+1$ is odd. Every integer is either even or odd, so no matter what integer $a$ I give you, you’re covered: you know how to pick a $b$ such that $a+b$ is odd. In other words, (a) is true: you do have a winning strategy.

(Of course there are other choices that work besides the ones that I’ve mentioned; mine are just the simplest.)

(b) There is an integer $b$ such that for all integers $a$, $a+b$ is odd.

Again you can think of this in terms of a game, with you picking $b$ and me picking $a$. The difference is that in this game you have to play first: you pick some integer $b$, then I pick my $a$, knowing what your $b$ is. You win if $a+b$ is odd, you lose if it isn’t, and (b) says that you have a winning strategy: there is some $b$ that you can pick that will make $a+b$ odd no matter what integer I pick for $a$.

But that’s clearly not true: if your $b$ is even, I’ll just let $a=0$, so that $a+b=b$ is even, and you lose. And if your $b$ is odd, I’ll pick $a=1$, so that $a+b=1+b$ is even, and again you lose. No matter how you play — no matter what integer you choose for your $b$ — I can beat you by choosing $a$ to make $a+b$ even.

Here the first true,because if $a=2k$( is even) choose $b=1$ if $a=2k+1$(odd) choose $b=2$ to complete proving the first fact.

In your proof i guess you have missed out the critical point "there exists".

• The first statement is not the negation of the second: the negation of the second is for each integer $b$ there is an integer $a$ such that $a+b$ is even. – Brian M. Scott May 25 '13 at 3:39
• Isnt this the contrapositive of the 2nd statement???? I am a bit confused. – Abhra Abir Kundu May 25 '13 at 3:42
• Only implications have contrapositives. Neither statement is of the form $\varphi\to\psi$, so neither has a contrapositive. (And negation and contrapositive are completely different things.) – Brian M. Scott May 25 '13 at 3:47
• Thanks @BrianM.Scott for clarifying . – Abhra Abir Kundu May 25 '13 at 3:56