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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

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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.

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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".

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  • $\begingroup$ 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. $\endgroup$ – Brian M. Scott May 25 '13 at 3:39
  • $\begingroup$ Isnt this the contrapositive of the 2nd statement???? I am a bit confused. $\endgroup$ – Abhra Abir Kundu May 25 '13 at 3:42
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    $\begingroup$ 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.) $\endgroup$ – Brian M. Scott May 25 '13 at 3:47
  • $\begingroup$ Thanks @BrianM.Scott for clarifying . $\endgroup$ – Abhra Abir Kundu May 25 '13 at 3:56

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