They statement is $:-$

For every rational number $x$, $x \lt x + 1$

At first glance my answer was $:-$

There exists a rational number $x$ such that $x \geq x + 1$

But then i saw this

$p : \sqrt{11}$ is rational

~$p$ : $\sqrt{11}$ is not rational

same as ~$p$ : $\sqrt{11}$ is irrational

I just wonder why not, For every irrational number $x$, $x \lt x + 1$ is a correct negation of the first statement ?

Sorry for this silly question i can't seem to find a answer in my textbook.

  • 2
    $\begingroup$ Nitpick: your "first glance" answer should have $\geq$. $\endgroup$ – MathematicsStudent1122 Jun 19 '16 at 21:08
  • $\begingroup$ Suppose that you know $p\in \mathbb{R}$ and $p$ is not rational, then you can say $p$ is irrational. But the negation of $p$ is rational is not $p$ is irrational if you don't know $p\in\mathbb{R}$. If $p$ is not a rational number, $p$ can be a cat! $\endgroup$ – Levent Jun 19 '16 at 21:14
  • $\begingroup$ @MathematicsStudent1122 sorry that was actually my first glance but i really dont know how to write that symbol in mahjax. so i skipped it $\endgroup$ – A---B Jun 19 '16 at 21:16
  • $\begingroup$ @Levent, isn't the set of rational numbers denoted as $\mathbb{Q}$? $\endgroup$ – Obinna Nwakwue Jun 19 '16 at 22:01
  • 1
    $\begingroup$ Difficulties with mathematical syntax are often relieved by paraphrasing, (carefully, that is,with no change in meaning) into the style of everyday speech. In common speech, no one says "For every whale x, x is big." We say "Every whale is big." The negation of this is obviously "There's (at least one ) whale that's not big." So the negation of "Every rational is less then (itself plus 1)" is "There's (at least one ) rational that's NOT less than (itself plus one.) ...As in Noble Mushtak's answer. $\endgroup$ – DanielWainfleet Jun 19 '16 at 23:25

This is a statement about rational numbers. Whatever properties irrational numbers have is irrelevant to the truth value of this statement. This statement is only talking about a property rational numbers have. Therefore, the negation of this statement is the existence of a rational number without this property because that would be a contradiction of the statement that all rational numbers have this property. Irrational numbers have nothing to do with the negation.

Now, to make this more clear, let's use your example: Clearly, the following statement is true:

For every rational number $x$, $x<x+1$

Now, by your logic, the negation of this is the following:

For every irrational number $x$, $x<x+1$

However, this statement is also clearly true. Therefore, by this logic, the statement and its negation are both true which can't be possible, so this is the wrong way to find the negation.

  • $\begingroup$ Wow that made sense, thanks $\endgroup$ – A---B Jun 19 '16 at 21:22

If you know the rules of predicate logic, you can prove that your initial answer is correct, as follows.

Suppose $\neg \forall x: [x \in Q \implies x\lt x+1]$

Changing the quantifier and removing the resulting double negation, we obtain:

$\exists x:\neg [x \in Q \implies x\lt x+1]$

Applying the definition of $\implies$ and $\ge$ and removing the resulting double negation, we obtain:

$\exists x: [x\in Q \land x\ge x+1]$

  • $\begingroup$ But i don't know those rules, sadly ... :( $\endgroup$ – A---B Jun 20 '16 at 18:47
  • $\begingroup$ You can learn them using my proof-checking freeware available at dcproof.com See the tutorial. $\endgroup$ – Dan Christensen Jun 20 '16 at 20:11
  • $\begingroup$ Hey, in your comment below the question, were you referring to my "whales" question ? sorry if you are confuse after reading this. $\endgroup$ – A---B Jun 20 '16 at 20:24

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