Negation of Atleast One

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Solution 1

My textbook says it is the negation is "for all elements"

Well, depending on what exactly it is the texbook claim about all those elements, that could work. For example, the negation of:

$$\text{"The statement is false for at least element"}$$

is (by your, correct, logic):

$$\text{"The statement is false for less than $1$ element"}$$

which is to say:

$$\text{"The statement is false for no elements"}$$

and that is the same as:

$$\text{"The statement is *true* for all elements"}$$

The important thing to remember that you can only negate claims. Thus, there really is no such thing as "the negation of 'at least one'", because 'at least one' is, by itself, not a claim.

Solution 2

"The average of three real numbers is greater than or equal to at least one of the numbers."

$$\forall x,y,z\in\mathbb{R},\underbrace{\exists t\in\{x,y,z\}}_{\text{at least one}},\underbrace{t\le\frac{x+y+z}{3}}_{\text{does}}$$

The negation is:

$$\exists x,y,z\in\mathbb{R},\underbrace{\forall t\in\{x,y,z\}}_{\text{all}},\underbrace{t>\frac{x+y+z}{3}}_{\text{doesn't}}$$

In another words, at least one does's negation is all doesn't

Solution 3

you are correct that the negation of “n is at least one” is “ n is less than 1” however your textbook is referring to an n that came from counting something, so n clearly can’t be negative, which means n=0: there are none of the thing being counted. This is often represented in math texts as saying that all of the elements are in the complement, so in your text book they’re negating “at least one of {x,y,z} is less than or equal to their average” with “all of {x,y,z} are greater than their average.

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Updated on November 10, 2020

Comments

  • mathguy
    mathguy almost 3 years

    What's the negation of “At least three of the sentence are false”?

    The negation of "atleast 3" or $n >= 3$ is $n < 3$ or "less than 3".

    But what is the negation of "atleast 1"?

    My textbook says it is the negation is "for all elements", but according to this logic, it should be "less than 1"?

    My textbook:

    Prove this statement by contradiction: "The average of three real numbers is greater than or equal to at least one of the numbers."

    Solution: Proof by contradiction. We assume that there are three real numbers $x$, $y$, and $z$ such that the average of the three numbers is less than each of the three numbers. That is (first term), (second term), and (third term).

    Each term is the average of the 3 real numbers $<$ a real number, but as you can see the statement says $and$ which means all elements.

    • Rushabh Mehta
      Rushabh Mehta about 4 years
      Can you write exactly what the textbook wrote down? This is not clear.
    • mathguy
      mathguy about 4 years
      sure give me a second
    • mathguy
      mathguy about 4 years
      @DonThousand updated
    • Manx
      Manx about 4 years
      $n\ge1$ negation is just $n<1$
    • Manx
      Manx about 4 years
      $$\neg(\exists^{\ge1}x,P(x))\equiv\exists^{<1}x,P(x)$$
    • Bram28
      Bram28 about 4 years
      "My textbook says it is the negation is "for all elements"" => For all elements.... what? Can you please be more precise what exactly it is your textbook claims?
    • Manx
      Manx about 4 years
      Also $$\exists^{<1}x,P(x)\equiv\forall x,\neg P(x)$$
    • mathguy
      mathguy about 4 years
      @Bram28 As you can see in the example, "atleast one of the numbers" is negated to "all of the 3 numbers"
    • Bram28
      Bram28 about 4 years
      @mathguy There is really no such thing as "the negation of 'at least one'" ... because 'at least one' is not a claim. This is what I was asking: what claim is the book making about 'all elements'?
  • mathguy
    mathguy about 4 years
    Thank you for breaking it down