How do you negate a for all statement?

Change it to 'there exists' and negate the claim. The negation of 'all x satisfy P' is 'some x does not satisfy P'.

What is the negation of 'there exists'?

It is 'for all', with the claim negated. Nothing satisfies the property.

Does the order of quantifiers matter?

Yes. 'For every x there is a y' lets y depend on x, whereas 'there is a y for every x' requires one y to work for all.

TMUA Quantifiers and Negation

Quantifiers say how many cases a statement covers: 'for all' means every case, and 'there exists' means at least one. Negating them means swapping the quantifier and negating the claim.

Paper 2 topic. Specification codes: Arg3, Arg4. Last checked 12 November 2026.

Negation rules

The negation of 'for all $x$ , $P(x)$ ' is 'there exists $x$ with not $P(x)$ '.
The negation of 'there exists $x$ with $P(x)$ ' is 'for all $x$ , not $P(x)$ '.
The negation of ' $A$ and $B$ ' is 'not $A$ or not $B$ ', and of ' $A$ or $B$ ' is 'not $A$ and not $B$ '.

Order matters

'For every $x$ there is a $y$ ' is different from 'there is a $y$ for every $x$ '. In the first, $y$ may depend on $x$ .

One case settles it

A 'for all' statement is false if one counterexample exists. A 'there exists' statement is true if one example exists.

Worked examples

Example 1

Write the negation of 'For every real $x$ , $x^2>x$ ', and say which of the pair is true.

Show the worked solution

Negation: 'there is a real $x$ with $x^2\le x$ '. Taking $x=\tfrac12$ gives $\tfrac14\le\tfrac12$ , so the negation is true and the original is false.

Answer: Negation: some real x has x squared at most x. The negation is true

Example 2

Negate: 'Every pupil is over 160 cm tall or plays chess.'

Show the worked solution

The negation of 'A or B' is 'not A and not B', and of 'every' is 'some'. So: some pupil is 160 cm or shorter and does not play chess.

Answer: Some pupil is at most 160 cm tall and does not play chess

Example 3

Over the real numbers, decide the truth of (i) 'for every $x$ there is a $y$ with $x+y=0$ ' and (ii) 'there is a $y$ such that for every $x$ , $x+y=0$ '.

Show the worked solution

(i) is true: take $y=-x$ . (ii) is false: a single $y$ cannot cancel every $x$ , for example $x=0$ forces $y=0$ while $x=1$ forces $y=-1$ .

Answer: (i) true, (ii) false

Common mistakes

Negating 'for all' as 'for all, not', instead of 'there exists, not'.
Forgetting to switch 'and' and 'or' in the negation.
Negating 'greater than' as 'less than' rather than 'less than or equal to'.
Swapping the order of 'for all' and 'there exists' without checking.

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TMUA Quantifiers and Negation

Negation rules

Order matters

One case settles it

Worked examples

Common mistakes

Quick answers

Practise it under timed conditions

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