$$ \forall n\exists m: m^{2}=n $$ How do I interpret this? This is part of my assignment question but I cant carry on without understanding this.
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This means: every natural number is a perfect square. $2$ is the smallest counterexample. – user26486 Mar 15 '15 at 21:43
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If $n$ is a natural number, the expression $n=m^2$ for some $n$ tells us that every natural number can be expressed as a product of a real number $m$ to itself. To see this note that:
$0=0^2$
$1=1^2$
$2=(\sqrt{2})^2$ Etc.
Bear in mind that $m$ is not necessary an integer.
It also works even if $n$ is a real number.
Jr Antalan
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It also works if $n$ is just a real number.
We'll have another real $m$ that satisfies the relation proposed.
Example: $\pi=(\sqrt{\pi})^2$
– Prasun Biswas Mar 15 '15 at 22:06 -
@PrasunBiswas yes it is :-). I put it as a note. Thanks for the comment. – Jr Antalan Mar 15 '15 at 22:08
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To make your statement complete (where "$\mathbb{N}$" indicates the natural numbers):
$\forall n \in \mathbb{N} \ \exists m \in \mathbb{N} \mid m^{2}=n $.
This is false, because $2$ is a counterexample.
To test your understanding:
Is this true or false?
$\forall m \in \mathbb{N} \ \exists n \in \mathbb{N} \mid m^{2}=n $.
marty cohen
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I take it that it's a negation of ∀n∃m:m^2=n so it's true. is that correct? – Curious_guy1111 Apr 15 '15 at 12:13
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It's not a negation; but it is true. My "|" means "such that", the same as your ":". – marty cohen Apr 15 '15 at 18:50