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I received the following equation. It is supposed to contain clues to something I have to solve. I am not familiar with the math symbols used here.

How do I read the following:

f: R → D, ∀d ∈ D, ∃!r ∈ R: f(r) = Lm = d
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    If you provide more context maybe we can decipher that `Lm`.2012-06-08
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    @AsafKaragila : Sir, I am extremely sorry to fork an off-topic discussion here. But I am really in need of some help sir. I wanted to prove that if $p^{2m-2n}-p^m+3p^{m-2n}+1$ is a perfect square then $m=2n$. Here the $p$ is a odd-prime and all the exponents are positive. I tried to post this but I am not clear myself. So I have deleted twice. If you can give some hints, it would be extremely useful. I tried using the Reductio ad absurdum . We then get 2 cases $m \lt 2n$ and $m \gt 2n$. Then I am stuck here.. I would be very happy to hear any suggestion from you sir..2012-06-09
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    @Iyengar: I cannot help you with number theory questions.2012-06-09

2 Answers 2

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It says that $f$ is a function from $R$ to $D$, and for each $d\in D$ there is exactly one $r\in R$ with the property that $f(r)=d$. I can’t help with the $Lm$: it isn’t a standard notation and appears to be something derived from the context.

The symbols $\forall$ and $\exists!$ are quantifiers meaning respectively ‘for each’ and ‘there exists exactly one’.

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    Also, something like $d \in D$ means "$d$, which is an element of the set $D$".2012-06-08
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Assuming that by $R$ you mean $\mathbb R$, the set of real numbers:

  • $f\colon\mathbb R\to D$ means that $f$ is a function from the real numbers to a set $D$.

  • $\forall d\in D$ is read "For every $d$ which is an element of $D$",

  • $\exists!r\in\mathbb R$ is to say "Exists a unique real number $r$"

  • $: f(r) = Lm = d$ such that $f(r)=Lm$, which as Brian said is not a standard notation, and also $f(r)=d$.

This means that $f$ is such function that for every $d\in D$ there is exactly one $r$ such that $f(r)=d$. This means that every two numbers are mapped to two distinct elements of $D$; such $f$ is called injective. We also deduce that every $d\in D$ is $f(r)$ for some $r$, and such $f$ is called surjective. If a function is both injective and surjective we say that it is bijective, or that it is a bijection.