Is this considered bad style

$$2 = \sqrt{4} < \sqrt{16} = 4?$$

It seems as though this is not strictly correct, since $2 = \sqrt{4}$ is a logical proposition which represents boolean value (true or false). A boolean value cannot be less than $\sqrt{16}$.

On the other hand, I am sure that most people will correctly interpret this as shorthand for $2 = \sqrt{4},$ $\sqrt{4} < \sqrt{16},$ and $\sqrt{16} = 4$

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    $\begingroup$ I use this notation, and seeing each relation was true would convince me that $2 \lt 4$. It is no worse than $\displaystyle {n \choose k}=\frac{n!}{k!(n-k)!}=\frac{n!}{(n-k)!(n-(n-k))!}= {n \choose n-k}$ $\endgroup$ – Henry Jan 1 '16 at 23:41
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    $\begingroup$ I have seen graduate-level math books do this, so I think it's quite acceptable. $\endgroup$ – vhspdfg Jan 1 '16 at 23:44
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    $\begingroup$ Looking at the argument you give for your concernthe point of mixing is really a red herring. If you have $2^2 = 4 = 8/2$ then $2^2= 4$ would represent true or false which does not equal $8/2$ etc. $\endgroup$ – quid Jan 1 '16 at 23:52
  • $\begingroup$ Yes, it's very standard throughout all the books I've read. $\endgroup$ – Daniel R. Collins Jan 2 '16 at 4:00
  • $\begingroup$ @Sam I think $2=\sqrt{4}$ is a proposition, but not itself being neither True or False. Proposition is proposition, whereas T/F is T/F. To indicate the truth value of a proposition, we use a function, called valuation function($V$), that takes the proposition to codomain $\{\mathbb{T},\mathbb{F}\}$. For example, $V(1+1=2)=\mathbb{T}$, but $1+1=2$ itself is $1+1=2$(proposition), neither $\mathbb{T}$ nor $\mathbb{F}$. $\endgroup$ – Eric Oct 8 '16 at 6:04

If $R_1, R_2, \ldots$ are binary relations, it is standard practice in mathematics to write:

$$a_1 \mathrel{R_1} a_2 \mathrel{R_2} a_3 \ldots a_{k} \mathrel{R_{k}} a_{k+1}$$

as a short hand for:

$$a_1 \mathrel{R_1} a_2 \mbox{ and } a_2 \mathrel{R_2} a_3 \ldots \mbox { and } a_{k} \mathrel{R_{k}} a_{k+1}$$

This convenient syntactic convention works because, in most mathematics, we usually write as if we are working in first-order logic, where boolean values aren't allowed as operands of relation symbols: $(1 < 2) = (3 < 4)$ isn't allowed. When you work in higher-order logic in mathematics, and in most programming languages, formulas like $(1 < 2) = (3 < 4)$ are allowed and so this convention doesn't work so smoothly.

  • $\begingroup$ This convention is so convenient that sometimes people want to get more out of this than it actually gives. Two examples. (1) when someone writes $a\neq b\neq c\neq d$, he probably thinks having expressed all four values are distinct, but that is not so. (2) I've once seen written, for an infinite sequence $(a_i)_{i\in\Bbb N}$, the condition $a_0=a_1=a_2=\cdots=0$ (which makes no sense, since the final relation $=$ has no left operand; on the other hand $0=a_0=a_1=a_2=\cdots$ might be considered acceptable). $\endgroup$ – Marc van Leeuwen Jan 2 '16 at 18:48
  • $\begingroup$ A notable exception on the programming front is Python, which does, for example, let w > x <= y < z be equivalent to w > x and x <= y and y < z. $\endgroup$ – wchargin Jan 2 '16 at 19:58
  • $\begingroup$ (This is widely considered to have been a Bad Idea precisely because of the higher-order capabilities you mention.) $\endgroup$ – wchargin Jan 2 '16 at 19:58

You will find such things often on this forum, and the meaning is usually clear. When I was an undergraduate we used to string $\implies$ between statements we had shown were dependent on previous statements - it was convenient shorthand rather than boolean logic.

The purpose of mathematical writing is to communicate clearly and accurately what you mean - that can depend on context and audience. If shorthand brings in an ambiguity, then write it a different way.

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    $\begingroup$ Sometimes space is at a premium... $\endgroup$ – vonbrand Jan 2 '16 at 0:34
  • $\begingroup$ Chaining binary relations is so universally done that I don't think it even makes sense to call it a shorthand. $\endgroup$ – David Richerby Jan 2 '16 at 12:45

It's fine - people write that way all the time. But don't ever do this: $$1\le b=c>d.$$

Edit: Various people have commented, saying that there's nothing wrong with the above. Perhaps not; it bothers me, but I'm not going to insist that it's wrong. If I claimed I didn't actually say it was wrong people would say I was being pedantic.

One person points out that if you write the above it certainly is wrong to deduce a relationship between $1$ and $d$. And that's the problem - in my experience in "beginning analysis" classes students who write things like what's above do tend to draw incorrect conclusions. So I'm going to just rephrase what I said: "Wrong or not, don't do that. It's a bad idea."

  • $\begingroup$ I think that's too strong: I would say don't write things like that unless you have a very good reason to, and then only accompanied by a careful explanation and justification. (Probably the only valid good reason would be if $b$ and/or $c$ is a very long expression, in which case there are probably better alternatives, which you will realise when you try to write the justification $\ddot{\smile}$.) $\endgroup$ – Rob Arthan Jan 2 '16 at 0:28
  • $\begingroup$ I would write this if you don't know if $d$ is larger or smaller than 1. However $1 \leq a, b \leq n$, while more common, can both mean that $1 \leq a$ and $b \leq n$, or that $a$ and $b$ are both in the interval $[1, n]$. $\endgroup$ – Ruben Jan 2 '16 at 11:32
  • $\begingroup$ I'm not sure there's anything wrong with $1\leq b=c>d$. The wrongness comes when you try to use those facts to assert anything about the relationship between $1$ and $d$. $\endgroup$ – David Richerby Jan 2 '16 at 12:47
  • $\begingroup$ @DavidRicherby, it implies that $1$ and $d$ have a common upper bound. (Not so useful for a total order, I agree ;) $\endgroup$ – Carsten S Jan 2 '16 at 19:27

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