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I recently showed a problem to someone and they stated the following:

If $a<b$ then $a\le b$.

I find this very confusing how can if we stated that $a$ is strictly less than $b$ how can we then say that $a$ is less than $b$ and could be equal to $b$? For example if I wanted to solve an equation where a strict inequality was stated putting $x\le b$ would be incorrect right?

J.G.
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The homeschooler
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    Do you see that it is true that $a < b \implies a \le b$ but it is not true that $a < b \Leftarrow a \le b$? In particular, it is not true that $a < b \Leftrightarrow a \le b$ – Benjamin Wang Dec 22 '20 at 09:00
  • What they state is true if you consider $a \ne b$. Otherwise consider the case where $a = b$ and get your own conclusions. – Gabrielek Dec 22 '20 at 09:01

2 Answers2

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$a\le b$ is an abbreviation for $a<b\lor a=b$ ($\lor$ denoting the inclusive or, which satisfies $p\implies(p\lor q)$).

J.G.
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  • But if I had an equation say $x+2<4$ then surely I cannot say x+2$\le$4 and then solve the equation giving x$\le$4 – The homeschooler Dec 22 '20 at 09:13
  • Sorry that should be x$\le$2 – The homeschooler Dec 22 '20 at 09:19
  • @Thehomeschooler I think you're referring to the fact that ${x|x\le2}\ne{x|x<2}$. This isn't an issue with inequalities so much as the fact that, when you determine which $x$ satisfy a condition, any unidirectional inference may introduce spurious solutions you remove by checking them later. – J.G. Dec 22 '20 at 09:22
  • So it is not incorrect to say this – The homeschooler Dec 22 '20 at 09:25
  • also what do you mean by satisfies $p$ $\implies $ (p$\lor$q) – The homeschooler Dec 22 '20 at 10:37
  • @Thehomeschooler I mean for ay true $p$ and for any proposition $q$, $p\lor q$ is also true. BTW, I advise you to just put one pair of dollar signs around the entire expression. – J.G. Dec 22 '20 at 11:47
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Define $\leq$ first say on the set of integers:

$a\leq b :\Leftrightarrow \exists c\in\Bbb Z_{\geq 0}:\; a+ c =b$.

Then define $<$ as follows

$a<b :\Leftrightarrow a\leq b \wedge a\ne b.$

Wuestenfux
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