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This seems like it is true since b is a real number it can be 10.5 for example, and a is 11.

I am thinking it could be false since it doesn't hold for every number? Like if b is 10.5, let a be 18. Do inequalities like this have to hold ALWAYS for them to be considered true regardless of the numbers assigned to the variables? I've searched around but can't find a clear answer on that, apologies if it is obvious. The phrasing of the question throws me off because it just says 'There exists.'

Peter
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MathPoser22
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    This is false if $b$ is an integer, say $b = 1$. Are you missing an "$\leq$" in "$b < a < b+1$"? – Eric Towers Oct 28 '21 at 15:39
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    In particular it is false, if $b+1$ is negative , since $a$ as a natural number can then not be smaller. – Peter Oct 28 '21 at 15:42
  • Assuming the statement is $b\leq a<b+1$ with $a\in\mathbb Z$, let $S\subset\mathbb Z$ be the set of integers less than $b$. $b$ is an upper bound, so there must be a least upper bound. It's easy to see that this least upper bound must be an integer. Call it $a-1$. Can you show that $a$ has the properties you claim? – Rushabh Mehta Oct 28 '21 at 15:42
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    You don't let $a=18$, rather you find an $a$ which works ("there exists" means you can find at least one such $a$, and does not mean that every value of $a$ must work). – Mark Bennet Oct 28 '21 at 15:43
  • I am not missing a ≤. the question is exactly as I submitted. b < a < b+1 – MathPoser22 Oct 28 '21 at 15:44
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    @MathPoser22 Then it is false, with the counterexample Eric provided. – Rushabh Mehta Oct 28 '21 at 15:44
  • @DonThousand That's what is confusing me a little Don. The question says, 'There exists.' Does that not indicate there only has to be 1 example of it being true for it to be true? A single counterexample would only be sufficient if the question is 'for all' ? – MathPoser22 Oct 28 '21 at 15:47
  • @MathPoser22 No, we need one $a$ for EVERY real number $b$. – Peter Oct 28 '21 at 15:50
  • Otherwise unquantified variables, $b$ in the sentence you are focussing on, are universally quantified. So, as a single sentence, "For all $b \in \Bbb{R}$, there exists an $a \in \Bbb{N}$ such that $b < a < b+1$." So, for each and every choice of $b$, there must be at least one $a$ making the inequality true. Is there a $b$ for which that $a$ does not exist? – Eric Towers Oct 28 '21 at 15:50
  • @EricTowers and Peter, thank you. That clears it up for me. I appreciate the help and patience. – MathPoser22 Oct 28 '21 at 15:54

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Typically, we write “Let $b\in \mathbb R$” in a proof to declare the variable $b$ as an arbitrary/dummy real number; because $b$ is just a placeholder for a general but no particular member of $\mathbb R,$ whatever ensuing derivation involving it generally applies to any real number in its place; we can then conclude by writing “For each $b\in \mathbb R,$ [the derived result involving $b$]”. See: How to interpret "let" in mathematics?

Let $b\in \mathbb R$. There exists $a\in \mathbb N$ such that $b < a < b+1$.

Guided by the above context where we convert “let $b\in \mathbb R$” (“let $b$ be an arbitrary real number”) to “for each $b\in \mathbb R$”, then the given two-sentence claim is false: $b=10$ is a counterexample.

On other hand, if we are intended to understand the given claim as true, then we can conclude that $b\in\mathbb R^+{\setminus}\mathbb Z.$

ryang
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