to make it clearer if it seems messy:
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Show that for ∀x > 0, ∃m ∈ N such that m - 1 ≤ x < m.
Let x ∈ (0,∞) be arbitrary.
As there is no greatest natural number, ∃m ∈ N such that m > x.
Specifically, set m = ⌊x⌋ + 1.
That is, m is the smallest natural greater than x.
Then m - 1 is the largest natural less than or equal to x.
So m - 1 ≤ x < m.
Hence, for ∀x > 0, ∃m ∈ N such that m - 1 ≤ x < m.
Let x ∈ (0,∞) be arbitrary.
As there is no greatest natural number, ∃m ∈ N such that m > x.
Specifically, set m = ⌊x⌋ + 1.
That is, m is the smallest natural greater than x.
Then m - 1 is the largest natural less than or equal to x.
So m - 1 ≤ x < m.
Hence, for ∀x > 0, ∃m ∈ N such that m - 1 ≤ x < m.
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keywords: an,that,039,natural,or,there,gt,such,Prove,for,in,all,numbers,lt,exists,Prove that for all x>0, there exists an 'm' in N (natural numbers) such that m-1<or= x < m