1
  1. $f(n) = (1+n)^2$
  2. $f(n) = \lfloor\log(n)\rfloor$
  3. $f(n) = \lfloor\sqrt{n}\rfloor$
  4. $f(n) = (1-n)^2$

I am having a hard time understanding what strictly increasing means. Aren't all of these functions increasing? I know the squared ones are decreasing at some point, but then would it be the log or square root, one? An explanation would be greatly appreciated.

Yiyuan Lee
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djhunterx
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  • For the function to be strictly increasing, as $x_1 < x_2 < \cdots$ in the domain of $f$, $f(x_1) < f(x_2) < \cdots$ in the range of $f$ as well – NasuSama Mar 04 '14 at 03:56
  • ... in particular, $f(x_1) \neq f(x_2)$ and so forth. –  Mar 04 '14 at 03:57
  • For ones you don't suspect to be true, find a counter example. For ones you believe do increas, follow Nasu Suma's advice. – MT_ Mar 04 '14 at 04:07

2 Answers2

2

A function is strictly increasing if whenever $a < b$, it is true that

$$f(a) < f(b)$$

For example, if $a < b$ in $\mathbb{N}$, and we consider $f(x) = (x + 1)^2$, we see that

\begin{align*} f(b) - f(a) &= (b + 1)^2 - (a + 1)^2 \\ &= b^2 + 2b + 1 - \Big(a^2 + 2a + 1\Big) \\ &= b^2 - a^2 + 2(b - a) \end{align*}

Convince your self that $b^2 - a^2 > 0$ and, and note that $b - a > 0$; it follows that $f(b) - f(a) > 0$, which is equivalent to checking that $f$ is strictly increasing.

As a different example, consider $$g(x) = \left\lfloor \frac x 2 \right\rfloor$$

Then $g$ is an increasing function, but it's not strictly increasing on $\mathbb{N}$: We have $g(2) = 1 = g(3)$.

So every strictly increasing function is increasing, but not the opposite.

1

Strictly increasing means that $f(n) \neq f(n+1)$ while just increasing allows that. The floor function tends to make things equal. You might compute some early values for each. The answer to 4 depends on whether $0 \in \Bbb N$

Ross Millikan
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