I am studying a physics course in college and I found that it is actually really difficult to remember long-phrased statements. Thus, I have been trying to determine those statements through the use of logic and mathematical reasoning and the formulas that are given in the textbook.
This problem is a concave mirror exercise that does not have any number. In fact, it is only based on a diagram and abstract variables.
Some formulas have been given in the textbook. They are: $$\frac 1f = \frac1s + \frac{1}{s'}$$
$$m = -\frac{s'}{ s}$$
$$2f = R$$ I have been told that all distances from the vertex made with the optic axis and a concave mirror are positive when going away from the mirror in the same side as the center of curvature of the mirror.
This is what I have done so far:
With all of this, I can prove that the image is inverted and real. However, It is highly difficult for me to determine whether |s'| is greater, equal or less than |s|, which is necessary to determine whether |m| is greater, equal or less than 1.
Let's start with real image, where $s$ and $s'$ are positive. And we should consider first the case $s'=s$. Then $$\frac1f=\frac2s$$or $s=2f$. So when the object is at the center of curvature, $R$ from the mirror, then the image is also formed at the same position. If you look now at the first equation, increasing $s$ will decrease $s'$. When $s\to \infty$, then $s'\to f$. If you decrease $s$, from $2f$ to $f$, then $s'$ will increase. Note now that at $s=f$, you get $\frac1{s'}=0$. This corresponds both to $s'=\infty$ and $s'=-\infty$.
Once $s<f$, then $\frac1s>\frac1f$. So $s'$ must be negative. So we rewrite the first equation as $$\frac1f=\frac1s-\frac1{|s'|}$$ Since $1/f>0$, then $$\frac1s>\frac1{|s'|}$$ or $$s<|s'|$$