This is a solution, I don't think it is the best. $M$ admits at least one element, let it be $x_0$, then we have two cases: 1) If $x_0 \geq 1$, then we found easily that : $3x_0 - 2 \geq x_0$, and $3(3x_0 - 2) - 2 \geq 3x_0 - 2 \geq x_0$ and so on. We have then : $3^n(x_0-1)+1 \geq x_0$, with $n$ to $\infty$, we see that $M$ doesn't accept a bigger element, so M is infinite. 2) If $x_0 \leq 1$, we proceed the same way : $-4x_0+5\geq x_0$, and we found : $(-4)^n(x_0-1)+1\geq x_0$ so $M$ doesn't admit a bigger element, so M is also infinite. Done.

oussama1305 wrote: This is a solution, I don't think it is the best. $M$ admits at least one element, let it be $x_0$, then we have two cases: 1) If $x_0 \geq 1$, then we found easily that : $3x_0 - 2 \geq x_0$, and $3(3x_0 - 2) - 2 \geq 3x_0 - 2 \geq x_0$ and so on. We have then : $3^n(x_0-1)+1 \geq x_0$, with $n$ to $\infty$, we see that $M$ doesn't accept a bigger element, so M is infinite. 2) If $x_0 \leq 1$, we proceed the same way : $-4x_0+5\geq x_0$, and we found : $(-4)^n(x_0-1)+1\geq x_0$ so $M$ doesn't admit a bigger element, so M is also infinite. Done. Your idea is good, but I think your solution should be fixed, because the problem says "either $3x-2\in M$ or $-4x+5\in M$."

vinhhop wrote: oussama1305 wrote: This is a solution, I don't think it is the best. $M$ admits at least one element, let it be $x_0$, then we have two cases: 1) If $x_0 \geq 1$, then we found easily that : $3x_0 - 2 \geq x_0$, and $3(3x_0 - 2) - 2 \geq 3x_0 - 2 \geq x_0$ and so on. We have then : $3^n(x_0-1)+1 \geq x_0$, with $n$ to $\infty$, we see that $M$ doesn't accept a bigger element, so M is infinite. 2) If $x_0 \leq 1$, we proceed the same way : $-4x_0+5\geq x_0$, and we found : $(-4)^n(x_0-1)+1\geq x_0$ so $M$ doesn't admit a bigger element, so M is also infinite. Done. Your idea is good, but I think your solution should be fixed, because the problem says "either $3x-2\in M$ or $-4x+5\in M$." Ah! So I need just to add one more case in each case, and discussing the smaller element, like discussing the fact that $-4x_0+5\leq x_0$ or $3x_0 - 2 \geq x_0$ in the first case for example, and it's done.

hi octav can you explain me with details your solution please?

To Octav you can't say that |3x-3|=3|x-1| counter example x=2.5

proof: f(x)=3x-2 g(x)=-4x+5 then we can see that if (not necessarily order of f,g...) A= f(g(...g(x)...))=(-1)^k*4^k*3^t*x - (-1)^k*4^k*3^t + 1 where * product of number and k number of g, t number of f. and we can se that if k+t don't equal to p+q where p,q number of f,g A not equal B where B=(-1)^q*4^q*3^p*x - (-1)^p*4^p*3^q + 1 so M is infinite.

akhan98 wrote: To Octav you can't say that |3x-3|=3|x-1| counter example x=2.5 In fact, you can say that and the counterexample is invalid: $|3x-3|=|3\cdot 2.5-3|=|4.5|=4.5$ and $3|x-1|=3|2.5-1|=3\cdot 1.5=4.5$. Actually, for any reals (or even complex) $a,b$ the following holds: $|a\cdot b|=|a| \cdot |b|$.

i thought that |a| means that the greatest integer less than or equal to x

Sorry)

to phantom and what do you think about my solution???

@akhan98: No problem . The integral part symbol is kind of confusing around here as far as I've noticed, so it is understandable. EDIT: Regarding the solution, I don't really get what you meant with "number of f".

@PhantomR: if A= f(g(...g(x)...)) in the expression there are k times g and t times f so A=(-1)^k*4^k*3^t*x - (-1)^k*4^k*3^t + 1 because we can see that coefficient near the x is (-1)^k*4^k*3^t*x and if x=1 A=1 what means that A=(-1)^k*4^k*3^t*x - (-1)^k*4^k*3^t + 1 and do you understand what I mean?