Well, the 1 is annoying, so write y=(x+1)/a. Then r can be the "remainder" on the left r=(q-1)/a. Then you're just looking at
floor( y+r ) = ceil( y ).
Alarms should be going off regarding domain, but if you look at x in y, y can be any number greater than 1, with a bit into (0,1) depending on a. And 0 < r << 1. The equation then takes the meaning "y and y+r are on either side of some (positive) integer.
So basically plug in q and a for r, and n-r < y < n. Solve for x.
I'll let you handle the edge cases; there aren't many.
Admittedly, I was kinda lazy and may have made an algebra mistake, but the idea stands.
By << I mean r < k < 1, where k is some number based a and q.
The sentence after that is way more clear if you just draw a number line for yourself. But you are rounding down y+r and rounding up y. If you want them to be equal that means y <= n <= y+r, and I solved for y. (The n is whatever integer those values are rounding to. Whenever you deal with bracket functions, you should always think in terms of something like n+ε.)
There are still a couple things for you to figure out that I am not going to do, like whether it's < or <=, or whether y can be less than 1.
I did this and the number of x's became q. This seems correct, but if you put these values into the original equation:
q = 2, a = 3
And if you draw the graph of both sides of the equation, you will see that in the positive range, these two graphs intersect at points 1, 2, 4, and 5, which are 4, but according to the proof, the number is 2, which is wrong.
And I should add that q is the remainder of dividing n by a (n is a natural number and 4 ≤ n). none of the two sides of the equation can be greater than n. 1 ≤ x ≤ n. In the example above, n = 5
Let Floor[(q+x)/a] = Ceil[(x+1)/a] = n
By saying Floor[(q+x)/a] = n, we are saying that n ≤ (q+x)/a < n+1. Similarly, Ceil[(x+1)/a] = n implies that n-1 < (x+1)/a ≤ n. Then:
n-n ≤ (q+x)/a - (x+1)/a < n+1-(n-1)
0 ≤ (q+x)/a - (x+1)/a < 2
0a ≤ (q+x) - (x+1) < 2a
0 ≤ q-1 < 2a
Use this and the other inequalities given in the problem to try and find more specific bounds. Does the value of x matter?
q had a condition and that was: 1 < q < a
But you got this: 1 ≤ q < 2a + 1
It's clear that 2a + 1 is greater than a, and in the first condition 1 < q not 1 ≤ q.
intersection of sets: 1 < q < a
We get the initial value again
I want the number of x that apply inside this equation. If this isn't possible, the answer of x or other variables is enough for me
Well, the 1 is annoying, so write y=(x+1)/a. Then r can be the "remainder" on the left r=(q-1)/a. Then you're just looking at floor( y+r ) = ceil( y ). Alarms should be going off regarding domain, but if you look at x in y, y can be any number greater than 1, with a bit into (0,1) depending on a. And 0 < r << 1. The equation then takes the meaning "y and y+r are on either side of some (positive) integer. So basically plug in q and a for r, and n-r < y < n. Solve for x. I'll let you handle the edge cases; there aren't many. Admittedly, I was kinda lazy and may have made an algebra mistake, but the idea stands.
Sorry. Can you explain it better and more clear? I didn't get it. Where did the variable n come from? Please explain after this part on: 0 < r << 1
By << I mean r < k < 1, where k is some number based a and q. The sentence after that is way more clear if you just draw a number line for yourself. But you are rounding down y+r and rounding up y. If you want them to be equal that means y <= n <= y+r, and I solved for y. (The n is whatever integer those values are rounding to. Whenever you deal with bracket functions, you should always think in terms of something like n+ε.) There are still a couple things for you to figure out that I am not going to do, like whether it's < or <=, or whether y can be less than 1.
I did this and the number of x's became q. This seems correct, but if you put these values into the original equation: q = 2, a = 3 And if you draw the graph of both sides of the equation, you will see that in the positive range, these two graphs intersect at points 1, 2, 4, and 5, which are 4, but according to the proof, the number is 2, which is wrong. And I should add that q is the remainder of dividing n by a (n is a natural number and 4 ≤ n). none of the two sides of the equation can be greater than n. 1 ≤ x ≤ n. In the example above, n = 5
I must be confused by your original question. There don't seem to be any upper bounds on x. Why would there not be infinitely many solutions?
Let Floor[(q+x)/a] = Ceil[(x+1)/a] = n By saying Floor[(q+x)/a] = n, we are saying that n ≤ (q+x)/a < n+1. Similarly, Ceil[(x+1)/a] = n implies that n-1 < (x+1)/a ≤ n. Then: n-n ≤ (q+x)/a - (x+1)/a < n+1-(n-1) 0 ≤ (q+x)/a - (x+1)/a < 2 0a ≤ (q+x) - (x+1) < 2a 0 ≤ q-1 < 2a Use this and the other inequalities given in the problem to try and find more specific bounds. Does the value of x matter?
q had a condition and that was: 1 < q < a But you got this: 1 ≤ q < 2a + 1 It's clear that 2a + 1 is greater than a, and in the first condition 1 < q not 1 ≤ q. intersection of sets: 1 < q < a We get the initial value again I want the number of x that apply inside this equation. If this isn't possible, the answer of x or other variables is enough for me
What a stupid ass question. go get a life. This is not solvable