00:01
So according to the question here, we will let that ab and c d be any two non -degrading, closed and abundant intervals.
00:24
Here, so it is having a and b here, this is 0 and here is the 1.
00:44
So from these, the step 1, we will do it the step twice.
00:50
So we are to show that ab is equal to the cd here.
01:02
So we actually show that a comma b is equal to the 0 .1 here.
01:11
We define as 1 equals to a and b by the equation fx is equal to b minus a plus a plus.
01:29
Be here so we can say that now show that f is the bijection here so right so f is the one one we will see first of all we will let here x1 x2 we're having 0 and 1 with which x1 is not equal to the x2 whereas b minus a x1 is not equal to the b minus a x2 here so we can say that b minus a x1 plus b is not equal to b minus a x2 plus b here so since x1 is not equal to x x2 so we can say that f x1 is not equal to f x2 here so therefore we can say that f is only one one here.
02:53
Now if f is on 2, we will see here if f is on 2, then what we will do is we will let m, a and b be the any point.
03:09
So we can say that m is equal to b minus a x minus 1 plus b here.
03:16
We can also say that m minus b and b minus a is equal to x minus 1 here right so we will see here we will put it into f x here is m minus v divided by b minus a plus 1 which will be equal to b minus a here and m minus b b minus a here and m minus b b minus a plus 1 minus 1 plus b here after which it will be m minus b plus b we will cut plus b minus b and it will form only m here so they show that for any point of the m right such that f is the onto and hence we can say that it is the bijective here so therefore we can say that 0 and 1 is equal to a, b, which is a first equation.
04:33
Similarly, we can say that 0, 1 is equal to cd.
04:40
It is a second equation.
04:45
So, by 1 and 2, we can say that a, comma, b is equal to c, d here...