00:01
The function is said to be injective when f of x1 is equal to f of x2, that implies x1 is equal to x2.
00:11
Then we say that the function is injected for all x1, x2 belonging to the domain of the function f.
00:18
Now here x1 and x2 are coming from the set integers, the set of integers.
00:24
So first of all, let's take case one.
00:27
Let x1 and x2 are both greater than are equal to 0.
00:32
And x1 and x2 are distinct, i mean, let's take x1 and x2 greater than equal to 0 and equate f of x1 equal to f of x2.
00:42
So f of x1 equal f of x2, then what do you get? what is f of x1 2 x1? f of x2, 2, 2, 2.
00:52
So that implies x2 equal to x2.
00:54
This is case 1.
00:55
So we got this condition, right? case 2.
00:59
Now suppose both x1 and x2 are negative.
01:02
In that case, f of x1 equal to f of x2 implies minus 2 x1 minus 1 is equal to minus 2 x2 minus 1 sorry minus 1 then again you'll get x1 equal x2 well angle now let's take the third case let's take whether third case is feasible or not what do you mean by third case when x1 is greater than equal to 0 and x2 is negative and f of x1 is equal to f of x2 let's check whether is it possible f of x1 is 2x1 and this is negative 2x2 minus 1 but this is always an even integer and this is an integer is this possible or integer can never be an even integer so that means case three itself is not feasible so this is not feasible it's not feasible so that means from case one and case two if images are equal the numbers exponent x2 should be equal so it is a injective function this is an injective function it's an injective function all right now for subjectivity what is the definition of subjective function for every element in the co -domain that exists, there should exist at least one pre -image in the domain.
02:16
Now if you can see the codomain is all natural numbers, then what is the range of this function? if you can see the first piece, the first piece covers all the natural numbers which are even, that is 0, 2, 4, 6, 8, so on.
02:37
Because 2 into x is an even number.
02:41
So if you take x is 0, 2 into 0, 0, when x is 1, 2 to 8, 1, 2, 2 like that.
02:46
And how about second piece? what numbers will be present in the range? when x is less than 0, what are the negative integers? negative 1, negative 2, negative 3, all those.
02:58
When you substitute, so when you substitute negative 1, what do you get? it's 1.
03:03
When you substitute negative 2, what do you get? it's 3 and 5 and so on, all the odd numbers.
03:09
So, all the odd numbers.
03:10
So, the even natural numbers, all the odd natural numbers are covered.
03:14
So that means what is the range? the complete set natural numbers.
03:20
So since the range is same as codomane, because codomane is also natural numbers, range is same as codomane.
03:27
The function is subjective.
03:31
So this function is bijective.
03:34
It is both injective and subjective.
03:40
So it is given that fmaps a to b, g, maps b to c and g oath maps a to c and is also given that x is bijective g circle f is also bijective so basically f is injective and subjective g circle f is also injective and subjective now we need to check whether g is bijecture two let's start something like this let x1 and x2 belong to set a and x1 is not equal to x2 when x1 is not equal to x2 f of x1 is also not equal to f of x2 why because f is injective f is injective because injectivity says that distinct elements will match the distinct images, right? that is objective.
04:25
Now this f of x1 and f of x2 are belonging to which set b because f of x1 and f of x2 will be lying here when x1 and x2 lies here.
04:35
Now when f of x1 is not equal to f of x2, let us suppose that, let us suppose that g is not injective.
04:51
I'm assuming g is not injective.
04:54
That means for some f of x1 and f of x2, i'll be having g of f of x1 is equal to g of f of x2 because that is the definition of not injective.
05:05
What do you mean by not injective when two elements are not equal but their images are equal? not injective means many to one.
05:12
Suppose two elements are not equal, a and b, but their images are equal.
05:18
That means f of a is same as f of b or g of a is same as g of b.
05:21
I'll use g.
05:22
G of is same as g of.
05:24
So is g injective? no, it's not injective.
05:27
So what do you mean by not injective? when two elements are not equal in the domain, so that means f of x1 is not equal to f of x2, because these two are in the domain of g, but their images should be equal.
05:38
That is the meaning of g is not injective.
05:41
So this implies g circle f of x1 is equal to g circle f of x2.
05:46
But, but, but, but g circle f is injective, but g circle f is injective.
05:53
G circle f is injective.
05:56
So this should imply x1 equal to x2.
06:00
That means we have arrived at a contradiction...