Is the set of all real numbers x such that 0 < x ≤ 1, with operation multiplication.
So I choose any elements x,y in the above set and considered their multiplication.
Since x and y are two different elements can their product be equal to 1? (Since if x = 1 then y cannot be equal to 1 right?)
So can we say the closure property holds for above set?
Thanks!
closure holds, but why do you say that x≠y? That is not necessary.
Isn't it required to show the closure property holds for two different number x,y?
closure means that if x and y are elements of the set, then x*y is also.
They need not be different.
To determine if the closure property holds for the set of all real numbers x such that 0 < x ≤ 1 with multiplication as the operation, we need to examine whether the product of any two elements in the set also belongs to the set.
Let's consider two different elements, x and y, from the given set:
1. If x is equal to 1, then y cannot be equal to 1, as you rightly pointed out. Since 1 multiplied by any non-zero real number is equal to that number, we have x * y = 1 * y = y. Since y must be a non-zero real number, and 0 < y ≤ 1, the product x * y still satisfies the condition of 0 < x * y ≤ 1. Therefore, if x = 1, the closure property holds.
2. If both x and y are less than 1 but greater than 0, their product x * y will also be less than 1. Since multiplication of two real numbers between 0 and 1 results in a number that is also between 0 and 1, the closure property holds in this case as well.
In both scenarios, we find that the product of any two elements from the set is still an element within the given set. Therefore, the closure property holds for the set of all real numbers x such that 0 < x ≤ 1 with multiplication as the operation.