prove why root 23 is irrational
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Here's the solution.
Notation:
a | b is read as "a divides b"
a | b means that "a" is a factor of "b"
Thus on dividing 'b' by 'a' , remainder is zero if a | b
Eg. 2 | 10 , 3 | 6 , etc.
Now, if a | b^2, then a | b.
Eg. 3 | 81 , so 3 | 9 is also true (as 81 = 9^2)
Next,
Significance of HCF (m,n) = 1
Here we want the least possible numbers that denote rational form.
For example, 3/4 can be written as
6/8, 9/12 , 12/16 , etc.
But in the latter ones, there are common factors which can be cancelled out.
Thus, for 3/4 , we want only 3/4. There is no common factor between 3 and 4. The statement HCF(3,4)=1 denotes just that.
So, HCF (m,n) = 1 tells us that there are no common factors between m and m.
First we assume √23 to be rational and express it in m/n form.
Later on in the proof, we create a contradiction that 23 divides both m and n. But as we had said that their HCF is 1 , our assumption that √23 is rational must be false.
So, √23 is irrational.
Notation:
a | b is read as "a divides b"
a | b means that "a" is a factor of "b"
Thus on dividing 'b' by 'a' , remainder is zero if a | b
Eg. 2 | 10 , 3 | 6 , etc.
Now, if a | b^2, then a | b.
Eg. 3 | 81 , so 3 | 9 is also true (as 81 = 9^2)
Next,
Significance of HCF (m,n) = 1
Here we want the least possible numbers that denote rational form.
For example, 3/4 can be written as
6/8, 9/12 , 12/16 , etc.
But in the latter ones, there are common factors which can be cancelled out.
Thus, for 3/4 , we want only 3/4. There is no common factor between 3 and 4. The statement HCF(3,4)=1 denotes just that.
So, HCF (m,n) = 1 tells us that there are no common factors between m and m.
First we assume √23 to be rational and express it in m/n form.
Later on in the proof, we create a contradiction that 23 divides both m and n. But as we had said that their HCF is 1 , our assumption that √23 is rational must be false.
So, √23 is irrational.
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