prove that √3 is an irrational number
Answers
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Answer:
The number 3–√ is irrational ,it cannot be expressed as a ratio of integers a and b. To prove that this statement is true, let us Assume that it is rational and then prove it isn't (Contradiction).
So the Assumptions states that :
(1) 3–√=ab
Where a and b are 2 integers
Now since we want to disapprove our assumption in order to get our desired result, we must show that there are no such two integers.Squaring both sides give :
3=a2b2
3b2=a2
(Note : If b is odd then b2 is Odd, then a2 is odd because a2=3b2 (3 times an odd number squared is odd) and Ofcourse a is odd too, because oddnumber−−−−−−−−−√ is also odd.
With a and b odd, we can say that :
a=2x+1
b=2y+1
Where x and y must be integer values, otherwise obviously a and b wont be integer.
Substituting these equations to 3b2=a2 gives :
3(2y+1)2=(2x+1)2
3(4y2+4y+1)=4x2+4x+1
Then simplying and using algebra we get:
6y2+6y+1=2x2+2x You should understand that the LHS is an odd number. Why?
6y2+6y is even Always, so +1 to an even number gives an ODD number.
The RHS side is an even number. Why? (Similar Reason)
2x2+2x is even Always, and there is NO +1 like there was in the LHS to make it ODD.
There are no solutions to the equation because of this.
Therefore, integer values of a and b which satisfy the relationship = ab cannot be found.
Therefore 3–√ is irrational