Question

prove that √3 is an irrational number​

Answer 1

Step-by-step explanation:

Let us assume to the contrary that √3 is a rational number. where p and q are co-primes and q≠ 0. It means that 3 divides p2 and also 3 divides p because each factor should appear two times for the square to exist. ... This demonstrates that √3 is an irrational number.

Answer 2

(1) √3 =a/b

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.

Hope it will helps you ☺️

# Mɪss Kʜᴀɴ

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