Question

show that √3 is an irrational

Answer 1

Let us assume that √3 is a rational number.

then, as we know a rational number should be in the form of p/q

where p and q are co- prime number.

So,

√3 = p/q { where p and q are co- prime}

√3q = p

Now, by squaring both the side

we get,

(√3q)² = p²

3q² = p² ........ ( i )


So,

if 3 is the factor of p²

then, 3 is also a factor of p ..... ( ii )


=> Let p = 3m { where m is any integer }

squaring both sides

p² = (3m)²

p² = 9m²

putting the value of p² in equation ( i )

3q² = p²

3q² = 9m²

q² = 3m²

So,

if 3 is factor of q²

then, 3 is also factor of q


Since

3 is factor of p & q both

So, our assumption that p & q are co- prime is wrong

hence,. √3 is an irrational number

Answer 2

The number 3–√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–√=ab3=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=a2b23=a2b2

3b2=a23b2=a2

(Note : If bb is odd then b2b2 is Odd, then a2a2is odd because a2=3b2a2=3b2 (3 times an odd number squared is odd) and Ofcourse a is odd too, because oddnumber−−−−−−−−−√oddnumber is also odd.

With a and b odd, we can say that :

a=2x+1a=2x+1

b=2y+1b=2y+1

Where x and y must be integer values, otherwise obviously a and b wont be integer.

Substituting these equations to 3b2=a23b2=a2gives :

3(2y+1)2=(2x+1)23(2y+1)2=(2x+1)2

3(4y2+4y+1)=4x2+4x+13(4y2+4y+1)=4x2+4x+1

Then simplying and using algebra we get:

6y2+6y+1=2x2+2x6y2+6y+1=2x2+2x

You should understand that the LHS is an odd number.

6y2+6y6y2+6y is even Always, so +1 to an even number gives an ODD number.

The RHS side is an even number. (Similar Reason)

2x2+2x2x2+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 = abab cannot be found.

Therefore 3–√3 is irrational

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