14.2.04
proof: sqrt(3) is irrational
A few days ago I woke up at around 2am to visit the rest room. I had left my MSN messenger on. There I had a message asking if I could prove the irrationality of sqrt(3). One of my college friends had sent the message. I tried to send her an outline of a proof by contradiction that I thought should work. She initially thought I was wrong, but later when I expanded each step in detail, she was convinced. Here is the proof:
Assume sqrt(3) is rational. Now since it is rational by assumption, it can be written as p/q (in its most reduced form). So we have,
sqrt(3)=p/q
or, 3=p^2/q^2
or, 3*q^2=p^2 ----(i)
q is even --> q^2 is even (even multiplied by even)
q^2 even -->3*q^2= p^2 is even (odd multiplied by even is even)
P^2 even --> p is even
So q is even --> p is even. But this would imply p/q is not in its most reduced form, since both p and q have a common factor of 2. So we cannot assume that q is even. Now let us take the other possibility: that q is odd.
q is odd--> q^2 is odd (odd multiplied by odd is odd) --> 3*q^2= p^2 is odd --> p is odd.
Hence, we must assume that both p and q are odd. So we can write p and q as:
p=2*m+1 and q=2*n+1, where m and n are any integers.
Now let us substitute this new expressions of p & q in eqn. (i).
3*(2*n+1)^2=(2*m+1)^2
or, 3*(4*n^2 + 4*n +1)= 4*m^2 + 4*m +1
or, 12*n^2 + 12*n + 3 = 4*m^2 + 4*m +1
Now divide both sides by 2. So we have,
6*n^2 + 6*n + 1 + 1/2 = 2*m^2 + 2*m +1/2
or, 6*n^2 + 6*n +1 = 2*(m^2 + m)
Now notice that on the left hand side the first term is even (since 6 which is an even is multiplying n^2). Similarly 6*n is even. But the last term is odd. So on the left hand side we have even + even + odd = odd. On the right hand side, the terms in the parentheses are multiplied by 2 (an even), and hence is even. So we have,
odd=even.
This is a contradiction that we arrive at if assume that sqrt(3) is rational. Hence, it cannot be rational, and hence must be irrational (since these two are mutually exhaustive possibilities: a number must either be rational or irrational). Q.E.D.
Assume sqrt(3) is rational. Now since it is rational by assumption, it can be written as p/q (in its most reduced form). So we have,
sqrt(3)=p/q
or, 3=p^2/q^2
or, 3*q^2=p^2 ----(i)
q is even --> q^2 is even (even multiplied by even)
q^2 even -->3*q^2= p^2 is even (odd multiplied by even is even)
P^2 even --> p is even
So q is even --> p is even. But this would imply p/q is not in its most reduced form, since both p and q have a common factor of 2. So we cannot assume that q is even. Now let us take the other possibility: that q is odd.
q is odd--> q^2 is odd (odd multiplied by odd is odd) --> 3*q^2= p^2 is odd --> p is odd.
Hence, we must assume that both p and q are odd. So we can write p and q as:
p=2*m+1 and q=2*n+1, where m and n are any integers.
Now let us substitute this new expressions of p & q in eqn. (i).
3*(2*n+1)^2=(2*m+1)^2
or, 3*(4*n^2 + 4*n +1)= 4*m^2 + 4*m +1
or, 12*n^2 + 12*n + 3 = 4*m^2 + 4*m +1
Now divide both sides by 2. So we have,
6*n^2 + 6*n + 1 + 1/2 = 2*m^2 + 2*m +1/2
or, 6*n^2 + 6*n +1 = 2*(m^2 + m)
Now notice that on the left hand side the first term is even (since 6 which is an even is multiplying n^2). Similarly 6*n is even. But the last term is odd. So on the left hand side we have even + even + odd = odd. On the right hand side, the terms in the parentheses are multiplied by 2 (an even), and hence is even. So we have,
odd=even.
This is a contradiction that we arrive at if assume that sqrt(3) is rational. Hence, it cannot be rational, and hence must be irrational (since these two are mutually exhaustive possibilities: a number must either be rational or irrational). Q.E.D.
Happy Valentine's Day
I saw this search tool bar on google.com site which is set as my default homepage (since I use it the most). I got attracted to the toolbar not because it would let me search from any page I am on, but because it had the pop-up blocking feature.
As I read about other features of this little toolbar, I happened to come across the "blogThis" botton and get introduced to blogging for the first time. So here I am giving it a try, and see if I can really keep up to posting something new every now and then.
Btw, today is Valentine's day. So let me wish you all a very HAPPY VALENTINE'S DAY. May you all find the true love, and be happy ever after.
As I read about other features of this little toolbar, I happened to come across the "blogThis" botton and get introduced to blogging for the first time. So here I am giving it a try, and see if I can really keep up to posting something new every now and then.
Btw, today is Valentine's day. So let me wish you all a very HAPPY VALENTINE'S DAY. May you all find the true love, and be happy ever after.
