Prove sqrt2 is an irrational number
1. Traditional method using contradiction
Assume sqrt2 is a rational number, so that sqrt2 = a/b, a and b are relatively prime integers.
sqrt2 = a/b => 2 =a2/b2 => 2b2 = a2
so that a2 is an even number, then a is an even number as well.
Therefore, we can let a = 2c => 2b2 =2(2C)2 => b2 = 4c2
so that b2 is an even number, then b is an even number as well.
Now both a and b are even number, they cannot be relatively prime, which is contrary to assumption: sqrt2 is a rational number, so that sqrt2 = a/b, a and b are relatively prime integers.
Therefore assumption is false, and sqrt2 is an irrational number.
2. Prove using strong induction
P(n) is the statement that sqrt2 =/= n/b for any positive integer b.
Basic step:
Prove P(1) is true
P(1) : 1/b <=1 <sqrt2, so that sqrt2 =/= 1/b, P(1) is true.
Inductive step:
We assume P(j) is true for all 1<=j<=k: sqrt2 =/= j/b, and we need prove P(k+1) is true as well: sqrt2 =/= (k+1)/b
We assume sqrt2 = (k+1)/b for some b. So that 2 = (k+1)2/b2 => 2b2 = (k+1)2
we can conclude that (k+1)2 is an even number and k+1 is an even number as well.
Let k+1 = 2c, 2b2 = (k+1)2 => 2b2 = (2c)2 => b2 = 2c2
we can conclude that b is an even number.
Let b = 2d, sqrt2 = (k+1)/b => sqrt2 = 2c/2d = c/d.
For sqrt2 = c/d, d is a positive integer, c<k for k>1 (since k+1 = 2c), this contradicts our previous assumption: We assume P(j) is true for all 1<=j<=k: sqrt2 =/= j/b.
Therefore our second assumption: We assume sqrt2 = (k+1)/b for some b. is false. We conclude sqrt2 =/= (k+1)/b for all positive integer b.
So that sqrt2 =/= n/b for any positive integer b, and sqrt2 is an irrational number.
2018/12/02
