Below is a proof by contradiction of the irrationality of √2.
I chose this to be my first post not only because I love maths, but also because steemit and the blockchain technology is an effective way to store knowledge and keep it available for those who need it.
A proof by contradiction
Let's assume √2 is a rational number.
By definition there must exist 2 real numbers, say a and b, with b non zero such that √2 = a/b.
a/b is the fraction simplified with the lowest terms i.e. an irreducible fraction.
√2 = a/b
-> 2 = a²/b²
-> 2b² = a²
Hence a² is even (because it is a multiple of 2).
a is also even because if it were odd, its square would be odd too.
a being even we can rewrite it as a multiple of 2, a = 2k.
We had 2b² = a², let's plug into this our new way of writing a.
2b² = (2k)²
-> 2b² = 4k²
-> b² = 2k²
b² must also be even since it is a multiple of 2.
As stated above, b² being even implies that b is also even.
Both a and b are even.
That means a/b could be reduced even further by diving the numerator and denominator by 2.
However a/b was supposed to be an irreducible fraction.
There is a contradiction.
The assumption must be false.
√2 is an irrational number.
Do let me know if you find any part of this proof unclear.
Thanks for reading.