I am setting out to find out why the square root of 5 is irrational. It may be obvious that there is no whole number or integer that when squared is equal to 5, but why the square root of 5 is irrational, is a different problem.
I found that there are many ways of proving that the square root of 2 is irrational including proofs by contradiction and by constructive proofs, so it may be easier to figure out how the square root of 5 is irrational.
Things that will be easy are that there are lots of resources to use in order to formulate my own proofs and gather ideas to figure out how to solve the problem.
The thing that will be hardest for me to solve this problem is that I am bad at proofs in general. I have a hard time understanding the logic behind why proofs are written the way they are and why certain strategies are used.
Here is my thought process on how I went about it. A rational number is one that can be written as m/n, m, n in the set of integers. So I let the square root of 5 equal m/n which means 5=m²/n². And in order to get an odd quotient, m² and n² must be odd. Thus m and n are odd because the square root of an odd square integer is odd since an odd integer times an odd integer is odd. So I tried substituting m for 2k+1 and n for 2j+1 and that did not work. Then I tried substituting m for 5x and n for 5y. That did not work either. I tried seeing if there was some kind of pattern with square numbers, but I decided that it would take too long to write and think about all of the square numbers and their relation to one another. Finally I tried substituting things for other known variables and assumed that m and n are coprimes because or else there would be a seemingly infinite amount of possible rational numbers that would equal the square root of 5.
Proof 1:
This is a proof by contradiction that will prove the square root of 5 is irrational.
Assume that the square root of 5 is rational.
Let sqrt(5)=m/n, m and n in the set of positive integers and m and n are coprime which means that m and n have no other common factors besides 1. Then 5=m²/n², 5n²=m². This means that 5 is a factor of m. Therefore 5c=m for some integer c → m²=25c². Then substitute 25c² for m² in 5n²=m², 5n²=25c² → n²=5c² therefore 5 is a factor of n as well as m. This is a contradiction because we stated that m and n only have a common factor of 1, but that is not true.
Conclusion: It was a lot easier to form a proof after looking at my notes. Discrete math was quite helpful while writing this and understanding others’ works.