Prove: \(\sqrt{2} \) is irrational. Note: In this proof, we assume the rules of algebra.  Furthermore, this is a reductio ad absurdum proof, or commonly known as, ''proof by contradiction.'' Proof: Suppose, on the other hand, that \(\sqrt{2} \) is rational.  If that is the case, then we can write \(\sqrt{2} \) as a ratio of two numbers or, in other words, as a fraction.  Then, \(\sqrt{2} \) = \(\frac {p}{q}\), where p is a positive integer and q is a natural number(this is the usual definition for a rational number), such that p and q share no common factors.  This essentially means that we've written \(\sqrt{2} \) in it's final reduced form(i.e., it cannot be reduced any further than here). Since \(\sqrt{2} \) = \(\frac {p}{q}\), this implies that \(2\) = \(\frac{p^{2}}{q^{2}}\).  Then, we can write the following by multiplying both sides by \(q^{2}\): (1)          \(p^{2}\) = \(2q^{2}\) Then, this implies that \(p^{2}\) is even.  Henceforth, we can make use of the lemma we were able to prove earlier(Aha!).  The lemma states that if \(k^{2}\) is even, then k is even.  Hence, we can assert that p is even.  Since p is even, then we can assert that p = 2a, where a is some integer.  Then, plugging this back into (1), we find: \((2a)^{2}\) = \(2q^{2}\) \(4a^{2}\) = \(2q^{2}\) \(2a^{2}\) = \(q^{2}\) Then, this implies that q is even by our lemma.  But, since p is even and q is even, this implies that they share a common factor of 2.  However, this is a contradiction!  This implies that \(\sqrt{2} \) =  \(\frac {p}{q}\) is not in it's final reduced form, which lies at the crux of our assumption.  Hence, our assumption was false and thus ... \(\sqrt{2} \) is not rational; it is irrational.  The proof is complete. .`, And we are done!  If you have any questions regarding this proof or you've spotted any errors, please feel free to message me. My next topic, which I shall explore tomorrow, shall be a primer on set theory and the mysteries concerning denumerabilities.  Is N infinite?  And if so, is it countably infinite?

Such a beautiful number!  But, it’s also highly underrated insofar as the general populace is concerned. The square root of 2 lies at the heart of teaching students the main concepts of irrationality.  In elementary school and high school, we are introduced to several number systems.  The basic and most utilized ones are the following: N, the natural number system. R, the real number system(otherwise known as just ‘the reals.’) Z, the integer number system. Q, the rational number system. C, the complex number system. Each number system is a set, which is a collection of objects — these objects being mathematical elements we refer to as ‘numbers — and some of these sets are subsets(think of a subset as a ‘child’ of a ‘parent’ set.  This definition is insufficient, but I shall expand more on set theory later on!) of other sets. For example, the integers Z are a subset of the rationals, Q.  The rationals are, in turn, a subset of R.  The ‘youngest child’ of them all, so to speak, is N.  The naturals form the basis of the vast majority of number sets and, as such, much work has been done on them.  The principle of induction is based primarily off of the natural numbers, and the well-ordering principle is concerned with N.  Despite being the youngest child, it is quite possibly one of the most powerful of the number sets. So now, let’s ask something interesting: What are the reals composed of?  What mathematical elements/numbers do they have?  Well, let’s see.. Since Q is a subset of R, as previously stated, all elements of Q are in R.  So, R contains all of the rational numbers.  But, the rational numbers contain the integers Z, which in turn include the natural numbers N.  So, what does R look like? A graphical image wouldn’t necessarily suffice to cover the whole spectrum of R — in fact, none exist! — but let’s actually take a peek into the set. R = {…1/3, 2/3, 1, 2, e, 3, sqrt(2), 4, 4.54, 5.924 ….} Something seems amiss; did I make a mistake of sorts?  Why the hell is the letter e in there?  Also, why is the square root of 2 involved?  That’s not involved in the rationals, nor is it involved in the integers.  Where are those two numbers coming from? Well, this is where it gets a wee-bit tricky(and interesting!).  The reals aren’t simply composed of those aforementioned number systems; they also include irrational numbers.  What’s an even bigger shock is that irrational numbers outnumber the amount of rational numbers in the set!  Isn’t that ridiculous?  The vast majority of the numbers that we see aren’t even the true majority their own set.  Irrational numbers are beasts within themselves and we can actually develop surds, which are almost entirely composed of irrational numbers.  Before we delve further, however, let’s actually define what a irrational number is: Definition: We say a number k is irrational if the decimal sequence of the number fails to terminate and the numbers consistently change throughout the decimal sequence.  In other words, it cannot be expressed as a ratio of two numbers(i.e., a fraction). This definition, though somewhat informal, is precise.  Consider the number 1/3.  This is equivalent to 0.33333… .  This number, however, is not irrational because the numbers do not change in the decimal sequence.  They don’t terminate at all and it goes on for infinity, but the numbers keep the same pattern.  Similarly, 1/9 = .11111… .  Again, this is a completely rational number, although the sequence of 1’s continue.  The pattern does not change, and the number is clearly a ratio of two numbers. What about \( \sqrt{2} \)?  Well, let’s try and see what happens when we try to analyze it. Putting it into a calculator, we see that \( \sqrt{2} \) is 1.41421356… .  The pattern within the decimal sequence changes for every other number and the sequence does not terminate.  Furthermore, we cannot express the square root of 2 as a ratio of two numbers(a fraction).  Henceforth, we shall cast this number as an irrational number. But now, we’ve come across a problem.  Sure, we can call it an irrational number, but what if some pattern exists somewhere in the sequence we can’t see?  We’re only human after all, and maybe some pattern arises and repeats itself, therefore making \( \sqrt{2} \) a rational number.  To combat against these uncertainties, we devise a mathematical argument called a proof.  A proof is what you think it is: it showcases the validity of a claim, or statement made.  The architecture of a mathematical proof rests upon the sturdy, concrete foundations of logic.  Without logic, mathematical proofs would be useless and wouldn’t be deemed fit for mathematics.  Hence, if you’ve ever been exposed to mathematicians and physicists alike, the word rigor may often come up during conversations in which they are discussing a proof.  Rigor essentially asks, how strong are the logical foundations of the proof?  Some proofs are weak, whereas others are strong.  Mathematicians are trained to make strong proofs.  Some are exceptionally strong and this adds to the ‘beauty’ in the proof: To create a proof in which there exists no refutation, even in other number systems, is astonishingly marvelous. Therefore, we need a proof which showcases beyond a shadow of a doubt that \( \sqrt{2} \) is irrational, in the off chance that our eyes fail us and we cannot foresee a pattern.  The next post shall showcase this proof.  The proof is not particularly long, but it is interesting how luxurious the proof flows.