<\/p>\n
The square root of n is the number x such that x(x)=n. Perhaps more to the point: if x is the square root of n, then<\/p>\n
x^2=n<\/p>\n
Put in everyday terms, the square root is the number you “multiply by itself” to arrive at the original one. Therefore, 5 is the square root of 25.<\/p>\n
I’ve written many posts that concern square roots: here<\/a> and here<\/a> are just two examples.<\/p>\n If you key √(2) into your calculator, you’ll find it’s 1.414213562…., which is an irrational number<\/a>. An irrational number is a decimal that neither ends, nor follows a repeating pattern. Such a number cannot be written as a fraction of integers<\/a>; in contrast, a rational number<\/a> can.<\/p>\n Let’s imagine we want to prove that √(2) is irrational. We can begin by assuming the opposite: that it is<\/em> rational. Exploring the implications, we might arrive at a contradiction. Said contradiction will prove that √(2) can’t be rational. Then, the remaining conclusion will be that it’s irrational. Such an approach – where you assume the opposite, then prove it can’t be true – is called indirect proof<\/strong>.<\/p>\n Let’s begin, therefore, by assuming √(2) is rational. Then, <\/p>\n √(2)=a\/b<\/p>\n where a and b are integers.<\/p>\n We’ll assume that a\/b is in reduced form (that is, that a and b have no factors in common).<\/p>\n We can follow along with<\/p>\n 2=a^2\/b^2<\/p>\n Which leads to<\/p>\n 2b^2=a^2<\/p>\n Since a^2 is 2 times b^2, a^2 must be even.<\/p>\n If a^2 is even, then a must be. After all, you can’t get an even number by multiplying two odds. Therefore,<\/p>\n a=2k,<\/p>\n for some integer k.<\/p>\n Furthermore, b must be odd, since a and b are assumed to have no factors in common.<\/p>\n Now, following along with a=2k,<\/p>\n 2=(2k)^2\/b^2=4k^2\/b^2<\/p>\n However,<\/p>\n 2=4k^2\/b^2<\/p>\n leads to<\/p>\n 2b^2=4k^2<\/p>\n which gives, when we divide both sides by two,<\/p>\n b^2=2k^2<\/p>\n Now we see that b^2 is even, implying that b must also be. Yet, we know a is even and that a and b share no common factors. Therefore, b can’t be even; otherwise, it will have 2 as a common factor with a.<\/p>\n The contradiction that b is proven to be even , while our earlier assumption prohibits its being so, proves that √(2) is not rational. It must, instead, be irrational.<\/p>\n As the school term develops, I’ll no doubt be drawn to more practical topics. Cheers:)<\/p>\nIndirect proof<\/h2>\n