Intuition behind Algebra

Question

When proving, or just straight up solving equations, we often manipulate variables until we get the results we want. For example, the square root of the discriminant is simply an algebraic manipulation to get the result of the distance (or difference) between the roots. However, is there intuition behind this manipulation? I get why Algebra works in the first place but when seeing it in action, manipulations as complex as that of involved in getting the discriminant, gets me lose sense of what's actually happening within the numbers.

Answer 1

The quadratic formula comes from completing the square. There are a number of places to read the proof that leads to it, but the step in the middle that leads to the discriminant is

$$\left(x - \frac{b}{2a}\right)^2 = \frac{ b^2 - 4 a c}{4a^2}$$

If we want to be able to solve for $x$ here, we're going to need to square root both sides (and then add $b/2a$). But what happens when we square root the right hand side depends on whether it is positive, negative, or zero. Since $4a^2$ is always positive, the sign of the right hand side is really just the sign of the numerator. That is why $b^2-4ac$ is the discriminant. If it's positive or zero, we can keep going as normal, and get either one or two real number solutions for $x$. If it's negative, square rooting will give us a complex number, so the quadratic will have no real solutions. (The crucial word there is "real", as it does have solutions in the complex numbers.)

Another way to look at it is that you're solving the quadratic by finding the midpoint between its roots, then going up and down by a fixed distance. That fixed distance is related to the discriminant, so if you think geometrically, maybe that can help put it into context for you. In the case of two real roots, $-b/2a$ is the midpoint between the roots, and $\frac{\sqrt{b^2-4ac}}{2a}$ is the (positive) distance between the midpoint and each root. In the case of one real root, $\frac{\sqrt{b^2-4ac}}{2a}$ is zero. (This makes sense-- the root is exactly the midpoint $-b/2a$, and you're moving up and down by $0$.) And in the case that there are no real roots, $\frac{\sqrt{b^2-4ac}}{2a}$ turns out to be a complex number, so this construction doesn't work on the real line.

Answer 2

In the particular example you mentioned, you can understand what's going on by looking at the graph of the parabola with equation $y = ax^2 + bx + c$. It's not hard to see that the vertex of the parabola is at the point where $$ x = - \frac{b}{2a} \quad ; \quad y = -\frac{b^2-4ac}{4a} $$ There's a better explanation here. So, you can understand the algebra by looking at the geometry. And (for me, anyway), developing intuition about geometry is not too hard because I can draw pictures.

On the other hand, there are algebraic processes for which it's difficult to draw pictures. Or, at least, I don't know how to draw them (I'd love to learn). In those cases, the algebra becomes (for me) a mysterious process of shuffling symbols around, which is devoid of any intuition.

Different people have different types of "intuition", and develop it in different ways. My way (drawing pictures) might not be the best way for you.

Answer 3

I don’t think the quadratic formula is the best place to “see” the intuition at work, because it is so many steps rolled into one. My intuition for quadratics comes from graph sense, when I can imagine “vertex form”. For example, consider the two equations:

$$y_1=x^2-6x+5=0$$

and

$$y_2=x^2-6x+13=0$$

I know that $y=x^2-6x+9$ would be a perfect square, with its vertex at $(3,0)$. Looking at our functions, $y_1$ is $4$ units too low, and $y_2$ is $4$ units too high. On a standard parabola $(a=1)$, four vertical units means $\sqrt4=2$ horizontal units away from that point $(3,0)$ is where we should find our intercepts.

By dropping $4$ units, our $x$-intercepts for $y_1$ spread out along the $x$-axis, so we’ll have solutions at $1$ and $5$, two units to either side of $3$. For $y_2$, on the other hand, the solutions spread the same distance off of the real line, so $3+2i$ and $3-2i$ should be our solutions.

What if that $+13$ had been instead $+12$? Then we’ve only lifted $3$ units off of the axis, so solutions will be $3\pm\sqrt3 i$.

What if $a\ne 1$? That complicates the picture, in ways that can be handled, but I won’t develop good intuition about that until I have quite good intuition about the $a=1$ case. Intuition builds like sedimentary rock, not like a skyscraper.

Answer 4

An equation like $x+4 = 3x$ returns True for some values of $x$ and False for others.

For example, this particular equation returns false for $x := 1$, because

$$(x+4 = 3x)(x:=1) \iff 1+4= 3\cdot 1 \iff 5 =3 \iff \mathrm{False}.$$

Whereas it returns True for $x:=2$, because

$$(x+4 = 3x)(x:=2) \iff 2+4= 3\cdot 2 \iff 6 = 6 \iff \mathrm{True}.$$

In summary, each equation is associated with a function that returns True for some values of $x$ and False for others.

When you solve equations, you perform manipulations that preserve this function.

For example:

• The equation $x+3 = 5$ is associated to a function that returns True for some values of $x$ and false for others.

• The equation $x = 5-3$ is also associated to such a function.

• Furthermore, these functions are equal.

Therefore:

• if $x+3=5$ returns True for some choice of $x$, then so too does $x=5-3$, and vice versa.

• if $x+3=5$ returns False for some choice of $x$, then so too does $x=5-3$, and vice versa.

This may not sound too deep, but it allows us to find the exact values for which an equation is true. If I write $3x+6 = 0$, it's a bit mysterious which values of $x$ make this function return True. But I write $x=5$, it's really obvious. For instance:

$$(x = 5)(x:=5) \iff 5 = 5 \iff \mathrm{True}$$

$$(x = 5)(x:=4) \iff 4 = 5 \iff \mathrm{False}$$

Surprise surprise, the equation $x=5$ seems to return True at $x:=a$ if and only if $a$ equals $5$.

What this means is that if I can repeatedly transform $3x+6= 0$ into a form like $x=5$, then I can work out which choices of $x$ make it true really easily. This is what solving equations is all about.

For example, if I write $$3x+6 = 0 \iff 3x = -6 \iff x = -6/3 \iff x = -2,$$

this instantaneously tells me that $3x+6=0$ returns True for $x:=-2$ and False otherwise.

That's really all algebra is; it's about manipulating conditions so that the condition hasn't changed, just the way we've written it has changed. The way we've written the final expression may tell us something useful about the original expression which wasn't obvious until we performed the manipulations.

Answer 5

I do not think it is always reasonable to expect intuition to play a role in every single step of an algebraic manipulation. Sometimes, it just works that way and there is no 'deeper' reason. For an analogy, if I asked you why $23$ is a prime and not $21$ or $25$, the only meaningful answer is that $23$ simply does not have any non-trivial factors. Algebraically, you can check the remainder when $23$ is divided by each smaller positive integer greater than $1$, and that is all there is to it.

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For another example, there is a curious algebraic solution to the general cubic equation, but as far as I can tell no 'deeper' reason why it should work:

To solve the general cubic equation it suffices to solve it for a monic cubic polynomial, namely to solve $x^3+ax^2+bx+c=0$. This first step is motivated by the idea of minimizing the degrees of freedom. That idea also motivates the next step of letting $y$ be such that $x = y-\frac{a}{3}$, since it will result in a monic cubic with no quadratic term (depressed cubic). So now it suffices to solve for $y$ in terms of $p,q$ given that $y^3+py+q=0$. The next step is one that I do not understand, in the sense that there is no obvious reason why it should work, but it just does. Let $z$ be such that $y = z-\frac{p}{3z}$, and such a $z$ can be found by solving the equivalent quadratic equation. One could say that this is motivated by trying to eliminate the linear term, but we know it does not really make sense since it will introduce negative-powers of $z$ as well, and there is no good reason for the extra cancellation in red in $\left( z^3 - 3z^2\frac{p}{3z} + \color{red}{3z(\frac{p}{3z})^2} - (\frac{p}{3z})^3 \right) + p \left( z - \color{red}{\frac{p}{3z}} \right) + q = 0$, or the incidental outcome of a quadratic equation in $z^3$. It seems to be just another instance of the strong law of small numbers.

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Another algebraic curiosity is the proof of the original group axioms from the apparently weaker versions that only guarantee that $1$ is a left-identity and that every element has a left-inverse, besides the usual associativity. It goes like this:

Take any $x$. Let $x',x''$ be such that $x'x=1$ and $x''x'=1$.

Then $xx' = 1xx' = (x''x')(xx') = x''(x'x)x' = x''1x' = x''x' = 1$. Thus $x'$ is a right-inverse for $x$.

Thus $x1 = x(x'x) = (xx')x = 1x = x$ and so $1$ is a right-identity for $x$.

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Even if one thinks there is a 'deeper' reason for such algebraic manipulations, I think it is pointless to expect any such reason in general. After all, by the same kind of information-theoretic reasoning as in the strong law of small numbers, an explanation has zero explanatory power if the complexity of its description exceeds the complexity of the original phenomenon it purports to explain! Since the above examples are short enough (and your example of the quadratic equation has an even shorter explanation), we do not expect anything 'deep'.

To understand why these examples differ from others where there is a deeper reason, one can look at say the intuition behind the first isomorphism theorem, which in suitable generality can both motivate and explain the various notions of normal subgroups for quotient groups and ideals for quotient rings and why every submodule yields a quotient module. There you see that one single explanation applies to every structure, so it can reasonably be considered a 'deeper' reason.

Answer 6

Sometimes you can get the same result without intuition but much more simply and directly with intuition. I can think of one situation where a complicated proof has been known for over a century but given the proper intuition it would be replaced in all subsequent textbooks by an absurdly simple proof. Unfortunately I sent this in to my coauthor $2$ years ago for editing and he still hasn't done anything with it so I assume the proof will not be published in my lifetime.

Answer 7

"Every discovery contains an irrational element or a creative intuition." - Karl Popper

If we look at modern derivations of the quadratic formula, we might see algebraic manipulations that prove it, but we might not feel the intuitive essence that brings it to light. Now, the chain "motivation-intuition-discovery" is a personal experience. But just for fun, using only an algebraic approach, let us set forth to 'discover' how to solve $ax^2+bx+c = 0 \text{ , } a \ne 0$ in two different ways.

Method 1:

It makes us happy that we can solve equations like $x^2 = 2$ and when we generalize, we see two solutions for

$\tag 1 ax^2 = -c \text{ with } a \text{ and } c \text{ of opposite signs}$

namely, $x = \pm \sqrt{\frac{c}{a}}$.

In a flash of insight (change of variable), we realize that we can 'manufacture' more interesting quadratic equations with solutions. For example,

$\tag 2 2 (x+3)^2 = 4$

has two solutions, $x = -3 \pm \sqrt2$, and (2) can be also be expressed as $2x^2 +12x +14 =0$.

With great hope, we now tell ourselves that we can somehow 'force' any quadratic equation into a '(1)/(2) pattern' and get the solution(s):

$\quad ax^2+bx+c = 0 \; \text{ iff }$

$\quad x^2+\frac{b}{a}x = -\frac{c}{a} \; \text{ iff }$

$\quad (x+\frac{b}{2a})^2 - (\frac{b}{2a})^2 = -\frac{c}{a} \; \text{ iff }$

$\quad (x+\frac{b}{2a})^2 = \frac{b^2}{4a^2} -\frac{c}{a} \; \text{ iff }$

$\quad x = -\frac{b}{2a} \pm \sqrt{ \frac{b^2}{4a^2} -\frac{c}{a}} \; \text{ iff }$

$\quad x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$

This method is known as 'completing the square'.

Method 2

There is nothing in the world that you love more than factoring quadratic equations, but sometimes the solution eludes you. Then one day you say

$\quad x^2+\frac{b}{a} x+\frac{c}{a} = (x-s)(x-t) \; \text{ iff }$

$\quad -(s + t) = \frac{b}{a} \text{ AND } st = \frac{c}{a}$

Ah, two equation in two unknowns - go for it! You see a 'trick' to get things going

$\quad (s - t)^2 = (s + t)^2 - 4st = \frac{b^2}{a^2} - \frac{4c}{a}$

So you are now one step away from finding that the distance between the two roots is equal to

$\tag 3 \frac{\sqrt{b^2 -4ac}}{|a|}$

You are now in 'mop up mode', and work out the details for your quadratic formula.

This method is known as the 'So you wanna see a factorization do you?'.