Seventh grader Brooke from Pennsylvania writes: "Professor Puzzler, I finished your article about stressed and unstressed syllables, and it has helped me a lot! I’m in the process of writing my first sonnet and trying to juggle rhyme, iambic pentameter, and the structure of the sonnet is super difficult—especially since I still struggle with finding the stressed syllable. My main questions are: how do you know if single syllable words are stressed or unstressed? Is there anyway to check? Thank you!"

Hi Brooke, that's a great question. I'm so pleased to hear of students working to develop sonnets; the hard work you put in on such a daunting task will serve you well in your future poetry endeavors. What seems like a struggle and a juggle now will eventually become more and more natural to you. Eventually the meter will flow with much less conscious thought on your part.

But, in the meantime, I'll attempt to answer your question. There are certain words in the English language which are deemed "less important." I don't mean that we can get by without them; many sentences would be incomprehensible without them. But they are words that we often don't even consciously notice*. They are words like articles (a, an, the) prepositions (on, of, in, etc.), and conjunctions (and, but, or, that, which, etc.). Linking verbs (is, are, etc.) can be added to this list. These "lesser" words, if they have only one syllable, will typically will be unstressed in the context of a sentence. In contrast, one-syllable words which play a significant role in the sentence (such as nouns and non-linking verbs) will most likely be stressed in a poem.

One of my favorite example poems to look at is "The Night Before Christmas" -- consider the first line with unstressed syllables capitalized:

'twas the NIGHT before CHRISTmas and ALL through the HOUSE

Notice those single-syllable words that are not stressed -- there are so many prepositions, articles, and conjunctions!

Now, I said that this is typically true, but that's not a hard-and-fast rule. Context plays a very key part in identifying the stressed syllables. Consider the following two sentences:

The CAR is ON the gaRAGE.
The CAR IS on the gaRAGE.

In the second sentence, the verb "is" has the stress, but in the first sentence, the preposition "on" is stressed. Why is that? Presumably in the first sentence, someone is startled because they didn't expect the car to be ON the garage (rather than IN it). But in the second sentence, it sounds as though someone has disagreed with them, so they are emphatically declaring the truth of the statement by emphasising the verb.

There's a bit of flexibility with these "lesser" words; the meaning you intend to convey can control whether they are stressed, but also, if you have two or three of them in a row, the natural rhythm of the sentence may dictate where the stress goes. Consider my rewrite of the first line of "The Night Before Christmas": 

it WAS the NIGHT beFORE the YULEtide, AND all THROUGH the HOUSE

What have I done to this line? I've rewritten it so it has an iambic meter. Read that out loud, putting the emphasis on my upper-case syllables. Now compare my line to the actual anapestic line. I've included many of the same words, but the rhythm dictated that they be emphasized differently. "Before" gets one of its syllables stressed in my line, but not the other. In my line, "and" and "through" get the stress, while "all" does not, and this is exactly the opposite of the original line where "all" gets the stress, while "and" and "through" do not. Why do I get away with doing this? Because I'm not messing with the stress of "important" words like "night," "yuletide," and "house."

How do you tell what you can get away with? You read your line aloud, and listen to hear whether you are naturally emphasizing certain syllables or not. Then read it with the stress you'd like the syllables to have, and see if it sounds awkward.

Eventually you'll stop analyzing stresses syllable-by-syllable and go straight to listening to how it sounds.

* Book Scrounger notes that many of these "lesser" words are words which are not capitalized in headlines and titles. If you look at the title of this blog post, you'll notice that there is one uncapitalized word in the title. It's the conjunction "and." The exception to this is linking verbs, which are always capitalized in headlines and titles.

Laina asks for an explanation of the puzzle shown below.

The text reads: "THE BATTLE OF ENGLISH AND MATHEMATICS.  

1 rabbit saw 6 elephants while going to the river.
Every elephant saw 2 monkeys going towards the river.
Every monkey holds 1 parrot in their hands.

How many animals are going towards the river???

Lets see who went to shcool..."

First, I doubt I'll be able to solve this, since I never went to shcool...

However, my answer is, simply, there is no solution, because there is not enough information given.

This is like a poor man's version of the "St. Ives" riddle. In case you don't know that riddle, it goes like this: "As I was going to St. Ives, I met a man with seven wives. Every wife had seven sacks, every sack had seven cats, every cat had seven kitts. Kitts, cats, sacks, wives, how many were going to St. Ives?"

With the St. Ives riddle, the narrator meets a man with seven wives. The word "meets" strongly implies* that the man with seven wives was going in the opposite direction.

Notice, though, that in our question here, it does not say that the rabbit "met" the elephants, only that he saw them, which could certainly mean that he overtook them as they were also on their way to the river. We simply don't know what the elephants are doing, and therefore there is not enough information to definitively answer the question.

I could end there, because the answer is indeterminate. However, I think it's worth exploring the question further.

The next question that arises is, how many monkeys are there? The two most common answers to this are two and twelve.

The first answer comes from assuming that each elephant saw the same two monkeys, and the second answer comes from assuming that each elephant saw two monkeys that none of the others saw. The most compelling argument for "two" as the answer is that the problem uses the word "every" instead of "each." The word "every" is a group word, implying the group as a whole did something. Thus, the group, as a whole, saw two monkeys. It's an interesting argument, but not all that convincing; grammarly states that each and every can be used interchangeable in some circumstances, and Cambridge Dictionary gives us such an example:

Each person has a bicycle.
Every person has a bicycle.

Both of these statements mean that there is a bicycle for each person. If we aren't going to assume that every person has the SAME bicycle (meaning, of course, that there's only one bicycle), then we can't assume that every elephant saw the same two monkeys.

Nor can we assume that every elephant saw two distinct monkeys that none of the others saw. 

Therefore, there must be somewhere between 2 and 12 monkeys, inclusive, but the exact quantity of monkeys is unknown. Giving us, once again, an indeterminate answer.

There was another argument I saw for two monkeys - some people were saying, "since the problem doesn't state explicitly, implicit rules apply, making it two." Implicit rules? What is that? Is it just a fancy way of saying, "Since they didn't specify, I get to pick whichever makes the most sense to me, and call it the 'default' meaning."? 

Regarding the parrots, it's interesting to me that the people who present the "each vs. every" argument are inconsistent; If "every elephant saw two monkeys" means that the group as a whole saw two monkeys, then "every monkey holds one parrot" means that the group as a whole is holding a parrot, and there is one parrot, instead of two parrots. It seems to me that you can't have it both ways.

Finally, there really is no battle between English and Mathematics. I've been hired to write competition math problems in at least three different states, and I can tell you that English and Mathematics work side-by-side to make a good problem. Ambiguity in the use of English makes the problem unsolvable. If this problem showed up in any math competition anywhere, the appeals committee for the competition would be inundated with appeals written by students arguing for different answers. The appeals committee would have to accept multiple answers or throw the problem out entirely (I saw this happen just a couple weeks ago at a state math competition; because of one missing word in a problem, the problem was thrown out, and no one received any points for it).

I asked my problem proofreader what he would do if I sent this problem to him to proofread. He assured me that he would throw it out. He also said: 

"I think this is a wonderful problem for a class discussion, but not for any type of grade or for a competition. There is far too much ambiguity for it to be a problem with a well defined answer."

* Even the St. Ives problem is ambiguous; there isn't actually a solid reason to conclude that the man with seven wives is going in the opposite direction; the narrator could have simply caught up with the man (all things considered, he's probably going fairly slowly!) and struck up a conversation with him. That is an appropriate usage of the word "meet," and the St. Ives problem is therefore ambigious.

Martha asks, "In a straight line equation, why is the slope change in y over change in x, instead of change in x over change in y?"

That's a great question, Martha. One simple answer would be to just say, "because that's the way it is." But there are better answers than that. Let me give you one. Consider the equation for a line, when written in slope-intercept form. It is:

y = mx + b, and we say that m is Δy/Δx. If you're not familiar with the Greek letter delta used here (Δ), it simply means "the change in", so we would read that fraction as "the change in y over the change in x." So the equation now is:

y = (Δy/Δx)x + b.

Now that we've established that, let's talk about units. In real-world situations, the axes would both have units. For example, if the linear equation was part of a word problem about a car traveling from place to place, the horizontal axis might have a time unit, such as hours, and the vertical axis might have a distance unit like miles.

So now let's look at how those units would fit into our equation:

y is a vertical value, so its units would be miles.

Δy is a vertical change, so its units would also be miles.

Δx is a horizontal change, so its units would be hours.

x is a horizontal value, so its units would also be hours.

b represents the y-intercept (which is a y-value when x = 0) so it also has the vertical unit, miles.

Therefore, the equation, written only in terms of its units, would look like this:

miles = (miles/hour)hour + miles

Notice what happens when you multiply out (miles/hour)hour - you get just miles, leading to:

miles = miles + miles.

This is exactly what we need; when you add a two mileages, you get another mileage.

But consider what this would look like if m was Δx/Δy instead of the other way around. We would have:

miles = (hour/miles)hour + miles
miles = hour²/miles + miles, which is about the most absurd conglomeration of nonsensical units I've ever seen.

So that's one way of looking at the question, "Why is the slope Δy/Δx?" Because if it was the other way around, it would produce absurd results.

"If a quadratic equation has roots 4 and -12, and the constant term is -144, what is the equation?" ~Anonymous

There are two ways you can go about solving this. I'll show you the method I like best. It's pretty straightforward, and makes use of two rules:

If the roots of a quadratic ax² + bx + c = 0 are x₁ and x₂, then x₁ + x₂ = -b/a, and x₁x₂ = c/a. These two rules will help you answer your question. We know the values of x₁ and x₂ (they are 4 and -12). We also know that c (the constant term) is -144. So we can rewrite these equations as follows:

4 + (-12) = -b/a; -8 = -b/a; b = 8a
4(-12) = -144/a; -48a = -144; a = 3

Plugging this into the first equation gives b = 8(3) = 24

Therefore, the equation is 3x² + 24x - 144 = 0

Sherry from Los Angeles asks, "I read that the method for subtracting without regrouping works all the time. But what about a problem like 6563 - 1998. You can take 1 away from each but still need to regroup. I’m a teacher and got so excited when I saw this, but it doesn’t always work??"

Hi Sherry, you're probably referencing this meme here. This meme shows subtracting a quantity from a number that has all zeroes after the first digit. Does this always work? Yes it does. Because if all the digits are zeroes, subtracting one from the number turns all the zeroes into nines, which means you don't ever have to regroup/borrow to finish the problem.

But that's only if those digits are all zeroes to begin with. What are the real world situations where this is likely to happen? If you're making change at a store. If you've paid with a 50 dollar bill, and the cost of the item is $24.32, your change would be $50.00 - 24.32. You can simplify this problem by rewriting it as $49.99 - 24.31 = $25.68.

Your question is, essentially, what if the first number doesn't end with all zeroes after the first digit?

And the answer is, it still works, but is not as likely to be practical. Here's an example: 1002 - 865. What we want to do is change the first quantity so it ends with nines. To do that, we subtract three instead of one. Which means, of course, that we have to subtract three from the second quantity as well: 999 - 862 = 137.

So now we get to your example: 6563 - 1998. We ask ourselves, what would we need to subtract from the first quantity to make it end in nines? Answer: 564. But that means we have to subtract 564 from the second quantity as well. In this particular case, we can do that subtraction without regrouping: 1998 - 564 = 1434. So 6563 - 1998 = 5999 - 1434 = 4565. But if regrouping had been required at the second step, we would have had to do our process again, and we'd end up with a cascading series of subtractions which would have been far more ugly than just regrouping in the first place.

As a general rule of thumb, even though it does (theoretically) work for problems like the one you suggested, in practice, it's very cumbersome, and I wouldn't recommend doing it that way.