Her second book, Kiss My Math, gets released on August 5th, 2008.

Danica was named "Person of the Week" by ABC World News with Charles Gibson on August 10, 2007.

As some of you already know, Danica took a hiatus from acting for college, and she graduated from UCLA with a Bachelor of Science in Mathematics! While she was there, she even co-authored a math proof - new research proving an original math theorem - highly unusual for an undergraduate. In fact, she was the only undergraduate invited to speak at Rutgers University's biannual Statistical Mechanics conference a few years back, and she was featured in the Science section of the NY Times on July 19th, 2005.

Danica loves math so much that she wants to share it with you, and help you get through some of your tough questions. Only a few questions each month will be forwarded to her for answering, and you can submit your questions to: MATH HELP

(Danica also answers questions for grades K-8 on the Danica's Corner page of the Mathathon website!)

If you need immediate, live help, Danica recommends Tutor.com, an on-demand tutoring service. If you own Danica's book "Math Doesn't Suck," you can get 75 minutes of FREE, live tutoring - now!

Says Danica: Let’s face it; by and large math is not easy, but that’s what makes it so rewarding when you conquer a problem, and reach new heights of understanding! I’ll be answering questions ranging from middle school math to Calculus and beyond, so skim along until you find something helpful or interesting to YOU. I challenge all of you to embrace the mind-sharpening qualities of practicing mathematics. Now let’s roll up our sleeves and do some!

Q: I come from a family of artists. Math skills are scarce. A couple of years ago I started taking math classes at Northwestern Michigan College. (At 40 years old...) It is my Mount Everest. Sunday morning 7:00. Coffee and Algebra. I ran into a procedural question. I remembered seeing your web site and knew you would have the answer.
In the "order of operations" when are square roots figured?

Respectfully,
Thom

Danica Answers:

Hi Thom,
Good question!
Believe it or not, square roots are actually exponents... in fact, they are fractional exponents. So the square root of a number, is actually that number, to the power "one half." The cube root of a number is that number to the "one third". So the order of operations being: Parentheses, Exponents, Multiplication, Division, Addition, Subtraction, then square roots fit in nicely at #2.

Hope that helps, and congrats on taking up math as a hobby!

Q: I am puzzled as to why the following problem works. It involves plugging in parts of anyone's phone number into a formula which ultimately yields the entire 7 digit number. To begin, you take the first three digits of the telephone number, multiply by 80, then add 1 to the sum, then multiply by 250. Then add the last 4 digits of the telephone number to this sum twice, then subtract 250 and then divide by 2. The result is the original phone number?? Can you explain?

Danica Answers: A popular internet "mystery," indeed- but it's really just a little math. :) So if your number is 555-1212, then the formula would yield the number: 5,551,212. And how does it get there? It first asks you to (essentially) multiply 555 times 10,000, so you get 5,550,000, and then it (essentially) asks you to add 1,212 to it, which yields 5,551,212. I'll break it down now; I just wanted to first let you know what was going on. I'm going to show you the most general case, for the phone number "abc-defg", where each letter stands for a whole digit 0-9, except the a, which cannot equal zero in a phone number, and which will yield the number a,bcd,efg = a(1,000,000) + b(100,000) + c(10,000) + d(1000) + e(100) + f(10) + g. Think about this "written out" version of the number- it's just like 5,551,212 could also be written as: 5,000,000 + 500,000 + 50,000 + 1000 + 200 + 10 + 2. See the parallel? Okay, let's start.

Using letters instead of numbers helps reveal the "tricks" that are used, so the numbers themselves don't muddy up the process.

Our phone number: abc-defg, and we want to see how we end up getting the number a,bcd,efg by the end of the process.

So: "Take the first three digits and multiply by 80 and add 1"

(80)abc + 1

"Then multiply by 250"

(250)[(80)abc +1] = (simplifying) = (20,000)abc +250

"Then add the last four digits of the phone number twice"

(20,000)abc + 250 + (2)defg

"Then subtract 250" (do you see how the "adding 1" was just a ploy to throw you off even more?)

(20,000)abc + (2)defg

"Then divide by 2"

(10,000)abc + defg = a,bcd,efg. The original "number!"

Remember, the defg is some number in the thousands, since it has four digits, and the abc number looks like: a,bc0,000. Does that make sense?

So when we add these two values together we get: a,bcd,efg, the original number again.

If they didn't want it to seem like a "miracle" but just wanted to result in the number a,bcd,efg, then they would have just said:

"Take the first three digits, and multiply by 40, and then by 250"

(or just by 10,000) which would give us a,bc0,000. Then we would have said, "now add the last four digits of the phone number ONCE."

Which would give us the very same answer. I probably "over-solved" this one for you, but since I was debunking a "mystery" I didn't want to leave anyone mystified. ;)

Hey, now try that other internet favorite- where you take your birthday and the current year or something.. it'll be a snap!

Q: I just saw you on the "Attack of the Show" on G4/Tech TV, and I saw you do what seemed to be simple mathematics, but when I try it myself, it's a little more complex that I remember. I was wondering if you could show how the repeating decimal of .9999... equals the whole number 1. None of my friends believe that it can be done. Thank you very much... you made it seem to easy. -Tom

Danica Answers: Hi Tom! Glad to help- you're not the only one who's asked me to repeat this. :)

First, we start by labeling:

x= .999999999999..... (repeating infinitely)

Then if you multiply both sides by 10, you get:

10x = 9.99999999999..... (repeating infinitely)

Well, what if we subtracted the top equation from the bottom one? We'll get another true statement. Before I do that, let me review why this will work. Say that a = b , and c = d. Then do you agree that "a-c = b-d" is also a true statement? After all, a = b , and c = d. So once you agree with that, let's continue and go ahead and subtract the top statement from the bottom one. We get, for the left hand side of the equations:

10x -x = 9x

And for the right hand side:

9.99999..... - .99999999...... = 9

And we conclude that "9x = 9" is also a "true statement." Divide both sides by 9, and you get x = 1. Since we started with x = .9999999 (repeating infinitely), we've now shown that the "infinite decimal of .9999 repeating" equals the whole number 1. Yes, I did it much faster on the TV show. :)

What I didn't talk about on the show is the somewhat philosophical issue this proof brings to light. Our mathematical system was invented by the human mind, which assumes that space (and the number line) is infinitely divisible and that there *is* such thing as a theoretical "point" that takes up no space, and has no volume or mass. So, this little proof shows that - according to a math system that assumes such things - indeed, .999999.... (repeating infinitely) equals 1. There are those who would say this isn't true, but then they are forgetting that we are dealing with a "language" that was invented by the human mind, and also that infinity is a slippery concept. :)

Q: I saw you on the "Attack of the Show" and decided to email you my problem. I've heard it many times before, and to this day, I was never able to solve it: Three men got to a motel. The receptionist tells them the room will be $30. Each man pays $10 before going to the room. The clerk then realizes that they were given a $25 room, and sends the bellhop to the room with $5. But the bellhop decides to pocket $2, and he only gives the three men $1 each. So each man paid $9 for the room. But three times $9 is $27, with the $2 the bellhop kept, that's only $29- where's the missing dollar?

Danica Answers: Ah yes, the so-called "missing dollar" problem. In the end, the men paid a total of $27, right? Well, $25 went to the room, and $2 went to the bellhop. That's it. There's no missing dollar. The $2 is PART of the $27, not something to be added to it. It actually works out just fine, it's just worded in a way that's meant to confuse you.

Q: Hi Danica, I've always had problems solving "absolute value" equations. Can you help me by doing one? How about |2x+6| = 10, and we have to "solve for x". Thanks!

Danica Answers: To find all the "x" values that satisfy your equation, we must solve two separate equations:

2x+6 = 10 and

2x+6 = -10

Do you see why? If the absolute value of a quantity must equal "something" (in this case, 10), then that quantity inside the absolute value signs can either be equal to that positive something, or to the negative of that "something," and the absolute value of it will always be that "positive something." So if the inside of the absolute value signs equals -10, the absolute value of it will still be positive 10.

So we just solve the above two equations with basic algebra and we get that x can equal 2 or -8, and both will satisfy |2x+6| = 10. As always, plug your final answers into the original equation to check your work. I like to think of absolute value as "the distance to zero" from whatever's inside. Hope that helps!

Q: Hi Danica, I heard a question from Mr. Feenie on a "Boy Meets World" episode which he claimed to be unanswerable. After hearing that, I decided to figure it out. If it takes Sam 6 minutes to wash a car by himself, and it takes Brian 8 minutes to wash a car by himself, how long will it take them to wash a car together?

Danica Answers: Hm, unanswerable? That's TV for you. :)

Let's do it: This is a "rates" problem. The key is to think about each of their "car washing rates" and not the "time" it takes them. A lot of people would want to say "it takes them 7 minutes together" but that's obviously not right, after you realize that it must take them LESS time to wash the car together than either one of them would take.

So, what is Sam's rate? How much of a car can he wash in one minute? Well, if he can wash one car in six minutes, then he can wash 1/6 of a car in one minute, right? (think about that until it makes sense, then keep reading). Similarly, Brian can wash 1/8 of a car in one minute. So just add their two rates together to find out how much of a car they can do together, in one minute, as they work side by side on the same car: 1/6 + 1/8 = 7/24 of a car in one minute. That's their combined RATE. (Note: that's a little bit less than 1/3 of a car in one minute). From this point, the way you want to think of it depends on your favorite way of dealing with fractions. You now have their rate. It's 7/24 cars per minute. You can either just take its reciprcal and say: 24/7 minutes for one car, and you're done.

Or, equivalently, you can think of the 7/24 cars/minute RATE as 24 minutes for 7 cars. (think about that until it makes sense, too) So just divide 24 by 7 to find out how many minutes it would take to do just one car. You get around 3.42 minutes for one car, just a little less than 3 and 1/2 minutes. Done! Yes, I think they should work together, it gets done much more quickly that way. :)

By the way, you said when you watched the TV show you decided that YOU would figure it out, right? How did you do?

Q: I'm an adult trying to pass the GED test and you do a lot of studying on your own, and no matter what I do, I keep getting wrong answers. I know the equations for parellelograms, the area of circles, etc. But I can't seem to get the right answers. They only grade you on the final answer you get, not on the work. I have no clue where I'm going wrong. I want to take this test so I can go to school and become a drug and alcohol counselor.

Thanks so much!

-Pat

Danica Answers: Hi Pat- First of all, congratulations on having the fortitude to go back and get your high school diploma! And I admire your goal of helping people by becoming a counselor. Good luck to you.

Now- Let me see if I can shed some light on your problem. Understanding and yet getting the wrong answers usually means one thing: You're making a lot of avoidable, careless mistakes. This is a VERY common problem these days, more than ever before. But it's totally solvable.

Believe it or not, I think this is a symptom of a larger issue that our society is facing, and it makes it that much more important that we deal with it. It seems that in our ever-increasingly fast paced information age, we are more and more often finding shorter and faster ways to do every day tasks: Abbreviating whole phrases in emails to "lol" and "imo", our computers type in our own addresses for us on websites, etc. We are accustomed to skipping steps. One of the unfortunate results of this, is that students are more and more often, skipping steps in solving math problems-- skipping the act of writing down every step along the way. And no matter how well they understand the concepts, they believe they cannot "do math" because they so often get the wrong answer.

You would be AMAZED, truly amazed, at how much more often you would get the *right* answer, if you took the time to write down every step when you are solving problems. We ALL make careless mistakes, but the chances of making careless mistakes goes up at least tenfold when we skip steps.

Other ways to help avoid making careless mistakes, in addition to writing down every step:

-Pay attention to WHAT is being asked for- you may have done a perfect job at answering a slightly different question. Did they ask for the radius or the diameter? etc.

-Restate the problem on your page; just write down all the info you're given. This is especially helpful for word problems. Trust me, it works.

-If parentheses are used in the problem, keep them during the solution until the last step. Especially if you've done any algebra, you know exactly what I'm talking about. :)

-Don't try to squeeze too much onto one page. When numbers get scrunched and small, mistakes are made.

-Once you've gotten the answer, then REREAD the problem to make sure you solved for exactly what it was asking for.

So listen, the next time you solve a problem that you know you have the right formula and understanding for, but you get the wrong answer-- do this:

Write down every step you can. Write down the formula. Then write down in information you're given, etc. So if the book says that the radius of a circle is 3, and you are supposed to get the area, then literally WRITE DOWN:

A=pi*r²

r=3

Then, beginning to solve it, don't do anything in your head without writing it down:

A=pi*3²

A=9pi

You'll get the right answer and you'll understand the value in making this a habit. What takes a few extra moments now, will save you many minutes of frustration and-- your grade. Another note on this: THIS IS HARDER TO DO THAT YOU THINK. We all think to ourselves, "Oh, I'm smarter than this. I don't need to write down every step. That's what you do in kindergarten, etc." So be patient with yourself if you can't discipline yourself to do it right away-- but rest assured that when you choose to practice this method of not skipping steps, your success will follow, I promise. :)

Q: I think you are great on "The West Wing"! Here’s my current problem, it’s in advanced finite math (I’m a high school senior): At the height of the Beatles’ popularity, it was estimated that every popular music station played their music 40% of the time. If you tuned through 10 such stations at any given moment, what is the probability that at least *one* of the stations would be playing a Beatles song? Thanks!

Danica Answers: A probability question! Okay, let’s call "x" the probability that "at least one of the 10 stations would be playing a Beatles song at that moment." That’s what we’re asked to find. Then let’s call "y" the probability that "none of those 10 stations would be playing a Beatles song at that moment." Notice that x+y = 1, since the two situations are mutually exclusive, but combined they make up ALL possible scenarios. And the probability that one of "ALL possible scenarios" occuring is of course 100%, which equals 1.

Okay, so we’ll now determine the value of "y" which is much easier than going through all the necessary calculations required to determine "x" directly. This is a common strategy in probability.

So, what is the probability that NONE of the stations are playing a Beatles song? That would be the (multiplicative) product each of the probabilities of each station NOT playing the Beatles. We know from the statement of the problem, that the probability of any given station, at any point in time, PLAYING a Beatles song is 40% = .4. This means that the probability of a station NOT playing a Beatles song is 60%, or .6. (After all, either a station is playing a Beatles song or it’s not: the two scenario’s probabilities must add up to 1 = 100%)

So, if the probability of 1 station NOT playing a Beatles tune at any particular moment is .6, then the probability of all 10 stations NOT playing a Beatles tune at that moment = [.6 raised to an exponent of 10]. Multiply .6 times itself 10 times and you will get a number like: .0060466176. This is our "y." To get "x" we must solve x = 1-y, and we get that x = .9939533824. Translated back into percentages, we get that, at any given time, there is a 99.39533824% chance that the Beatles are playing on at least one of the ten stations. Wow! Did you expect it to be that high? :)

Q: Hi Danica, I am working on equivalent fractions. I forget the formula but once I relearn it I’m usually ok. Do you know any tricks for this? Thanks! Dena.

Danica Answers: Yes, there is a trick. It’s called "cross-multiplication." Say you have the two fractions, a/b and c/d. To determine whether or not they are equivalent, you can multiply "a" times" d," and also "b" times "c", and if their products (ad and bc) are equal, then you have equivalent fractions. For example, 2/3 and 8/12. You could "cross multiply" and see that 2 times 12 = 24, and 3 times 8 = 24, so the two fractions must be equivalent.

Tricks are a good shortcut when you already understand the concepts, but tricks can get you in trouble if you’re fundamentally confused. So let’s make sure you understand the concept of equivalent fractions.

First of all, what does it mean for fractions to be equivalent? It means that they represent the same VALUE. Just like if I wrote down the expressions "5-3" and "10-8". They are *different* ways of writing the same VALUE. They both equal "2." Equivalent fractions do the same thing. They are different ways of writing the same number. 2/3 and 8/12 HAVE THE SAME VALUE. If you were to cut a pie, and you said, "give me 2/3 of that pie" or if you said "give me 8/12 of that pie", you’d end up with the *same* amount of pie. (That’s a lot of pie!)

So the "real" way of determining if two fractions are equivalent would be to determine if they represent the same value. After all, it is clear that 2/3 does NOT equal 24. We simply used a trick, and "24" has little to do with the *value* of either fraction. A good way to determine if two fractions are equivalent is to REDUCE them. That is, take out common factors in the numerator and denominator. So, with 8/12, you might notice that both 8 and 12 have a factor of "4" in them. So you can reduce the numerator and denominator by 4. Then the fraction becomes 2/3. And certainly, we can see that 2/3 = 2/3. No tricks needed. Hope this helped, and Good luck!

Q: I think you are great on "The West Wing" and I just saw you on NYPD Blue! This may be more of a physics question, but I was curious- a friend of mine was talking about an outfielder who could throw a ball from the outfield, have it go no higher than head height, and reach the catcher at home. It seemed IMPOSSIBLE but then I started thinking about the viability of this being possible. Can you help?

Here are the assumptions:

1. The outfielder (pt. A) and home plate (pt. B) are 180 feet apart (roughly twice the distance from home plate to 1st base).
2. The ball is released from a height of 6 feet.
3. The ball travels along a curved path (pulled down by the force of gravity (32 ft/sec2)).
4. It reaches home plate at ground level after not traveling at any time above 8 ft above the ground (roughly head height for the tallest human).
5. An official baseball weighs 5 ozs. (although I’m not sure if that’s relevant.) Thanks!

Danica Answers: A Hi there! Alright, let's solve this. Some physics and algebra knowledge is definitely needed to make it through this proof. I'm going to skip most of the algebra steps, assuming you can do those on your own if you like. So don't be discouraged if you don't follow it all- I answer all sorts of levels of problems on this site. :)

First we will assume that there is no wind drag-- just to simplify things. You are right that (with no wind velocity) the weight of the ball does not matter.

What we will do is find out what the velocity of the ball would have to be in order for this hypothetical situation to be possible, and then see if a human is capable of it. So, the way we do this is to first find out how long the ball would be in the air. (it will be clear "why" later) I recommend drawing a diagram just to make it clear to yourself. One thing to remember is that we can treat the up/down component of velocity separately from the side to side component of velocity.

First, looking only at the up/down motion: The ball gets thrown in the air from 6 ft, goes to 8ft, and then down to 0ft. (the ground). The equation for the height change of the ball (when it starts with zero velocity) is:

H = (1/2)gt²

You can find this equation in any elementary physics book. g= the acceleration of gravity, which is 32ft/sec². (or 9.8 meters/sec²) First let's see the time it takes for a ball to reach the ground, when dropped from a height of 8ft. Since it starts with zero velocity, we can use this formula. Solving for "t" when "H" = 8ft, using basic algebra, we get approx. .707 of a second. Since when the ball is arcing across the baseball field, its up/down velocity is zero at the point when it hits 8ft, this .707 of a second also represents the time it takes the ball to go from the highest point of its arc, to the ground. If that sounds completely foreign, there's a great lesson on this concept at:

http://www.glenbrook.k12.il.us/gbssci/phys/Class/vectors/u3l2b.html

So now we need the first part: the time it takes the ball to go from the pitcher's hand to the highest point of its arc (8ft). Since the ball reaches an up/down velocity of zero at its highest point (meaning that the vertical component of its velocity is zero at that moment), and because the only external force acting on the ball's up/down velocity is gravity, this would be the same amount of time it would take for the ball to be dropped from 8ft, and have it caught by someone at the height of 6ft.

So now, using the same above formula, we'd say that the change in height, H, equals 2ft. And solving for "t" we get approx. .354 of a second.

So now we know, that in this hypothetical situation, the ball is in the air for approx. 1.061 seconds and travels for 180 feet. So how fast would the ball have to be going? It's a simple rt = d problem. Solving for r, the rate, we get approx. 169.65ft/sec. Using the conversions 3600sec = 1 hour, and 5280ft = 1 mile, we get the rate of approx. 115.7 mph.

115.7 mph? Hm. And that's WITHOUT drag. If there were drag, the ball would be slowing down throughout its journey, so the initial throw would have to be FASTER than this. I've checked the Guiness Book of World Records and it seems that the fastest anyone's ever thrown a baseball was 100.9 mph by Lynn Nolan Ryan (California Angels) at Anaheim Stadium in California on August 20, 1974.

As was pointed out by one reader, (Thanks, Alan!) if we assume that the thrower dropped his arm down as he let go of the ball at a height of 4ft, then the thrower would only have to throw the ball with a speed of approx. 101.9 mph. I would have assumed it to be harder to throw a ball with incredible speed from far below shoulder height, but perhaps it's easier. I certainly know more about math than I do about baseball! Of course, you could also run these numbers using a very short (but fast) pitcher. Assume the guy (or gal) is only 5 ft. tall; then a 4 ft release point becomes even more feasible. Experiment, and have fun discovering what math can tell you!

Q: I have a calculus question for you: Gravel is being dumped from a conveyor belt at a rate of 30 ft-cubed/min and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 10 ft high?

Danica Answers: Okay, this is a typical "related rates" problem, and it’s a good problem to understand for ALL related rates problems in first year calculus. We need the RATE of the changing height at a certain point in time. We’re told the RATE of the changing volume (30 ft-cubed/min). So we will need to "relate" the "rates" of the height and the volume. So we need to FIRST write down an equation that determines:

1) The relationship between the VALUES of the heights and volumes, h and V.

And then we’ll take the DERIVATIVE of this equation, which will then give us:

2) The relationship between the RATES of these values, dh/dt and dV/dt.

When determining this first, important, equation between the VALUES of height and volume, always start with what you know.

Well, we know that for every cone,
V = (1/3)h(pi)r².
Additionally, we are told that for THIS cone, the diameter, which equals 2r, is always equal to the height. So we know that r = h/2. Plug this in for r, and we get:
V = (1/12)(pi)h³.

This is our important equation #1 relating VALUES. Now, to get the #2 "related rates" equation, we must take the derivative of the entire thing with respect to time, t. Don’t forget to use the chain rule!

dV/dt = (1/12)(pi)3h²dh/dt

Now remember that this equation, as it’s written, is true for ALL moments in time. And now let’s consider the moment in time that we were asked about: the moment when the height = 10ft. So, at that moment, we can plug in h = 10. We also know dV/dt; we were told in the problem that the "rate the volume is increasing" is constant. It’s 30 ft-cubed/min. So we can certainly plug that value in for this moment in time. Now the only variable left is dh/dt—the rate that the height is growing. And when we solve for it with simple algebra, we’ve solved the problem! (You should get dh/dt = 6/(5pi) ft/min.)

Q: How do you figure out percentages using word problems such as: To finish a certain job, John made 35 parts while Allen made 15. So, how much MORE of the job did John do than Allen? Express your answer in terms of percentages. Thank you!

Danica Answers: Percentages rely on parts of a total. So what’s the total number of parts made? 35+15 = 50 total parts. How many more parts did John make? 35-15 = 20 more parts than Allen made. So John did 20/50 more of the job than Allen, which equals 2/5 = .4 = 40% more. The answer is: John did 40% more than Allen. As a check, figure out what percentage of the job that each of them did. So John did 35 parts total out of 50. That’s 35/50th’s of the job, which equals .7 = 70% of the job. And Allen did 15/50th’s of the job, which equals .3 = 30% of the job. Thus, you can see that John did 70%-30% = 40% MORE than Allen did. It’s always a good idea to check your work like this. Hope that helped!

Q: Danica, This is a problem for Calculus II
f(x)= x²+2x-3
Problem:Subdivide the function's domain into subintervals on which each function has an inverse, and find the inverse function for each subinterval.

Now I found the critical #'s by first finding the derivative which is 2x+2 thus the critical # is x= -1 Then I found the intervals which would be [negative infinity, -1] & [-1, infinity]. But now I can't seem to find the inverse of f(x).

So I'm asking;
What is the inverse of f(x)?
What are the two inverse functions for each interval?

Danica Answers: You started correctly! If f(x) = y, then you must solve for x, in terms of y. That will give you the inverse function. [hint: currently the function is in ax² + bx + c form; subtract the “y” from both sides and make it look like ax² + bx + (c-y) and then solve it with the quadratic formula]
Then, you’ll get an answer— a “function” that says “x = (something with Y in it) with a “+/-” in it, and that will show you that you have two functions represented. And then it is up to you to determine which of the functions is applicable for which domain. [hint: substitute values and see what makes sense] I highly recommend sketching a crude graph of the function. Then turn your head sideways and get a feel for the inverse functions that will come from it. Just do a simple drawing based on the critical value, and what the “zeros” of the function are. Remember, the definition of a function includes the rule that for each value in its domain, there must be a unique function value, not more than one. Sketching graphs can be a GREAT aid in solving many of these problems, and it's very fast to do once you get the hang of it! This function is especially easy to get zero values from. (that means where x=0 on the graph of f(x)). A grouping solution will show you that the zeros are 1 and –3. Good luck!

Q: Danica, I teach high school math and just read about your “Figure This!” campaign to promote mathematics. How can I find out more about this?

Danica Answers: You can visit www.figurethis.org; it has all sorts of fun math puzzles, and can give you more information about this organization. Best of all, you can get ideas about how to bring math into the “real world” from this site. Have fun!

Q: Hi Danica, my advanced calculus prof asked us to prove that the square root of 5, Sqrt(5), is not a rational number. Any suggestions about where to start? Also, I really loved your Wonder Years show, and I’ve seen you on West Wing, too. Thanks a bunch.

Danica Answers: Thanks! Okay- as with most “disprove this” proofs, start by writing down the hypothesis (as if the thing you are trying to disprove were true) and then work with the equation until you get a contradiction. Here the hypothesis is that the square root of 5 is a rational number, and we’re going to show that it’s a faulty hypothesis. In “math language” this is equivalent to saying that you can write the square root of 5 as a fraction of whole numbers; that's in fact the definition of a rational number. We can assume that this fraction looks like p/q where p and q do not divide each other; that is, they share no common factors (except 1). In other words, we are assuming the fraction is written in reduced form, and we shall also assume that q is greater than 1. (And why can we do this? Here's a "mini sub-proof": This is really the same as proving that Sqrt(5) itself is not a whole number; although this seems obvious, we can easily prove it for those who like the detail: So- let's say that q=1, then Sqrt(5) = p/1, where p is a whole number. This is the same as saying that Sqrt(5) = p. But because Sqrt(5)>2 and Sqrt(5)<3 and no whole number p satisfies those conditions, we arrive at a contradiction and we may now assume that q is GREATER than 1 for the rest of the proof.)

Now it’s time to work with the original expression and hope for a contradiction to appear, that expression being: Sqrt(5) = p/q. Remember that p/q is a fraction in reduced terms. That is, q does NOT share any factors with p. Since every fraction can be written in reduced form, we are "allowed" to make this assumption.

So, let’s square both sides of the equation:
Sqrt(5) = p/q and we get: 5=p²/q².

But if q does not share any factors with p, then q² certainly cannot be a factor of p². Then p²/q² cannot be a whole number, so it can’t be equal to 5! There’s the contradiction we needed, which tells us that our original hypothesis was false. We proved it!

Q: Hi Danica, my teacher (loves puzzles) has given us a problem, which, can easily be solved using algebra. However, I have trouble grasping the question, please help… here it is: “I am three times the age that you were when I was your age. When you get to be my age, our ages will equal 63. How old will we be? Thanks, Glynis, The Netherlands.

Danica Answers: This was not easy! Think of three timelines: Before, Now, and Future. Let’s call “I” Sam, and let’s call “you” Daisy. I’ll reword the question here, with added timelines to help make the problem easier to understand: “Sam is NOW three times the age that Daisy was (BEFORE) when Sam (BEFORE) was Daisy’s (NOW) age. When (FUTURE) Daisy gets to be Sam’s (NOW) age, THE SUM OF their (FUTURE) ages will equal 63. How old will they be?

Let’s label these ages. For the BEFORE time, Daisy’s age = y, and Sam’s age = x. Let’s say there are m years between BEFORE and NOW, and there are n years between NOW and FUTURE. Since NOW Daisy’s age is Sam’s age BEFORE, we know that Daisy’s age NOW = x, and that Sam’s age NOW = 3y. (stop and go over this, make sure you’re following so far) We also know that y+m=x, and that x+m = 3y. This is how Sam’s and Daisy’s ages have progressed from BEFORE to NOW. Let’s say that there are n years between NOW and FUTURE. Then their ages will be: Daisy FUTURE = x+n, and Sam FUTURE = 3y +n. We also know that the future ages add up to 63, which means that x+n+3y+n=63. One more thing we know is that in the FUTURE, Daisy will be Sam’s age NOW, so x+n = 3y. Put all of our known equations together that will need to be reconciled:

y + m = x

x +m = 3y

x + n +3y + n = 63

x + n = 3y.

This is a set of four equations with four unknown variables, which can be solved by regular algebra steps of substitution, etc. I highly recommend making a chart with the timeline and ages data: Put across the top: BEFORE m NOW n FUTURE and along the left write Daisy, and (below that) Sam. This is how I came to better understand the age relationships. The hardest part of most word problems is simply being able to translate English into Math!
By the way, you’ll get that their future ages are 27 and 36 for Daisy and Sam, respectively.