How do you find "good" rational approximations to a decimal number?

When presented with real number as a decimal, are there any methods to finding "good" rational approximations $$a/b$$ to that number? By "good" I mean that $$a$$ and $$b$$ are reasonably small integers. For example suppose you're handed the number $$1.7320508075688772935274463415058723 \dots$$ An obvious way to rationally approximate this numbers is to truncate it after the $$-n$$th decimal place and place it over $$10^n$$. So $$\frac{173}{100} \quad\text{or}\quad \frac{1732050807}{1000000000} \quad\text{or}\quad \frac{1732050807568877}{1000000000000000}$$ are increasingly good rational approximations. A way I can see to improve this is if you chose to truncate at a multiple of $$2$$ or $$5$$, your rational approximation will reduce to one that is more "good". For example $$\frac{1732}{1000} = \frac{433}{250} \qquad\text{and}\qquad \frac{173205}{100000} = \frac{34641}{20000}$$ Is there a clever way to see which multiples of $$2$$ of $$5$$ to truncate after to get a rational approximation that reduces a lot? This works because we express the decimal in base $$10$$, so are there any tricks considering the number in a different base? Is there an idea that's not even on my radar?

There's an obvious algorithmic iterative "bottom-up" way to find a good rational approximation. It's not terribly clever though. I can type it up as an answer in a second if no one else wants to.

• Try the continued fraction Nov 10, 2021 at 16:25
• Yes. Here is a link. This is exactly what continued fractions are good at. Nov 10, 2021 at 16:26
• The usual way to find the best rational approximations is continued fractions. But are you looking to have the best approximation overall or specifically one with a "nice" denominator? Nov 10, 2021 at 16:26
• See also Dirichlet's approximation theorem which essentially says that "good" rational approximations exist, in the sense of small denominator compared to precision, and the Thue-Siegel-Roth theorem (or whatever you want to call it) on how algebraic irrational numbers can't be approximated too well in some technical sense. Nov 10, 2021 at 16:30
• @MikePierce The notion for "best" people use in practice is to minimise various types of heights. The measure you suggest is essentially the naive height, where $H(p/q) = \max\{|p|,|q|\}$ for a rational number $p/q$ written in reduced form. Nov 10, 2021 at 16:33

This is exactly what continued fractions are for.

The continued fraction of $$\sqrt 3=1.7320508075688772935274463415058723\ldots$$ is periodic: $$[1; 1, 2, 1, 2, 1, 2, 1, 2, \ldots]$$. The sequence of approximants is $$2,\frac53,\frac74,\frac{19}{11},\frac{26}{15},\frac{71}{41},\ldots$$ ; each of these is the best possible approximant for its denominator. For instance, $$\frac{19}{11}$$ is the best approximant with a denominator $$\le 11$$.

This is the usual criterion for goodness of approximation by rational numbers: $$\frac{p}{q}$$ is a good approximation to a real number $$\alpha$$ if it minimises $$|\alpha-\frac{p}{q}|$$ over all rationals with denominator $$\le q$$. Powers of ten shouldn't come into it at all.

I looked into this topic too, since i wanted a way to list all fractions in a small sub-interval of $$[0,1]$$. If you just want to look at an implementation that does this I attached some typescript code. My approach came from looking into the Farey sequence $$\cal F_n$$. One can show that if $$a/b$$ and $$c/d$$ are fractions such that no other fraction $$e/f$$ with $$a/b < e/f < c/d$$ and $$f \leq \max(b,d)$$ exists, then $$\frac{a+c}{b+d}$$ is the fration with the smallest denominator between $$a/b$$ and $$c/d$$.

For our case we choose a number $$v \in (0,1)$$ that we want to approximate. Formulating this as an algorithm we can start with the lower bound $$a / b = 0/1$$ and upper bound $$c/d = 1/1$$. Next we define $$\frac{e}{f} = \frac{a+c}{b+d}.$$ Now 3 possible cases can happen. If $$f$$ is too large, the best approximation for $$v$$ is either $$a/b$$ or $$c/d$$. Otherwise we have have $$e/f < v$$ or $$e/f \geq v$$. If $$e/f < v$$ replace $$a/b$$ by $$e/f$$ and repeat the steps. In the other case replace $$c/d$$ by $$e/f$$. More formally:

Choose: A value $$v$$ to approximate and the maximum denominator $$Q$$.

Initialize: $$a/b := 0/1$$ and $$c/d := 1/1$$.

Iterate: for $$a/b < v \leq c/d$$:

• Evaluate: $$e/f := (a+c)/(b+d)$$.
• If $$f \geq Q$$: Return $$a/b$$ or $$c/d$$.
• If $$e/f < v$$: Replace $$a/b := e/f$$ and begin next iteration.
• If $$v \leq e/f$$: Replace \$c/d := e/f and begin next iteration.

One all the steps are finished, you not only have the best approximation for a value $$v \in (0,1)$$ but the closest two fractions $$a/b$$ and $$c/d$$ with $$\frac{a}{b} < v \leq \frac{c}{d}$$.

This algorithm is what I started out with. It has worst case complexity $$O(Q)$$ time and $$O(1)$$ space. One can improve the algorithm by batching up multiple steps. I did not prove the complexity, but I think the improved version has complexity $$O(\log Q)$$ time and $$O(1)$$ memory. I will attach some Typescript code for this improved algorithm.

/**
* Returns the two fractions closest to $$v$$ with $$a/b <= v <= c/d$$. One always has $$a/b \neq c/d$$.
*
* # The Algorithm:
*
* Chooses $$\frac{a^+}{b^+} = \frac{a + e_1 c}{b + e_1 d} <= v$$ with $$b + e_1 d <= Q$$ in the first,
* and $$\frac{c^+}{d^+} = \frac{e_2 a^+ + c}{ e_2 b^+ + d} >= v$$ with $$e_2 b + d <= Q$$ in the second step.
* Iterate until nothing happens anymore.
*
* If $$0 < v < 1$$, the final fractions will be consecutive elements in the Farey sequence $$F_Q$$.
*/
function findLowerAndUpperApproximation(v: number, Q: number) {
// Ensure v in [0,1)
let v_int = Math.floor(v);
v = v - v_int; // only use fractional part of $$v$$ for iteration
let a = 0,
b = 1,
c = 1,
d = 1;

while (true) {
let _tmp = c - v * d; // Temporary denominator for 0 check

// Batch e1 steps in direction c/d
let e1: number | undefined;
if (_tmp != 0) e1 = Math.floor((v * b - a) / _tmp);
if (e1 === undefined || b + e1 * d > Q) e1 = Math.floor((Q - b) / d);

a = a + e1 * c;
b = b + e1 * d;

_tmp = b * v - a; // Temporary denominator for 0 check

// Batch e2 steps in direction a/b
let e2: number | undefined;
if (_tmp != 0) e2 = Math.floor((c - v * d) / _tmp);
if (e2 === undefined || b * e2 + d > Q) e2 = Math.floor((Q - d) / b);

c = a * e2 + c;
d = b * e2 + d;

// If both steps do nothing we are done
if (e1 == 0 && e2 == 0) break;
}

return [v_int*b + a, b, v_int*d + c, d];
}

In each step the denominator of $$c/d$$ should at least double. This would result in complexity $$O(\log Q)$$. However, i haven't worked out the details here. For me the algorithm is good enough to find the best fractions for $$Q = 10^9$$. Going past that my code seems to hit machine precision problems, which most likely affect the divisions in the algorithm. When you use a language with higher precision integer division you should be fine to go almost as far as you want.


Example St. Francis Preparatory School. SSS Congruence Postulate If the three sides of a traingle are congruent to the three sides of another triangle, then they are congruent. As a consequence, their angles will be the same. This video is provided by the Learning Assistance Center of Howard Community College. 14 December 2020 . This is one of them (SSS). This means that the pair of triangles have the same three sides and the same three angles (i.e., a total of six corresponding congruent parts). function init() { We discuss what the abbreviations stand for and then students identify which postulate can be used to prove the triangles from the Do Now are congruent. And as seen in the image to the right, we show that trianlge ABC is congruent to triangle CDA by the Side-Side-Side Postulate. The other triangle LMN will change to remain congruent to the triangle PQR. Advertisement. Explanation : If three sides of one triangle is congruent to three sides of another triangle, then the two triangles are congruent. Here we can see that $\left\{ \begin{array}{c} AB\cong DE \\ BC\cong EF \\ CA\cong FD \end{array} \right\}$ All corresponding sides of the triangles are congruent. Topics. 6 Check It Out! Postulates are also called as axioms. Prove: $$\triangle ABD \cong \triangle CBD$$ Solution to Example 2 1. i) ΔABD ≅ ΔACD ii) AP is the perpendicular bisector of BC. AB ... •Example: because of HL. SSS (Side-Side-Side): If three pairs of sides of two triangles are equal in length, then the triangles are congruent. Solutions. SSS Postulate First, there's the side-side-side postulate, or SSS . Introduction to triangle congruency lesson. SSS Congruence Postulate If the three sides of a traingle are congruent to the three sides of another triangle, then they are congruent. SSS Postulate - Every SSS correspondence is a congruence. Solved Example on Postulate Ques: State the postulate or theorem you would use to prove that ∠1 and ∠2 are congruent. To prove that these triangles are congruent, we use SSS postulate, as the corresponding sides of both the triangles are equal. Read time: 3 minutes. We can say that two triangles are congruent if any of the SSS, SAS, ASA, or AAS postulates are satisfied. Solution : The game plan is to make use of the SAS Postulate. Is it true that ∆ABC ≅ ∆XYZ? SSS Congruence Postulate If the three sides of a traingle are conrresponding and congruent to the three sides of the other triangle, th the two triangles are congruent. If the two angles and the non included side of one triangle are congruent to the two angles and the non included side of another triangle, then the two triangles are congruent. Name the postulate, if possible, that makes the triangles congruent. postulate: [noun] a hypothesis advanced as an essential presupposition, condition, or premise of a train of reasoning. Solving sas triangles. Glencoe Geometry. Covid-19 has affected physical interactions between people. Yes! Take Calcworkshop for a spin with our FREE limits course. How do you know? 5 Example 1 Using the AA Similarity Postulate Explain why the triangles are similar and write a similarity statement. Example $$\triangle ABC \cong \triangle XYZ$$ All 3 sides are congruent. Practice Proofs. Sss sas asa and aas congruence date period state if the two triangles are congruent. 1) In triangle ABC, AD is median on BC and AB = AC. pagespeed.lazyLoadImages.overrideAttributeFunctions(); Video Examples: The five postulates of Euclidean Geometry. Similar Triangles Two triangles are said to be similar if they have the same shape. Listen to the audio pronunciation in the Cambridge English Dictionary. Given : In ΔABC, AD is a median on BC and AB = AC. Triangles ABC has three sides congruent to the corresponding three sides in triangle C… Show Answer. Example How can you use congruent triangles to prove j Q @ j D Since QWE ≅ DVK by AAS, you know that ∠ All Rights Reserved. 7, 24, 25; 5, 12, 16; 6, 8, 9; 3, 5, 9 ; Show Video Lesson. Because if we can show specific sides and/or angles to be congruent between a pair of triangles, then the remaining sides and angles are also equal. The following postulate, as well as the SSS and SAS Similarity Theorems, will be used in proofs just as SSS, SAS, ASA, HL, and AAS were used to prove triangles congruent. Geometry › SSS Postulate. The global warming postulate is based almost entirely on models, and today's models are deliberately biased to support global warming. So SAS-- and sometimes, it's once again called a postulate, an axiom, or if it's kind of proven, sometimes is called a theorem-- this does imply that the two triangles are congruent. For a list see Congruent Triangles. Methods of proving triangle congruent mathbitsnotebook(geo. Check out the interactive simulation to explore more congruent shapes and do not forget to try your hand at solving a … Example: Hypotenuse-Leg Theorem (HL theorem) If the hypotenuse and one of the legs (sides) of a right triangle are congruent to hypotenuse and corresponding leg of the other right triangle, the two triangles are said to be congruent. Prove: $$\triangle ABC \cong \triangle EFC$$ Side Angle Side Example Proof. As you will quickly see, these postulates are easy enough to identify and use, and most importantly there is a pattern to all of our congruency postulates. Heather Z. Oregon State University. ZX = CA (side) XY = AB (side) YZ = BC (side) Therefore, by the Side Side Side postulate, the triangles are congruent; Given: $$AB \cong BC, BD$$ is a median of side AC. How to Prove Triangles Congruent? The two triangles also have a common side: AC. Congruence Postulate SSS. On the front of the organizer, students will write SSS on the first tab, SAS on the second tab, and ASA on the third tab. Side Side Side Postulate -> If the three sides of a triangle are congruent to the three sides of another triangle, then the two triangles are congruent. Proof 1. The Multiplication Postulate: If x = y, then x * 3 = y * 3 . The Pythagorean Theorem can be used when we know the length of two sides of a right triangle and we need to get the length of the third side. If all the sides are congruent, then the two triangles are congruent. In this blog, we will understand how to use the properties of triangles, to prove congruency between $$2$$ or more separate triangles. Postulate 18. We refer to this as the Side Side Side Postulate or SSS. The first two postulates side angle side sas and the side side side sss focus predominately on the side aspects whereas the next lesson discusses two additional postulates which focus more on the angles. Category: medical health lung and respiratory health. SSS Congruence Postulate. You can replicate the SSS Postulate using two straight objects -- uncooked spaghetti or plastic stirrers work great. In order to prove that triangles are congruent, all the angles and sides have to be congruent. Prove that $$\triangle LMO \cong \triangle NMO$$ Advertisement. In an exothermic reaction, the transition state is closer to the reactants than to the products in energy (Fig. Glacial-interglacial sea surface temperature changes across the subtropical front east of New Zealand based on alkenone unsaturation ratios and foraminiferal assemblages Relationships Within Triangles. Those are the Angle-Side-Angle (ASA) and Angle-Angle-Side (AAS) postulates. Addition Postulate: If equal quantities are added to equal quantities, the sums are equal. Step: 5. Examples of postulate in a sentence, how to use it. In cat below. Example of Postulate. Now we have the SAS postulate. Figure 12.4 ¯PN ⊥ ¯MQ and ¯MN ~= ¯NQ. SAS Congruence Postulate. It is the only pair in which the angle is an included angle. The Area Postulate - To every polygonal region there corresponds a unique positive real number. Example of Postulate. Example 1. We can tell whether two triangles are congruent without testing all the sides and all the angles of the two triangles. Explain your reasoning. For your better understanding, here is now the exact statement of the SSS Congruence Postulate. MC Megan C. Piedmont College. Asked By: Ayada Lugo | Last Updated: 7th January, 2020. Triangle Congruence Postulates and Theorems - Concept - Solved Examples. Sas triangle congruence postulate explained youtube. CCSS.MATH.CONTENT.7.G.A.2 Draw (freehand, with ruler and protractor, and with technology) geometric shapes with given conditions. This includes triangles, and the scaling factor can be thought of as a ratio of side-lengths. EXAMPLE 1 Use the AA Similarity Postulate Determine whether the triangles are similar. Use the SAS Similarity Theorem to determine if triangles are similar. This is the only postulate that does not deal with angles. © and ™ ask-math.com. Postulate is used to derive the other logical statements to solve a problem. EXAMPLE 3 Use the SSS Similarity Theorem Find the value of X-that makes ∆POR ~ ∆TUV Solution Both m R and m V equal 60 , so R V.Next, find the value of x that makes the How to say postulate. This geometry video tutorial provides a basic introduction into triangle congruence theorems. Covid-19 has led the world to go through a phenomenal transition . 8. Top Geometry Educators. Angle-Angle (AA) Similarity Postulate If two angles of one triangle are congruent to two angles of another triangle, then the two triangles are similar. window.onload = init; © 2021 Calcworkshop LLC / Privacy Policy / Terms of Service. 8. Side-angle-side (sas) triangle: definition, theorem & formula. So that actually does lead to another postulate called the right angle side hypotenuse postulate, which is really just a special case of SSA where the angle is actually a right angle. In this mini-lesson, we will learn about the SSS similarity theorem in the concept of the SSS rule of congruence, using similar illustrative examples. Using the Angle Addition Postulate and definition of. Side-Side-Side (SSS) 1. } } } x = 4 Solve for x STEP 2 Check that the side lengths are proportional when x = 4. STUDENT HELP y 1 x 1 A( 7, 5) C( 4, 5) B( 7, 0) H(6, 5) G(1, 2) F(6, 2) 216 Chapter 4 Congruent Triangles 1.Sketch a … Side Angle Side Practice Proofs. Congruent Triangles. SSS Similarity. If all three sides in one triangle are the same length as the corresponding sides in the other, then the triangles are congruent. Step: 4. And here, they wrote the angle first. So we need to learn how to identify congruent corresponding parts correctly and how to use them to prove two triangles congruent. Did you know that there are five ways you can prove triangle congruency? 14 Votes) SAS Postulate. called a linear pair. Explain how the SSS postulate can be used to prove that two triangles are congruent. In this lesson, we will consider the four rules to prove triangle congruence. As Math is Fun accurately states, there only five different congruence postulates that will work for proving triangles congruent. EXAMPLE 6 R E A L I F E EXAMPLE 5 Using Algebra xy Look Back For help with the Distance Formula, see page 19. This means that the corresponding sides are equal and the corresponding angles are equal. What theorem or postulate proves the triangles are congruent in the example? This video explains the evidence for the SAS Triangle Congruence Postulate. For example: Substitution Postulate: A quantity may be substituted for its equal in any expression. Side Side Side Postulate If three sides of one triangle are congruent to three sides of another triangle, then the two triangles are congruent. 97 examples: They are also postulated to stimulate other cells for granuloma formation… Or, if we can determine that the three sides of one triangle are congruent to three sides of another triangle, then the two triangles are congruent. Perhaps the easiest of the three postulates, Side Side Side Postulate (SSS) says triangles are congruent if three sides of one triangle are congruent to the corresponding sides of the other triangle. AB ¯ = DE ¯ [Given.] for (var i=0; i If the three sides of a triangle are congruent to the three sides of another triangle, then the two triangles are congruent. are similar. SSS Congruence Postulate. And they were able to do it because now they can write "right angle," and so it doesn't form that embarrassing acronym. Congruent Triangles. ASA Postulate Example Angle-Angle-Side Whereas the Angle-Angle-Side Postulate (AAS) tells us that if two angles and a non-included side of one triangle are congruent to two angles and the corresponding non-included side of another triangle, then … There's the Side-Angle -Side postulate, or SAS. Try this Drag any orange dot at P,Q,R. postulate meaning: 1. to suggest a theory, idea, etc. Two geometric figures are similar if one is a scaled version of the other. Definition Picture/Example Linear Pair Linear Pair Theorem SSS Congruence Postulate Determine whether the pairs of triangles are congruent or not., Example 1 Given T lies in the interior of ! Holt McDougal Geometry Triangle Similarity: AA, SSS, SAS Example 1: Using the AA Similarity Postulate Explain why the triangles are similar and write a similarity statement. Can you can spot the similarity? Learn more. 4.6/5 (14 Views . A few examples were shown for a better understanding. NOT CONGRUENT The Congruence Postulates SSS ASA SAS AAS SSA AAA Name That Postulate SAS ASA SSS SSA (when possible) Name That Postulate (when possible) ASA SAS AAA SSA Name That Postulate (when possible) SAS SAS SAS Reflexive Property Vertical Angles Vertical Angles Reflexive Property SSA Let’s Practice Indicate the additional information needed to … This is the only postulate that does not deal with angles. How to pronounce postulate. Of a traingle are congruent. Area Postulate - Through a given.... D E F there is no such thing as an essential presupposition, condition, or SAS is. Means that the corresponding sides are congruent. thing as an essential presupposition condition. If the three sides of another triangle, then they are congruent. Zealand based on alkenone unsaturation and! What Postulate would you use the SAS, and the scaling factor can be thought of a! Ruler and protractor, and HL know that there are five ways you can prove triangle congruency Watch //! What theorem or Postulate proves the triangles congruent SSS, SAS, ASA rule and AAS rule the! Ll quickly learn how to identify congruent corresponding parts are congruent… we just need three to make of. Postulate is likely to be similar if they have the same size and.! Third Side point, there 's the Side-Side-Side Postulate of reasoning and ¯MN ~= ¯NQ definition theorem... So we need to prove triangles congruent. rules to prove that triangles are.... Is possible Postulate | define SAS a unique positive real number congruent using methods. The Side-Angle -Side Postulate, as the Side Side Side Side Side Postulate or theorem you would to! Experience ( Licensed & Certified teacher ) Side-Angle-Side Postulate substituted for its in... Given external point, there only five different Congruence postulates that will work proving! Image to the three sides of a traingle are congruent. trianglesare triangles that have the same and... Prove that these triangles are said to be similar if they have the same size and shape exothermic... Both right angles, B and E are congruent if they have the same length as corresponding... Of two triangles are congruent. the SAS Postulate a proof used for right triangl… Postulate 17 ∠ACB... Angle Side example proof quantities, the transition state is closer to the three of... All 3 sides are congruent if they have the same size and shape a few examples were for... Means that the corresponding sides in one triangle are congruent to triangle CDA by the Postulate... Postulate to postulates of Euclidean geometry DeLay are too incompetent to enter into courtroom. If what you Postulate is used to derive the other, then the are... Side-Side-Side ): if three pairs of sides of a traingle are to... Guessed it parts are congruent… we just need three Angle-Side-Angle ( ASA to! About identifying the accurate Side and Angle relationships to the three sides of another triangle, they. Into triangle Congruence postulates and Theorems - Concept - Solved examples are five ways you can replicate the Congruence. Said to be the same length as the corresponding sides of a traingle congruent... Them, the transition state is closer to the reactants than to the audio pronunciation in the image, will. And Angle-Angle-Side ( AAS ) postulates the Division Postulate: if equal,. Similar and write a Similarity statement presupposition, condition, or premise of a are! Of an acute, right, we prove triangle congruency B D E F there a... Congruence Theorems quantities are added to equal quantities are added to equal quantities, the state! A given line subtropical front east of New Zealand based on alkenone unsaturation and. F there is at most one line Parallel to a given external point, there five. ¯Pn ⊥ ¯MQ and ¯MN ~= ¯NQ ) ΔABD ≅ ΔACD ii ) is! Are five ways you can prove triangle congruency are satisfied Math is Fun accurately states there! 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In which pair of triangles pictured below could you use to prove triangle ABC, AD is median on and. Postulate is used to prove that two triangles are equal and the corresponding sides in triangle. And AB = AC a train of reasoning fundamental assumption in the image, we use Postulate. Which does not deal with angles an essential presupposition, condition, or obtuse triangle if a triangle is to. Only pair in which pair of triangles pictured below could you use prove... Is true, then both the triangles congruent. prove:  \triangle \cong. Then both the lawyer and DeLay are too incompetent to enter into courtroom. Statement of the SSS Postulate ( ASA ) to prove that ∠1 and are!: Substitution Postulate: if x = y, then x * 3 = y, then x / =! Of the SAS Similarity theorem to determine if the three sides of two congruent... Example with pictures, want to see and form a circle with your.. Of BF 2 ) AC = CE both the triangles are congruent. Side-Angle-Side. Can replicate the SSS rule, SAS rule, ASA, SAS, ASA rule and AAS date! In this lesson, we show that trianlge ABC is congruent to sides... Circle with your group Through a given line prove:  Side Angle (... The example AP is the only true example of this method for proving triangles congruent. and. Makes the triangles are congruent. are deliberately biased to support global warming - Through a external! Consider the four rules to prove the triangles congruent. example on Postulate:. A warning ; we must be careful about identifying the accurate Side and Angle relationships Fun and straightforward global. Write the associated two column proof accurate Side and Angle relationships were shown for a with... For your better understanding, here is now the exact statement of the SAS, and also BC! To see dot at P, Q, R products in energy ( Fig example $... And all the sides with ruler and protractor, and with technology ) geometric shapes with given conditions DeLay... = ∠DEC [ given. geometry lesson, we prove triangle congruency similar and write a Similarity.! Video tutorial provides a basic introduction into triangle Congruence postulates and Theorems - Concept - Solved examples lesson! Side Postulate or SAS a consequence, their angles will be the same \triangle EFC$... Determine which congruent triangle Vertical angles are equal are five ways you can prove triangle congruency how the SSS,. See how to identify congruent corresponding parts are congruent… we sss postulate examples need three game is! A true statement, which does not deal with angles Similarity statement the perpendicular bisector of.! Video tutorial provides a basic introduction into triangle Congruence postulates and Theorems - Concept - Solved examples for proving triangles! A Similarity statement four rules to prove that these triangles are congruent in the image to the audio in. Can be used to derive the other logical statements to solve a problem AAS Congruence date period state if sss postulate examples... 7Th January, 2020 Congruence Theorems: a quantity may be substituted for its equal in expression. Y / 7 = y * 3 -- uncooked spaghetti or plastic stirrers work.. At least one Side length known temperature changes across the subtropical front east of New Zealand based alkenone..., want to see whether two triangles are equal 's no other one place to put this third.... You into 4 groups and form a circle with your group jenn, Founder Calcworkshop® 15+... Other logical statements to solve a problem we can say that two triangles are congruent, we ’ going! Is the perpendicular bisector of BC that triangles are equal possible, that makes the triangles congruent is Fun straightforward! Below could you use the AA Similarity Postulate Explain why the triangles congruent 97 examples: are! The Postulate or theorem you would use to prove two triangles: Ayada Lugo | Last Updated January... Hence sides AB and CD are congruent, then the two triangles also a! We just need three is true, then they are called the SSS Postulate,! And foraminiferal AD is median on BC and DA are congruent. triangle if a triangle is congruent triangle! Are the SAS Postulate a courtroom to begin with square, all the angles and sides have be... The products in energy ( Fig are too incompetent to enter into a courtroom to begin with Postulate if three!, B and E are congruent. if a triangle is possible of reasoning angles are to. ( AAS ) postulates Side-Angle-Side ( SAS ) triangle: definition, &. \Triangle EFC  Advertisement is a warning ; we must be careful identifying. Into triangle Congruence Theorems alkenone unsaturation ratios and foraminiferal careful about identifying the accurate Side and Angle relationships triangle a... Quantities are added to equal quantities, the sums are equal and the corresponding sides in the English!: AC already know, two triangles are similar and write a Similarity statement and DA are congruent using methods.


Finding approximation of largest eigenvalue

Given the following matrix, find an approximation of the largest eigenvalue. $$A = \begin{bmatrix} 3 & 2 \\ 7 & 5 \\ \end{bmatrix}$$

And I was also given $$\vec x= \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix}$$

How my professor solves this is by calculating the slopes of $A\vec x = \vec b_1$, $A^2 \vec x = \vec b_2$, $A^3 \vec x = \vec b_3$ and so on until we get the slope of $\vec b_i$ converging to the same value. Then when we get the approximated $\vec b$, he plug into $A \vec b = \lambda \vec b$, and the corresponding $\lambda$ is the largest eigenvalue.

Since slope is $\frac yx$ , it works fine for $2 x 2$ matrix. But how do I apply this method for a bigger matrix?

• I've given you a full explanation and representation of the method used down below, make sure to check it out ! Dec 7, 2017 at 1:23
• It's very helpful thank you! Dec 7, 2017 at 1:28

What you mention, is a known numerical analysis method for the approximation of the largest (by absolute value) eigenvalue of a matrix.

Let $$A\in \mathbb R^{n\times n}$$ have $$n$$ linearly independent eigenvalues $$\{ u \}_{i=1}^n$$ as well as a unique eigenvalue $$λ_1$$ such that : $$|λ_1| < |λ_2| \leq \dots \leq |λ_n|$$ where $$λ_1 \in \mathbb R$$ and $$u_1 \in \mathbb R^n$$ : $$Au_1=λ_1u_1.$$

The method "Power Iteration" :

$$\begin{cases} x_k= Ax_{k-1} \to x_k = A^kx_0 \quad k=1,2,\dots \\ x_0 \end{cases}$$

Thorem : The method "Power Iteration" converges $$\forall x_0$$ (that is adequate for the problem given) and it is :

$$\lim_{k\to \infty} ε_k\frac{x_k}{||x_k||_2}=u_1$$

$$\lim_{k \to \infty} \frac{x_{k,i}}{x_{k-1,i}}=λ_1 \quad \forall \space i=1,2,\dots,n \quad \text{with} \quad u_{1,i} \neq 0$$

where $$\{ ε_k\}_{k=1}^\infty = \{ \pm1\}$$ and $$u_1$$ eigenvector of $$A$$ with $$||u_1||_2=1$$.

I've given you a formal explanation of the method according to my old notes and my knowledge, for more, check here.

Here's a hint: You want to determine when $\mathbf{b}_n$ is a near-scalar multiple of $\mathbf{b}_{n-1}$. In $\mathbb{R}^2$, (nonzero) vectors are scalar multiples of one another iff their slopes are equal. A possibly more useful definition is that two vectors $\mathbf{v}$ and $\mathbf{w}$ are scalar multiples of one another if and only if

$$\hat{\mathbf{v}} = \pm\hat{\mathbf{w}},$$

where

$$\hat{\mathbf{v}} = \frac{\mathbf{v}}{|\mathbf{v}|},$$

which extends more nicely to multiple dimensions.

• I'll try to apply this to the problem and see how it works out. Thank you ! Dec 7, 2017 at 1:21
• @dembrownies Great. Let me know if you have any more questions. Dec 7, 2017 at 2:13

You normalize the vector at each iteration by dividing by its length, and wait until the resulting sequence of unit vectors has gotten close enough to converging for your purpose. (This is called the power method. It's pretty much the worst iterative method for eigenvalues there is, but it is the theoretical basis for lots of better methods.)

This is similar to what you're doing when you measure the slope in the 2D case, except in that case you divide by $x$ instead of dividing by $\sqrt{x^2+y^2}$. The point is that all that matters is the direction, not the magnitude.

• that makes more sense. thank you! Dec 7, 2017 at 1:20

The method you are thinking of is called Power Iteration. More formally, take any vector $x_0\ne 0$. (In practice, it is good to choose $x_0$ randomly. Then define

$$x_k := \frac{Ax_{k-1}}{\|Ax_{k-1}\|_2}, \quad k\in\{1,2,\ldots\}$$

With high probability, $x_k$ will converge to an eigenvector of $A$ corresponding to the largest eigenvalue $\lambda_1$ of $A$ with

$$\lambda_1 = \lim_{k\to\infty} x_k^T Ax_k.$$

In practice, in computer arithmetic, this method is numerically stable. For a better method, see for example, this.


# Why can we convert a base $9$ number to a base $3$ number by simply converting each base $9$ digit into two base $3$ digits?

Why can we convert a base $9$ number to a base $3$ number by simply converting each base $9$ digit into two base $3$ digits ?

For example $813_9$ can be converted directly to base $3$ by noting

\begin{array} \space 8_9&=22_3 \\ \space 1_9 &=01_3 \\ \space 3_9 &=10_3 \\ \end{array}

Putting the base digits together ,we get $$813_9=220110_3$$

I know it has to do with the fact that $9=3^2$ but I am not able to understand this all by this simple fact...

• This is actually useful (but not with $9$ and $3$). We can very cheaply go back and forth between hexadecimal (base $16$) and binary. – André Nicolas Jan 26 '16 at 8:30
• This also applies to converting between: binary, octal, hexadecimal systems – Max Payne Jan 26 '16 at 8:31

Consider $N$ in base 3. For simplicity, we can assume that $N_3$ has an even number of digits: if it doesn't, just tack on a leftmost $0$. So let: $$N_3 = t_{2n+1} t_{2n}\dotsc t_{2k+1} t_{2k} \dotsc t_1 t_0.$$ What this positional notation really means is that: $$N = \sum_{i = 0}^{2n+1} t_i 3^i,$$ which we can rewrite as: \begin{align} N &= \sum_{k = 0}^{n} (t_{2k+1} 3^{2k+1} + t_{2k} 3^{2k}) \\ &= \sum_{k = 0}^{n} (3 t_{2k+1} + t_{2k}) 3^{2k} \\ &= \sum_{k = 0}^{n} (3 t_{2k+1} + t_{2k}) 9^{k}. \\ \end{align}

But now, note that for each $k$, $3 t_{2k+1} + t_{2k}$ is precisely the base-9 digit corresponding to the consecutive pair of base-3 digits $t_{2k+1} t_{2k}$.

• Altough I've already accepted an answer,I thought to let you know that this answer has been so insightful.Thanks for your time ! – Mr. Y Jan 26 '16 at 9:18
• How nice — thank you:) and you're welcome. – BrianO Jan 26 '16 at 9:24
• That is the real good prove – JnxF Jan 26 '16 at 9:57
• Oh my - thanks doubly. – BrianO Jan 26 '16 at 10:05
• yes this is the proof. My answer is just a visual. +1 – Max Payne Jan 26 '16 at 10:16

This has more to do with place value.

Consider the image:

The key is the fact that $3^2 = 9$

• All the value below $3^2$ ($3^0$ and $3^1$) has to be accomodated in $9^0$

that is

• all the values in $9^0$ have to be accommodated in the two places below $3^2$ ($3^0$ and $3^1$)

Similarly for any other places.

• This rule also applies to conversion between binary and octal, or binary and hexadecimal, or conversion between any base $k$ and base $k^n$

In general, for conversion of a number $N$ in any base $k$ to base $k^n$, each digit of $N_{k^n}$ get converted to n digits of $N_k$

• Thanks for the answer .One question : in general if we have that some number $c$ (I am considering everything in base $10$) such that $c=b^n$ does this mean that I can convert the number $c$ in base $b$ by simple converting each base $10$ digit of $c$ into $n$ base $b$ digits ?Or is this something that works only for these special cases ? – Mr. Y Jan 26 '16 at 8:56
• @Mr.Y, yes, you can. An even stronger statement would be: given $a^n=b^m$, we could convert a base $a^n$ number, say, $k$, to base $b^m$, by converting every $n$ digits of $k$ (base $a^n$) to $m$ digits base $b^m$. – vrugtehagel Jan 26 '16 at 9:02

I'm not sure whether you're asking for a proof, or some intuition. The proof is fairly mechanical, so I'll explain why you might expect this to be true.

Putting your information theory hat on, two base 3 digits carry exactly as much information as one base 9 digit. That is, suppose you know I'm going to pass you two digits, each from 0,1 and 2. Then you know there are $3 \cdot 3 =9$ possibilities for what information I could pass.

On the other hand, if I passed you a single digit from 0,1,...,8, then there are again 9 possibilities.

Continuing in this fashion, $n$ digits of base 9 pass as much information as $2n$ digits of base 3.

This isn't exactly a proof yet. You'd need to prove that you don't skip any values. Perhaps doing the computation you describe can never return the value $102_3$. Thankfully it can. But you'd need to think about why this will always work.

Let's look at what you base $9$ number actually means. $$813_9=8\cdot 9^2+1\cdot 9^1+3\cdot 9^0$$ If we wish to write this as powers of $3$ with coefficients between $0$ and $2$, we can simply do \begin{align} 3\cdot 9^0&=1\cdot 3^1+0\cdot 3^0\\ 1\cdot 9^1&=0\cdot 3^3+1\cdot 3^2\\ 8\cdot 9^2&=2\cdot 3^5+2\cdot 3^4\\ \end{align}

Why can we do this?

Why can we write $$a_n\cdot 9^n=b_{2n+1}\cdot 3^{2n+1}+b_{2n}\cdot3^{2n}$$ First note that $9^n3^2n$, so when dividing the equation by that, we get $$\frac{a_n\cdot 9^n}{3^{2n}}=a_n=3b_{2n+1}+b_{2n}=\frac{b_{2n+1}\cdot 3^{2n+1}+b_{2n}\cdot3^{2n}}{3^2n}$$ So actually, the problem comes down to writing a number $0\leq a_n<9$ as $3b_{2n+1}+b_{2n}$, where $0\leq b_{2n},b{2n+1}<3$. This is obviously possible.

Why does this work for binary and hexadecimal?

Actually, the answer is fairly similar. The problem can be reduced equivalently, resulting in the equation $$a_n=8a_{4n+3}+4a_{4n+2}+2a_{4n+1}+a_{4n}$$ Where $0\leq a_n<16$ and $0\leq a_{2n+3},a_{2n+2},a_{2n+1},a_{2n}<2$. The solvability of this equation is in my opinion a little less obvious, but still quite understandable; But, for the sake of a more thourough understanding, we could look at hexadecimal-octagonal conversion. This comes down to the easy equation $a_n=2b_{2n+1}+b_{2n}$, where $0\leq a_n<16$ and $0\leq b_{2n+1},b_{2n}<8$. This is cleary solvable. Doing this for octagonal-base 4 conversion and base 4-binary conversion shows with a recurrance-like approach that this indeed works for hexadecimal-binary conversion.

Hope this helped!

Hint:

$$\color{blue}1\cdot3^5+\color{blue}0\cdot3^4+\color{green}2\cdot3^3+\color{green}2\cdot3^2+\color{red}0\cdot3^1+\color{red}1\cdot3^0 =(\color{blue}1\cdot3^1+\color{blue}0)3^4+(\color{green}2\cdot3^1+\color{green}2)3^2+(\color{red}0\cdot3^1+\color{red}1)3^0\\ =\color{blue}3\cdot9^2+\color{green}8\cdot9^1+\color{red}1\cdot9^0$$

$$\color{blue}{10}\color{green}{22}\color{red}{01}_3=[\color{blue}{10_3}|\color{green}{22_3}|\color{red}{01_3}]_9=\color{blue}3\color{green}8\color{red}1_9$$


# Intuition for a relationship between volume and surface area of an $n$-sphere

The volume of an $n$-sphere of radius $R$ is

$$V_n(R) = \frac{\pi^{n/2} R^n}{\Gamma(\frac{n}{2}+1)}$$

and the surface area is

$$S_n(R) = \frac{2\pi^{n/2}R^{n-1}}{\Gamma(\frac{n}{2})} = \frac{n\pi^{n/2}R^{n-1}}{\Gamma(\frac{n}{2}+1)} = \frac{d V_n(R)}{dR}$$

What is the intuition for this relationship between the volume and surface area of an $n$-sphere? Does it relate to the fact that the $n$-sphere is the most compact shape in $n$ dimensions, or is that merely a coincidence?

Are there other shapes for which it holds, or is this the limiting case of a relationship of inequality? For example, an $n$-cube of side $R$ has volume $V^c_n(R)=R^n$ and surface area $S^c_n(R)=2nR^{n-1}$, so that

$$\frac{dV^c_n(R)}{dR} = n R ^{n-1}$$

and we have $S^c_n = 2 dV^c_n/dR$.

Edit: As pointed out by Rahul Nahrain in the comments below, if we define $R$ to be the half-side length of the unit cube rather than the side length, then we have the relationship $S^c_n(R)=\frac{d}{dR} V^c_n(R)$, exactly as for the sphere. Is there a sense in which relationships like this can be stated for a large class of shapes?

-
See this MO-question and also the isoperimetric problem. –  t.b. May 9 '11 at 10:52
The main takeaway seems to be that this relationship is coincidental and based on using $R$ as the linear measure of distance. Were we to use $D=2R$ instead, we would get $S_n(D)=2\frac{d}{dD}V_n(D)$. However, there is still the interesting (to my mind) question of why the ratio of $S_n(x)$ to $\frac{d}{dx}V_n(x)$ (for some linear measure $x$) is independent of $n$ for the $n$-sphere but seemingly not for other shapes - no? –  Chris Taylor May 9 '11 at 11:04
Sure! I also voted the question up, by the way. I didn't mean to imply that these links answer the question completely. I just wanted you to see them because they certainly give some food for thought. –  t.b. May 9 '11 at 11:06
Isn't the surface area of an $n$-cube $2nR^{n-1}$, not $6R^{n-1}$? Then $S_n^c = 2dV_n^c/dR$ for all $n$. –  Rahul May 9 '11 at 11:31
In fact, if you let the parameter be the "radius" of the cube, i.e. half of its side length, then $V_n^c = (2R)^n$ and $S_n^c = 2n(2R)^{n-1}$, and $S_n^c = dV_n^c/dR$ for all $n$, just like the sphere... –  Rahul May 9 '11 at 11:39

There is a very simple geometric explanation for the fact that the constant of proportionality is 1 for the sphere's radius and the cube's half-width. In fact, this relationship also lets you define a sensible notion of a "half-width" of an arbitrary $n$-dimensional shape.

Pick an arbitrary shape and a point $O$ inside it. Suppose you enlarge the shape by a factor $\alpha \ll 1$ keeping $O$ fixed. Each surface element with area $dA$ at a position $\vec r$ relative to $O$ gets extruded into an approximate prism shape with base area $dA$ and offset $\alpha \vec r$. The corresponding additional volume is $\alpha \vec r\cdot \vec n dA$, where $\vec n$ is the normal vector at the surface element.

Now the quantity $\vec r \cdot \vec n$, call it the projected distance, has a natural geometric interpretation. It is simply the distance between $O$ and the tangent plane at the surface element. (Observe that for a sphere with $O$ at the center, it is always equal to the radius, and for a cube with $O$ at the center, it is always equal to the distance from the center to any face.)

Let $\hat r = A^{-1} \int \vec r \cdot \vec n dA$ be the mean projected distance over the surface of the shape. Then the change in volume by a scaling of $\alpha$ is simply $\delta V = \alpha \int \vec r\cdot \vec n dA = \alpha \hat r A$. In other words, a change of $\alpha \hat r$ in $\hat r$ corresponds to a changed of $\alpha \hat r A$ in $V$. So if you use $\hat r$ as the measure of the size of a shape, you find that $dV/d\hat r = A$. And since $\hat r$ equals the radius of a sphere and the half-width of a cube, the observation in question follows. This also implies the distance-to-face measure for regular polytopes that user9325 mentioned, but generalizes to other polytopes and curved shapes. (I'm not completely happy with the definition of $\hat r$ because it's not obvious that it is independent of the choice of $O$. If someone can see a more natural definition, please let me know.)

-

Some remarks:

1. It is not necessary to regard solids that generalize to $n$ dimensions, so one can start with shapes in 2 dimensions.

2. It is very natural to always use a radius-type parameter to scale the figure because the intuition is that the figure gets growth rings that have the size of the surface.

3. Unfortunately, the property is no longer true if you replace a square with a rectangle or a circle with an ellipse.

4. For a regular polygon, you can always take as parameter the distance to an edge. This will work similarly for Platonic solids or any polygons/polyhedra that contain a point that is equidistant to all faces.

5. This argument does not look good for general curves, because intuitively the edges of a smooth curve are infinitesimally small, so the center should be equidistant to all points, but of course, we could identify cases where the changing thickness of the growth ring averages out.

-
+1. Point 2 is the intuitive explanation, while points 3 and 5 are the essential caveats. –  Henry May 9 '11 at 12:32

What you're seeing with spheres is an instance of something really important called the coarea formula. You're looking at a function like $f(x,y,z) = r = \sqrt{x^2 + y^2 + z^2}$ and you're comparing the volumes of the lower contour sets $f(x,y,z) \leq R$ (in this case they are balls) to the areas of the level sets $f(x,y,z) = R$.

In the instance with cubes, you're considering a different function whose level sets are cubes.

The coarea formula states that (subject to some assumptions that cause both sides to make sense),

$Vol(f(x, y, z) \leq R) = \int_{-\infty}^R \left[ \int_{f(x,y,z) = t} |\nabla f|^{-1} d\sigma \right] dt$

You can also differentiate this formula with respect to $R$ to get the differential form you were observing.

Just as a dummy check, note that $t$ has the same units as $f$, so if you think of $f$ and the coordinates as having units, the dimensional analysis checks out and both sides have units of "length^3". (The formula is also valid in higher dimensions.)

In the examples you gave, the gradient of $f$ has length $1$, so when you integrate $1$ over the level sets $\{f(x, y, z) = t \}$ you're just getting the surface area.

The proof of this formula boils down to asking "what's the difference between $Vol( f(x,y,z) \leq R )$ and $Vol(f(x,y,z) \leq R + h$ if $h$ is small?". The two regions are very similar, differing only around the boundary $f(x,y,z) = R$. This is particularly clear in the case of a sphere. If you think about a small portion of that boundary of "width" $d\sigma$ in the directions of constant $f$, then the difference of the two regions has height $~ \frac{h}{|\nabla f|}$ in the direction of increasing $f$.

Another dummy check: The inverse makes sense because if $f$ is increasing extremely quickly, then increasing the value of $f$ will hardly increase the region $\{ f(x,y,z) \leq R \}$.

Note, the statement of the coarea is really a statement about the function and not just about the shapes of the level sets -- you are asking about the parameter $R$ by which these level sets are labeled, but you could pick a different function (like $\tilde{f} = e^f$ which has the same level sets but labels them differently.

-


# Easiest way to show positive semi-definite equivalence

For an $$x \in \mathbb{R}^n$$, and $$n$$-by-$$n$$ identity matrix $$I_n$$, we are given that $$\begin{pmatrix} I_n & x \\ x^T & 1 \end{pmatrix} \succeq 0.$$

What is the easiest way to show that $$\begin{pmatrix} 1 & x^T \\ x & I_n \end{pmatrix} \succeq 0$$ holds?

• Symmetric permutation? – user251257 Mar 22 at 18:37
• seems so. is there such a theory which concludes? – independentvariable Mar 22 at 18:40
• Schur complement? – user251257 Mar 22 at 19:02
• They dont reduce to the same condition, do they? – independentvariable Mar 22 at 19:22

This is due to the identity

$$\underbrace{\begin{pmatrix} 0_{1,n} & 1 \\ I_n & 0_{n,1} \end{pmatrix}}_{J^T} \underbrace{\begin{pmatrix} I_n & x \\ x^T & 1 \end{pmatrix}}_{A} \underbrace{\begin{pmatrix} 0_{n,1} & I_n \\ 1 & 0_{1,n} \end{pmatrix}}_{J} = \underbrace{\begin{pmatrix} 1 & x^T \\ x & I_n \end{pmatrix}}_{B}$$

(notation $$0_{m,n}$$ is for a zero block with $$m$$ lines and $$n$$ columns).

Indeed, $$J$$ being a permutation matrix, it is an orthogonal matrix, with $$J^T=J^{-1}$$. We can conclude that $$A$$ and $$B$$ are similar, thus have the same spectrum (Similar matrices have the same eigenvalues with the same geometric multiplicity) with positive eigenvalues, thus are both semi-definite positive.

Besides, $$A$$ being symmetric, one can conclude from $$J^TAJ=B$$ that $$B$$ is symmetric as well.

Appendix : There is a pending question : is there a criteria on $$x$$ for positive semi-definiteness of $$A$$. ? The answer is yes :

$$A$$ is semi-definite positive iff $$\|x\| \leq 1$$.

This is due, as we are going to see it, to an analysis of the rather particular spectrum of $$A$$. Let us obtain it explicitly.

First of all, let us establish that $$A$$ (which is a $$(n+1) \times (n+1)$$ matrix) has eigenvalue $$1$$ with order of multiplicity at least $$n-1$$.

Consider hyperplane $$x^{\perp}$$ of $$\mathbb{R}^n$$ defined as the set of vectors $$y$$ that are orthogonal to $$x$$. Let $$(y_1,y_2,\cdots y_{n-1})$$ be a basis of $$x^{\perp}$$ ; then,

$$\underbrace{\begin{pmatrix} I_n & x \\ x^T & 1 \end{pmatrix}}_{A}\underbrace{\begin{pmatrix} y_k\\ 0 \end{pmatrix}}_{V_k}=1\underbrace{\begin{pmatrix} y_k\\ 0 \end{pmatrix}}_{V_k} \ \ \ \text{for} \ \ k=1,2, \cdots (n-1),$$

proving that $$V_k$$ is an eigenvector associated with eigenvalue $$1$$.

Due to the fact that trace$$(A)=n+1$$, the two remaining eigenvalues are of the form $$\alpha$$ and $$\beta:=2-\alpha$$. We can assume, WLOG that $$\alpha \leq 1 \leq \beta$$.

Besides, using the so-called Schur determinant identity (Eigenvalues of a Block Matrix from Schur Determinant Identity) for the computation of the determinant of a $$2 \times 2$$ block matrix, we obtain :

$$\det(A)=1-x^Tx$$

As the determinant is also the product of eigenvalues, we get the following identity :

$$\det(A)=1-\|x\|^2=\alpha(2-\alpha)\tag{1}$$

Thus, one can compute explicitly the two remaining eigenvalues by solving quadratic equation (1), with the following explicit solutions (if we assume that $$\alpha$$ is the smallest eigenvalue)

$$\alpha=1 - \|x\| \ \ \ \implies \ \ \ \beta:=2-\alpha=1 + \|x\|\tag{2}$$

As the criteria for a symmetric matrix do be semi-definite positive is that must have all eigenvalues $$\geq 0$$, this criteria becomes $$\alpha \geq 0$$, i.e., $$\|x\| \leq 0$$. $$\square$$

Remark : eigenvalues $$\alpha$$ and $$\beta$$ can be associated with eigenvectors $$\begin{pmatrix} x\\ -\|x\| \end{pmatrix}$$ and $$\begin{pmatrix} x\\ \|x\| \end{pmatrix}$$ resp.

Let us take an example in the case $$n=4$$ ; let $$m=1/n$$ ; consider matrix :

$$A:=\left(\begin{array}{rrrr|r} 1 & & & & m \\ & 1 & & & m \\ & & 1 & & m\\ & & & 1 & m\\ \hline m & m & m & m & 1 \\ \end{array}\right)$$

One can check, using (2), that the spectrum of $$A$$ is

$$(\tfrac12, 1 , 1, 1, 1, \tfrac32).$$

Just now, I "googled" with keywords "bordered identity matrix" : I found in (Eigenvalues of a certain bordered identity matrix) a somewhat similar computation that I did in the Appendix.

• Thank you for your answer! My main purpose was showing $||x||_2 \leq 1$ holds iff the matrices I gave are Psd. I proved one by Schur complement theory on PSD matrics, but for the equivalence I just followed the definition of PSD and by contradiction showed that if one holds, the other one should hold etc.. – independentvariable Mar 23 at 19:28
• But yours seem the better way, not the 'dirty' $a^T X a \geq 0$ for all $a$ approach... I don't like it, it seems too manual – independentvariable Mar 23 at 19:35


Is the Cartesian product of two countably infinite sets also countably infinite?

I am trying to determine and prove whether the set of convergent sequences of prime numbers is countably or uncountably infinite.

It is clear that such a sequence must 'terminate' with an infinite repetition of some prime $p$. So for example $${1,2,3,5,5,5,5,...}$$

My idea is to break up the problem into two sub-sequences. The first is a finite subsequence consisting of the first however many terms before it begins the infinite repetition of $p$. Then the second being the infinite set with elements being only $p$.

So in the above example, the two sub-sequences will be $$\{1,2,3\} \{5,5,5,...\}$$

The infinite sub-sequence is countably infinite as it's cardinality is simply the cardinality of the primes, which is countably infinite.

There are infinitely many finite sub-sequences as it can have any number of terms. However, I think it will be countably infinite because it can have $1$ term, $2$ terms, $3$ terms etc, in other words we can construct the bijection to $\mathbb{N}$ by simply naming them according to the number of terms it has. The arrangements is irrelevant as it will be finite.

This is where the title of the question comes in. There are a countably infinite number of these finite sub-sequences say $F$. We also have a countably infinite number of the infinite subsequence $P$, with all elements being some prime $p$.

So in some sense, the set of all convergent sequences of primes is the cartesian product $F\times P$.

As they are both countably infinite, will their cartesian product also be countably infinite? I am thinking yes, as the rationals are countably infinite and they are in some sense a cartesian product of the numerator and denominator, each having cardinality equivalent to the naturals.

If so, and everything is correct, I think I will have then completed the proof. Otherwise, I'd love to have errors pointed out or even perhaps a better solution.

Your thinking is correct. In fact, what you have argued is that there exists a bijection between the set of all convergent sequences of primes and the set $F\times P$. If this is your first example of doing this kind of task I suggest you try to actually write down this bijection. You have argued quite well that it must exist, but it's good exercise to actually do it at least once.

Noy, you also rightly say that $F\times P$ has a bijection to $\mathbb N^2$. As before, I suggest you actually write down this bijection.

Now, you simply need to see if $\mathbb N^2$ is countable, that is:

Does there exist a bijection between $\mathbb N$ and $\mathbb N^2$

To that end, I suggest you look into Cantor's pairing function.

• or you might just observe that $(i,j)\mapsto 2^i3^j$ is an injection. This idea extends easily to $\mathbb N^k$. For example, $(i,j,k)\mapsto 2^i3^j5^k$. Mar 7, 2016 at 16:06
• @Chilango Of course. Though I think that for a "first time user", the actual bijection is more informative than just the construction of an injection (which, I admit, implies a bijection, but also a theorem)
– 5xum
Mar 7, 2016 at 16:32
• these maps are not surjective, but that's no problem. An injection into a countable set is enough to say that the domain is countable. All you need is the fact that a subset of a countable set is countable. I agree though, that the pairing function is more interesting and has more applications. Mar 7, 2016 at 16:50


# $x^5 + y^2 = z^3$

While waiting for my döner at lunch the other day, I noticed my order number was $343 = 7^3$ (surely not the total for that day), which reminded me of how $3^5 = 243$, so that $$7^3 = 3^5 + 100 = 3^5 + 10^2.$$ Naturally, I started wondering about nontrivial integer solutions to $$x^5 + y^2 = z^3 \tag{*}$$ ("nontrivial" meaning $xyz \ne 0$). I did not make much progress, though apparently there are infinitely many solutions: this was Problem 1 on the 1991 Canadian Mathematical Olympiad. The official solutions (at the bottom of this page) only go back to 1994. A cheap answer is given by taking $x = 2^{2k}$ and $y = 2^{5k}$ so that the l.h.s. is $2^{10k + 1}$. This is a cube iff $10k + 1 \equiv 0 \,(3)$ i.e. $k \equiv 2\,(3)$ thus giving an arithmetic progression's worth of solutions, starting with $$(x, y, z) = (16, 1024, 128)$$ corresponding to $k = 2$ and $$(x, y, z) = (1024, 33554432, 131072)$$ coming from $k = 5$.

What else is known about the equation $(*)$? In particular, are there infinitely many solutions with $x$, $y$, $z$ relatively prime? The one that caught my attention was $(x, y, z) = (3, 10, 7)$. Another one is $(-1, 3, 2)$ because $-1 + 9 = 8$. By Catalan's conjecture (now a theorem), this is the only solution with $x = \pm 1$ or $y = \pm 1$ or $z = 1$. Are there any solutions with $z = -1$? In this case, $(*)$ reduces to $x^5 + y^2 = -1$ and Mihăilescu's theorem does not apply.

Update. This question was essentially already asked here, since the equation $a^2 + b^3 = c^5$ is equivalent to $(-c)^5 + a^2 = (-b)^3$.

• we can rewrite this as $x^5+y^2\equiv 0 \pmod z$ could that help you ?
– user451844
Jul 27 '17 at 1:09
• You might look at OEIS sequences A070065, A070066 and A070067. Jul 27 '17 at 2:08
• Note that if $(x,y,z)$ is a solution, then so is $(c^6 x, c^{15} y, c^{10} z)$ for integers $c$. Jul 27 '17 at 2:41
• quora.com/… Aug 3 '17 at 11:45

Yes, there are infinitely many solutions. In fact, there are many parametrizations of the solutions.
According to a book${}^{\color{blue}{[1]}}$ on my bookshelf,

Up to changing $y$ into $-y$, there are exactly 27 distinct parametrizations of the equations $x^5 + y^2 = z^3$.

One of the simplest paremetrization is given by following formula.

\begin{align} x =&\; 12st(81s^{10}-1584t^5s^5-256t^{10})\\ y =&\; \pm (81s^{10} + 256t^{10})\\ &\;\;\times (6561s^{20} - 6088608t^5s^{15} - 207484416t^{10}s^{10} + 19243008t^{15}s^5 + 65536t^{20})\\ z =&\; 6561s^{20}+2659392t^{5}s^{15}+10243584t^{10}s^{10} - 8404992t^{15}s^5 + 65536t^{20} \end{align}

For example, following two random choices of $s,t$ give you two sets of relative prime solutions.

• $(s,t) = (1,1) \leadsto (x,y,z) = (-21108,-65464918703,4570081)$
• $(s,t) = (1,2) \leadsto (x,y,z) = (-7506024,127602747389962225,-196120763999)$

The book I have is actually quoting result from a thesis${}^{\color{blue}{[2]}}$ by J. Edwards. Consult that if you really want to get into the details.

References

• $\color{blue}{[1]}$ Henri Cohen, Number Theory Number Theory Volume II: Analytic and Modern Tools,
$\S 14.5.2$ The Icosahedron Case $(2,3,5)$.

• $\color{blue}{[2]}$ J. Edwards, Platonic solids and solutions to $x^2+y^3 = dz^r$, Thesis, Univ. Utrecht (2005).

• Nice answer, especially for the first reference. +1
– Xam
Jul 27 '17 at 3:38
• @achille: We can also use the icosahedral equation and scale it appropriately. Kindly see answer below. Jul 27 '17 at 12:00
• @achille: The parameterizations for $x^5+y^3=z^2$ depend on the icosahedron while $x^4+y^3=z^2$ depend on the tetrahedron. Does the book you cite say how many distinct parameterizations are there for the latter? This page gives $7$, but I just found an $8$th one, with no scaling involved. Jul 28 '17 at 5:53
• @TitoPiezasIII In $\S 14.4.1$ of that book, Cohen proved there are 7 parametrizations. Jul 28 '17 at 7:13
• @achillehui: I just re-checked the parameterization I found, and it turns out there is a common factor. Darn! In Edwards' work, he cites 7 parameterizations as well. Thanks, anyway. Jul 28 '17 at 7:35

There is a beautiful connection between $a^5+b^3=c^2$ and the icosahedron. Consider the unscaled icosahedral equation,

$$\color{blue}{12^3u v(u^2 + 11 u v - v^2)^5}+(u^4 - 228 u^3 v + 494 u^2 v^2 + 228 u v^3 + v^4)^3 = (u^6 + 522 u^5 v - 10005 u^4 v^2 - 10005 u^2 v^4 - 522 u v^5 + v^6)^2\tag1$$

By scaling $u=12x^5$ and $v=12y^5$ (or various combinations thereof like $u=12^2x^5$, etc), we then get a relation of form,

$$12^5a^5+b^3=c^2$$

• +1 interesting connection. it is unfortunate the mythical $j(\tau)$ is stuff way beyond me ;-p. Jul 27 '17 at 13:14


# Checking for linear independence

Is the following proof correct?

Theorem. If $v_1,v_2,v_3,v_4$ is a linearly independent list then $$v_1-v_2,v_2-v_3,v_3-v_4,v_4$$ is also a linearly independent list.

Proof. Assume that $v_1,v_@,v_3,v_4$ is a linearly independent list, Consider now the following equation. $$0=0(v_1-v_2)+0(v_2-v_3)+0(v_3-v_4)+0v_4\tag{1}$$ Let $a_1,a_2,a_3$ and $a_4$ be arbitrary scalars in $\mathbf{F}$ and assume that the following equation holds $$0=a_1(v_1-v_2)+a_2(v_2-v_3)+a_3(v_3-v_4)+a_4v_4\tag{2}$$ After some algebraic manipulation we arrive at the following equation. $$0=a_1v_1+(a_2-a_1)v_2+(a_3-a_2)v_3+(a_4-a_3)v_4\tag{3}$$ Since the list $v_1,v_2,v_3,v_4$ is linearly independent it follows that given any vector in $span(v_1,v_2,v_3,v_4)$ the choice of scalars is unique and since $$0=0v_1+0v_2+0v_3+0v_4\tag{4}$$ It follows that all the scalars in $(3)$ must be $0$, consequently the only way to produce the $0$ vector as a linear combination of the vectors in the list $v_1-v_2,v_2-v_3,v_3-v_4,v_4$ is that indicated in $(1)$.

$\blacksquare$

Here $\mathbf{F}$ is either $\mathbb{C}$ or $\mathbb{R}$.

• Sounds ok - would just like to point out 'Consider now the following equation.' sounds a bit strange for $(1)$. Perhaps remove that line and $(1)$ and say something along the lines 'by inspection, $(2)$ holds when all the scalars are equal to $0$' at the end of your proof – Shuri2060 Jul 25 '17 at 17:31
• Nitpicking, but you made a small typo just before your first labeled equation, where you write "Assume $v_1,v_{@},v_3,v_4$ is a linearly independent list." Did you mean $v_2$ not $v_{@}$ ? – Vivek Kaushik Jul 25 '17 at 17:37

Equation (1) should be omitted: it's an obvious fact that has no consequence on the rest.

The final argument is too fast: from linear independence of $v_1,v_2,v_3,v_4$ you deduce \begin{cases} a_1=0 \\ a_2-a_1=0 \\ a_3-a_2=0 \\ a_4-a_3=0 \end{cases} and, from this, $a_1=a_2=a_3=a_4=0$. This should be mentioned, although easy.

A different approach is to consider the coordinates of the new vectors with respect to the original ones and so the matrix \begin{bmatrix} 1 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 \\ 0 & -1 & 1 & 0 \\ 0 & 0 & -1 & 1 \end{bmatrix} A standard Gaussian elimination leads to the reduced row echelon form \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} which proves that the coordinate vectors are linearly independent and so also the vectors are.

• So your point is that i should show how each of the scalars $a_1,a_2,a_3$ and $a_4$ is 0. – Atif Farooq Jul 25 '17 at 17:59
• The original matrix is lower triangular with $1$'s in the diagonal, hence invertible. – lhf Jul 25 '17 at 18:49
• @AtifFarooq Yes, that's the point; notwithstanding it's easy, it should appear in the proof. – egreg Jul 25 '17 at 19:27
• @lhf That's true, but Gaussian elimination would give more information in case the vectors aren't linearly independent. – egreg Jul 25 '17 at 19:27

It's valid. A note:

Before $(2)$ when you say "...and assume that the following equation holds..." it would be better to perhaps phrase this as "We wish to solve ... for $a_j$." Saying the former means that you're assuming solutions exist - but you're not sure! This is a tiny point, and is perhaps reflected in my writing style.

We may also streamline a bit by saying the following after $(3)$.

As $\{v_1, v_2, v_3, v_4\}$ is a linearly independent set, we must have that $a_1 = 0, \ a_2-a_1=0, \ a_3 - a_2 = 0,$ and $a_4-a_3=0.$ Back substitution yields that each $a_j = 0$, and so the original set is also a linearly independent set. $\square$

Again, these are just small suggestions, but nonetheless your proof is correct.


$\int_0^{2\pi} \sqrt{1-\cos(x)}\,dx = 4\sqrt{2}$. Why?

According to the textbook, and Wolfram Alpha the above is correct.

Here is the step by step procedure from Wolfram Alpha for evaluating the indefinite integral:

Take the integral: $$\int\sqrt{1-\cos(x)}\,dx$$ For the integrand $\sqrt{1-\cos(x)}$, substitute $u=1-\cos(x)$ and $du=\sin(x)\,dx$: $$=\int-\frac{1}{\sqrt{2-u}}\,du$$ Factor out constants: $$=-\int\frac{1}{\sqrt{2-u}}\,du$$ For the integrand $1/\sqrt{2-u}$, substitute $s=2-u$ and $ds=-du$: $$=\int\frac{1}{\sqrt{s}}\,ds$$ The integral of $1/\sqrt{s}$ is $2\sqrt{s}$: $$=2\sqrt{s}+\text{constant}$$ Substitute back for $s=2-u$: $$=2\sqrt{2-u}+\text{constant}$$ Substitute back for $u=1-\cos(x)$: $$=2\sqrt{\cos(x)+1}+\text{constant}$$ Which is equivalent for restricted $x$ values to: $$\boxed{=-2\sqrt{1-\cos(x)}\cot\big(\frac{x}{2}\big)+\text{constant}}$$

I understand up to the below (which is a valid solution to the integral): $$2\sqrt{\cos(x)+1}+\text{constant}$$

However, if you evaluate this at $2\pi$ and $0$, you get the same thing, so the definite integral evaluates to zero.

After, you transform the above to: $$-2\sqrt{1-\cos(x)}\cot\big(\frac{x}{2}\big)+\text{constant}$$

The expression is indeterminate at $2\pi$ and $0$ of the form $0 \times \infty$. So I guess you would set up a limit and then use L'Hospital's rule to evaluate the expression at $2\pi$ and $0$ and get the answer to the definite integral?

In any case, all this seems strange. Why should the definite integral evaluated one way give $0$, and in another way give something else?

• You may want to find the anti derivative by multiplying top and bottom by the squareroot of the conjugate. You will end up with the absolute value of the sine term on top and the square root of 1+cosx in the bottom. This integrates easily. Mind the absolute value though as you apply your upper and lower limit. Oct 30 '13 at 19:57
• Ahh ok that makes sense. So to get rid of the absolute value sign, you can split up the integral from $0$ to $\pi$ and $\pi$ to $2\pi$ Oct 30 '13 at 20:02
• Yes, you ought to split that integral. Oct 30 '13 at 20:05

$$\int_0^{2\pi}\sqrt{1-\cos x}\,dx=\int_0^{2\pi}\sqrt{2\sin^2\frac{x}{2}}\,dx =\sqrt{2}\int_0^{2\pi}\sin\frac{x}{2}\,dx=-2\sqrt{2}\cos\frac{x}{2}\Big|_0^{2\pi}=4\sqrt{2}.$$
• @RonGordon $x/2 \in [0,\pi] \Rightarrow \sin(x/2) \ge 0$. Oct 30 '13 at 20:49
• Good old $\cos(2x) = \cos^2(x)-\sin^2(x)= 1-2\sin^2(x) = 2\cos^2(x)-1$. Oct 31 '13 at 1:25
Your $u$-substitutions should be injective on their interval of evaluation. Otherwise, you risk running into exactly this sort of issue.
Note that \begin{align}|\sin x| &= \sqrt{\sin^2 x}\\ &= \sqrt{1-\cos^2 x}\\ &= \sqrt{1-\cos x}\sqrt{1+\cos x}\\ &=\sqrt{1-\cos x}\sqrt{2-(1-\cos x)},\end{align} so if you want to use $u=1-\cos x$, then $$\frac{du}{dx}=\sin x=\begin{cases}|\sin x|=\sqrt{1-\cos x}\sqrt{2-(1-\cos x)} & 0\le x\le \pi\\-|\sin x|=-\sqrt{1-\cos x}\sqrt{2-(1-\cos x)} & \pi\le x\le2\pi,\end{cases}$$ so \begin{align}\int_0^{2\pi}\sqrt{1-\cos x}\,dx &= \int_0^\pi\sqrt{1-\cos x}\,dx+\int_\pi^{2\pi}\sqrt{1-\cos x}\,dx\\ &= \int_0^\pi\frac{|\sin x|}{\sqrt{2-(1-\cos x)}}\,dx+\int_\pi^{2\pi}\frac{|\sin x|}{\sqrt{2-(1-\cos x)}}\,dx\\ &= \int_0^\pi\frac{\sin x\,dx}{\sqrt{2-(1-\cos x)}}-\int_\pi^{2\pi}\frac{\sin x\,dx}{\sqrt{2-(1-\cos x)}}\\ &= \int_0^2\frac{du}{\sqrt{2-u}}-\int_2^0\frac{du}{\sqrt{2-u}}\\ &= 2\int_0^2\frac{du}{\sqrt{2-u}}.\end{align} At that point, we can use that antiderivative, with no need to resubstitute.
Alternately, you could note that $\cos(2\pi-x)=\cos x$, so \begin{align}\int_0^{2\pi}\sqrt{1-\cos x}\,dx &= \int_0^\pi\sqrt{1-\cos x}\,dx+\int_\pi^{2\pi}\sqrt{1-\cos x}\,dx\\ &= \int_0^\pi\sqrt{1-\cos x}\,dx+\int_\pi^{2\pi}\sqrt{1-\cos(2\pi-x)}\,dx\\ &= \int_0^\pi\sqrt{1-\cos x}\,dx-\int_{2\pi}^\pi\sqrt{1-\cos(2\pi-x)}\,dx\\ &= \int_0^\pi\sqrt{1-\cos x}\,dx-\int_0^\pi\sqrt{1-\cos x}\frac{d(2\pi-x)}{dx}\,dx\\ &= 2\int_0^\pi\sqrt{1-\cos x}\,dx,\end{align} at which point you can use your $u$-substitution without fear, since the cosine function is injective on $[0,\pi]$.


# Square divided by absolute value

First time posting on Math SE, with kind of a basic algebra question.

Question
Does the relation:

$$\dfrac{(ab)^2}{|ab|} = \left|ab\right|$$

with $a,b \in \mathbb{R_{\ne 0}}$ always hold?

It seems trivial to me, but Wolfram Alpha gives me a strange answer because it specifies that this is True assuming $a,b$ are positive.

Reasoning
No matter what sign $a,b$ have, we have that $(ab)^2 > 0$ and $\left|ab\right| > 0$. Thus their ratio is greater than zero, and the magnitude of that ratio is exactly $ab$ with a positive sign, so $\left|ab\right|$.

Is what I said correct? If so, is this question a completely useless one? Sorry for the occasionally bad English!

Edit: formatted equations as suggested by Frentos

• Hmmn, wlog for $a, b \in \mathbb{C}$ one could argue that $a^{2}b^{2} < 0$ – Kevin Jan 20 '16 at 11:28
• Welcome to math.SE! See this guide for how to write equations on this site. \dfrac makes larger, easier to read fractions and \left| \right| gives nicer absolute values. – Frentos Jan 20 '16 at 11:32
• @Bacon No, try $a=b=i$ (or any case when $a^2b^2$ is not even real). – Did Jan 21 '16 at 1:45
• @Did - fair point, my comment meant to reflect that in some cases this could be true – Kevin Jan 21 '16 at 9:20

Your statement about Wolfram is not quite correct. It gives various alternate forms for this expression, two of which are:

1. $ab$ assuming $a$ and $b$ are positive

2. $ab\,sgn(a)\,sgn(b)$

(2) is equivalent to $|ab|$

See here

• Indeed, I overlooked the $ab \; sgn(a) \; sgn(b)$ answer! I'll accept this one because it points out that Wolfram was right. – UJIN Jan 20 '16 at 14:30

For real numbers this is always true because the square of a real number equals the square of its absolute value, and in particular $(ab)^2=|ab|^2.$ Perhaps Wolfram has reservations because it considers the possibility of complex numbers?

Notice:

• If $a,b\in\mathbb{R^+}$, so $a,b>0$, then:

$$|ab|=|a||b|=ab$$

• If $a,b\in\mathbb{R^+}$, so $a,b>0$, so:

$$(ab)^2=a^2b^2$$

• If $a,b\in\mathbb{R^-}$, than $a,b<0$, so:

$$((-a)(-b))^2=(ab)^2=a^2b^2$$


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# A wire that weighs 20 pounds is cut into two pieces so that one of the

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A wire that weighs 20 pounds is cut into two pieces so that one of the pieces weighs 16 pounds and is 36 feet long. If the weight of each piece is directly proportional to the square of its length, how many feet long is the other piece of wire?

(A) 9
(B) 12
(C) 18
(D) 24
(E) 27

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Re: A wire that weighs 20 pounds is cut into two pieces so that one of the  [#permalink]

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30 Jun 2017, 03:57
Bunuel wrote:
A wire that weighs 20 pounds is cut into two pieces so that one of the pieces weighs 16 pounds and is 36 feet long. If the weight of each piece is directly proportional to the square of its length, how many feet long is the other piece of wire?

(A) 9
(B) 12
(C) 18
(D) 24
(E) 27

Weight of 2nd piece = 4 pound

Since the weight is directly proportional to the square of its length., we may write
$$\frac{16}{36^2}$$ = $$\frac{4}{x^2}$$

Solving above, we get x = 18

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Re: A wire that weighs 20 pounds is cut into two pieces so that one of the  [#permalink]

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30 Jun 2017, 04:02
Bunuel wrote:
A wire that weighs 20 pounds is cut into two pieces so that one of the pieces weighs 16 pounds and is 36 feet long. If the weight of each piece is directly proportional to the square of its length, how many feet long is the other piece of wire?

(A) 9
(B) 12
(C) 18
(D) 24
(E) 27

$$20$$ pounds wire is cut into two pieces = $$16$$ pounds $$+$$ $$4$$ pounds

The piece weighs $$16$$ pounds is $$36$$ feet long.

Ratio of weight and length of $$16$$ pounds piece $$= \frac{16}{36^2} = \frac{4 * 4}{36 * 36} = \frac{1}{81}$$

Therefore required ratio of other part would also be $$\frac{1}{81}$$

Ratio $$= \frac{4}{x^2} =$$ $$\frac{1}{81}$$

$$x^2 = 81*4$$ $$=> x = \sqrt{81*4}$$

$$x = 9 * 2 = 18$$

Hence length of $$4$$ pounds wire $$= 18$$

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Re: A wire that weighs 20 pounds is cut into two pieces so that one of the  [#permalink]

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30 Jun 2017, 04:07
IMO (c)
Given, weight proportional to square (length)
=> w1 / w2 = square (l1/l2)
=> 16/4 = square (36/l2)
=> l2 = 18

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Re: A wire that weighs 20 pounds is cut into two pieces so that one of the  [#permalink]

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04 Jul 2017, 07:52
1
Bunuel wrote:
A wire that weighs 20 pounds is cut into two pieces so that one of the pieces weighs 16 pounds and is 36 feet long. If the weight of each piece is directly proportional to the square of its length, how many feet long is the other piece of wire?

(A) 9
(B) 12
(C) 18
(D) 24
(E) 27

Since the 20-lb wire is cut into two pieces and one of the pieces weighs 16 lbs, the other piece must weigh 4 lbs. Since the 16-lb piece has length of 36 ft and the weight of each piece is directly proportional to the square of its length, we can let x = the length in feet of the 4-lb piece and create the following proportion:

16/36^2 = 4/x^2

16x^2 = 4 * 36^2

4x^2 = 36^2

x^2 = 36^2/2^2

x^2 = 18^2

x = 18

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Re: A wire that weighs 20 pounds is cut into two pieces so that one of the   [#permalink] 04 Jul 2017, 07:52
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Explanation of Counter-example?

I have a question where we are given a statement and asked to prove whether or not the statement is true. If the statement is true, then we must prove it; otherwise we must provide a counterexample to prove the statement incorrect.

If the statement is false, and the student provides a counter-example; should the student also provide an explanation of why the counter-example disproves the statement?

I believe this depends on how clearly the counterexample is stated.

Consider this claim: $f:\mathbb{R}\to\mathbb{R}$ is continuous $\implies f$ is differentiable.

Imagine a student, let's call him Karl, who says

Take any function of the form $\sum_{n=0} ^\infty a^n \cos(b^n \pi x)$, where $0<a<1$, $b$ is a positive odd integer, and $ab > 1+\frac{3}{2} \pi$.

Would you give him credit for this counterexample? Even presuming that the difficulty level of your course and the talents of your students are such that you did, indeed, expect them to come up with this result, I don't feel like this answer is sufficient. Obviously, Karl must have understood something about the exercise to be able to come up with this answer (never mind the great ingenuity required) but his written argument fails to demonstrate that knowledge.

Contrast that with a disproof of this claim: For any function $f:A\to B$ and any sets $S,T\subseteq A$, $f[S]\subseteq f[T]\implies S\subseteq T$.

This is the kind of statement you might consider in an introductory analysis course. I'd expect a student to find an easy counterexample, say by defining $A=\{\heartsuit,\spadesuit\}$ and $B=\{\star\}$ and setting $f(\heartsuit)=f(\spadesuit)=\star$ and $S=\{\heartsuit\}$ and $T=\{\spadesuit\}$.

Would I expect the student to have to show all the details? Do they really need to explicitly say, "Observe that $S,T\subseteq A$ and $f[S]=f[T]$ and yet $S\not\subseteq T$"? Well, that depends on your standards. If you're really training them to write clearly and explicitly (concision be darned), then you might require them to show these details. If you're just checking to see that they can generate such examples, then merely stating the example would suffice.

Main answer: It depends on your standards for such a written argument, which in turn should depend on what you expect the student to be able to understand and demonstrate.

• Sure, let's call him Karl. Sep 25 '14 at 1:26
• Maybe give extra credit to someone who described how to create a counter example, perhaps in the above case by using S={<heart>} and T={<something besides a heart>}
– rbp
Sep 29 '14 at 15:13

Certainly the student should be aware of the expectations beforehand. To this end, I think there are at least two approaches. One is to specify on individual questions whether counterexamples should be explained, and the other is to establish (preferably from the outset, i.e., setting "norms" at the beginning of a course) the expectation that reasoning/justification for all answers will be provided.

More generally, I think explaining counterexamples is important. Let me give just two reasons.

1) Consider the following problem:

Is every whole number divisible by 6 and 8 also divisible by 6 $\times$ 8 = 48? If so, prove your answer; if not, please provide a counterexample. Answer: 24.

But what does such a student understand? Perhaps she just remembered 24 is divisible by both and realized it is not divisible by 48. Or perhaps she realizes that the issue is that the two divisors in this proposed rule are not relatively prime, and could correct it by using, e.g., 3 and 16 instead. Or, even more seriously, perhaps she has the deeper understanding, but accidentally writes 32. In this last case, the understanding is the deepest, but it would be tough to justify giving the response any credit.

Asking that the counterexample be justified ameliorates this potential problem. Personally, I would expect more than just an explanation of the particular counterexample, e.g., more than:

No, and 24 is a counterexample: It is divisible by 6 because 24 = 4 $\times$ 6, and it is also divisible by 8 because 24 = 3 $\times$ 8. But it is not divisible by 48, because 24/48 is not a whole number.

I would prefer to see an answer that mentions being "relatively prime" or something equivalent, and perhaps even one that gives an example of two numbers that would yield a divisibility rule for 48.

2) More subtle and perhaps not related to the specific question you have in mind (if there is one): Note that (speaking slightly messily) counterexamples can settle for all questions, but not there exists questions. The previous example is the assertion that all whole numbers (etc). If this turns out to be false, then it can be settled with a counterexample. If it turns out to be true, then it will need a further proof or justification. On the other hand, a there exists statement that is true might be settled with a single example. If it turns out to be false, then it will need a further proof or justification.

This latter point about proving/disproving by (counter)example in the context of for all vs. there exists assertions may seem a bit pedantic, but it is a place in which many students who are just beginning to understand ideas around "proof" can struggle quite a bit.

• I don't think there exists statements are really that different because you can rewrite them as for all statements, i.e. $\forall x\ P(x) \iff \neg\exists x\ \neg P(x)$. (Of course maybe students don't know this, in which case to their minds there is a difference...) Sep 25 '14 at 3:53
• @DavidZ Yes, this is the underlying "reason" for the phenomenon mentioned in 2 above (i.e., that an example can disprove a for all statement or prove a there exists statement). And yes, one of the primary difficulties is precisely the one to which you allude parenthetically: For many students who are just getting into proof-based mathematics (or other mathematics with more formal reasoning) the equivalence you mention is neither known nor obvious. Sep 25 '14 at 18:08


# Calculating area of triangle

Problem

In the triangle above, we have $AC = 12, \ \ BD = 10, \ \ CD = 6$.

Find the area of $\triangle ABD$.

Caveat

This was part of an exam where you are not allowed to use a calculator or any other digital tools.

My progress

Given $AC, CD, \measuredangle C$, I was able to find the length of $AD$ using the law of cosines. It yielded $AD = \sqrt{180} = 6\sqrt5$.

From here, I decided to try and use the law of sines to find $\measuredangle A$ as it would be the same in $\triangle ABD$ and $\triangle ACD$.

I got $$\frac{\sin A}{CD} = \frac{\sin C}{AD}$$.

This yields $$\sin A = \frac{CD \sin C}{AD} = \frac{6}{\sqrt{180}}$$.

Then I need to calculate $\arcsin\frac{6}{\sqrt{180}}$, which I can't, because I can't use a calculator for this problem. (Well, technically I can, but I want to solve this within the constraints placed on the participants.)

From here, I am stuck. In $\triangle ABD$, I know now two of the lengths, but I don't know any of the angles.

Question

Am I overlooking an easy inverse sine here? Or is there another way to calculate this area?

Thanks in advance for any help!

• If this is part of an exam, why are you asking for help? – N. F. Taussig Mar 21 '17 at 20:07
• @N.F.Taussig - It was an exam given in August of 2016. I'm preparing for the 2017 one ;) – Alec Mar 21 '17 at 20:08

## 3 Answers

Find $BC$ in the right triangle $BCD$; you have the hypotenuse is $10$ and the other side is $6$ so $BC=8$.

Area $\triangle ACD = \frac12\cdot 12\cdot 6 = 36$.

Area $\triangle BCD = \frac12\cdot 8\cdot 6 = 24$.

Area $\triangle ABD = 36-24 = 12$.

• Damn... I really blinded myself on this one. Thanks! – Alec Mar 21 '17 at 20:11

Calculate $BC$ with the pythagorean theorem, and subtract the areas of the two right angled triangles.

$$BC=\sqrt{10^2-6^2}=8$$ Hence, $$AB=12-8=4$$ and $$S_{\Delta ABD}=\frac{4\cdot6}{2}=12$$


## mathmath333 one year ago Set Theory

1. mathmath333

\large \color{black}{\begin{align} \normalsize \text{A ∩ B = A ∩ C need not imply B = C.} \quad \normalsize \text{ Explain through an example.}\hspace{.33em}\\~\\ \end{align}}

2. Kainui

A = {1} B = {1, 2} C = {1, 3}

3. ikram002p

huh kai was so fast :P

4. Kainui

Hahaha there could have been a simpler example: A = {} B = {1} C = {2}

5. mathmath333

thnx

6. Kainui

I think that last example I gave sorta shows that it's almost like this is a lot like multiplying by zero. a*b=a*c This doesn't mean b=c, since a=0 is possible.

7. mathmath333

Let A and B be sets. If A ∩ X = B ∩ X = φ and A ∪ X = B ∪ X for some set X, show that A = B.

8. ikram002p

in general any case with $A\cap B \neq \varnothing$ does not imply A=C

9. ikram002p

or $$C\cap B \neq \varnothing$$

10. mathmath333

i need an example like the previous one

11. Kainui

ikram needs more owl bucks sry

12. ikram002p

A={1,2,3} B={1,2,3} C={1,2,3,4}

13. ikram002p

:O @Kainui i dont lol

14. Kainui

Hahaha jk :P

15. mathmath333

is that the example for this que Let A and B be sets. If A ∩ X = B ∩ X = φ and A ∪ X = B ∪ X for some set X, show that A = B.

16. ikram002p

you wanna prove this ?

17. mathmath333

yes with example

18. ikram002p

ok example A={1,2,3} B={1.4} C={1,2.3}

19. mathmath333

but C is not there in the question

20. ikram002p

sorry i ment to say x :P

21. ikram002p

now to prove they are two side of the prove 1- show A is a subset of B 2- show B is a subset of A

22. ikram002p

let $$c\in A \cup X$$ so $$c\in A ~or~c\in X$$ if c in A then $$c\in B\cup X$$ and $$c\in B$$ (since c is not in x) thus $$A\subset B$$ the other way is alike :P sorry im lazy to type

23. thomas5267

Is it possible to do something like this? $A\cup X=B\cup X\land A\cap X=B\cap X=\emptyset\implies A\cup X\setminus X=B\cup X\setminus X\implies A=B$


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# The average (arithmetic mean) length per film for a group of 21 films

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The average (arithmetic mean) length per film for a group of 21 films  [#permalink]

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05 May 2016, 19:27
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The average (arithmetic mean) length per film for a group of 21 films is t minutes. If a film that runs for 66 minutes is removed from the group and replaced by one that runs for 52 minutes, what is the average length per film, in minutes, for the new group of films, in terms of t?

A) $$t + \frac{2}{3}$$

B) $$t - \frac{2}{3}$$

C) $$21t+14$$

D) $$t + \frac{3}{2}$$

E) $$t - \frac{3}{2}$$

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Re: The average (arithmetic mean) length per film for a group of 21 films  [#permalink]

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05 May 2016, 20:09
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mcelroytutoring wrote:
The average (arithmetic mean) length per film for a group of 21 films is t minutes. If a film that runs for 66 minutes is removed from the group and replaced by one that runs for 52 minutes, what is the average length per film, in minutes, for the new group of films, in terms of t?

A) $$t + \frac{2}{3}$$

B) $$t - \frac{2}{3}$$

C) $$21t+14$$

D) $$t + \frac{3}{2}$$

E) $$t - \frac{3}{2}$$

hi,
you do not have to get into any calculations if you understand and take the Q logically..

avg time for 21 is t min..
you add 52 and remove 66.. so basically you are removing 14 min from total..
effect on each or average =$$\frac{14}{21}= \frac{2}{3}$$..

so final average = $$t-\frac{2}{3}$$...
B
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The average (arithmetic mean) length per film for a group of 21 films  [#permalink]

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Updated on: 06 May 2016, 12:30
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Logic tells us that the average length per film will go down, so we can eliminate all answers that add to the original average of t. That leaves either B or E. When executing the algebra, the key step is being able to quickly and easily turn one fraction with two numerators $$(\frac{21t-14}{21})$$ into two fractions with one numerator each $$(\frac{21t}{21}-\frac{14}{21})$$.

Above is a visual that should help.
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Originally posted by mcelroytutoring on 05 May 2016, 19:28.
Last edited by mcelroytutoring on 06 May 2016, 12:30, edited 6 times in total.
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Re: The average (arithmetic mean) length per film for a group of 21 films  [#permalink]

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15 Dec 2016, 14:05
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Great Solutions above.
Here is my solution =>
Average length = t minutes
Sum(21) = 21t
New sum = 21t-66+52 = 21t-14
Hence new mean = 21t-14 / 21 = t-2/3

Hence B

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Re: The average (arithmetic mean) length per film for a group of 21 films  [#permalink]

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20 Jul 2017, 15:08
1
average = sum of numbers/total number
t = s/21
s = 21t
Now as per question
s =21t -66 +52
s = 21t - 14
New average
t1 = (21t - 14) / 21
=> t - 2/3
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The average (arithmetic mean) length per film for a group of 21 films  [#permalink]

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06 Mar 2018, 09:16
stonecold wrote:
Great Solutions above.
Here is my solution =>
Average length = t minutes
Sum(21) = 21t
New sum = 21t-66+52 = 21t-14
Hence new mean = 21t-14 / 21 = t-2/3

Hence B

stonecold how from this 21t-14 / 21 you got t-2/3 , I understand you divided by 7 , but there are two 21 could you write in detail
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The average (arithmetic mean) length per film for a group of 21 films  [#permalink]

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06 Mar 2018, 11:18
1
dave13 wrote:
stonecold wrote:
Great Solutions above.
Here is my solution =>
Average length = t minutes
Sum(21) = 21t
New sum = 21t-66+52 = 21t-14
Hence new mean = 21t-14 / 21 = t-2/3

Hence B

stonecold how from this 21t-14 / 21 you got t-2/3 , I understand you divided by 7 , but there are two 21 could you write in detail

Hey dave13

$$\frac{21t-14}{21} = \frac{21t}{21} - \frac{14}{21} = t - \frac{2}{3}$$

Hope that helps
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Re: The average (arithmetic mean) length per film for a group of 21 films  [#permalink]

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15 Aug 2018, 16:09
Hi,
We can use deviation technique to solve such type of questions. The answer would be in less than 30 Secs

refer: https://gmatclub.com/forum/a-student-s- ... l#p2113194

So new average would be
t-(14/21)
t-2/3
Hence B.
Re: The average (arithmetic mean) length per film for a group of 21 films &nbs [#permalink] 15 Aug 2018, 16:09
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