# Mathematical field, show that 0·a = 0, (-1)·a = -a, …

Based on the axioms for a mathematical field, the wiki article states that 0·a = 0 and (-1)·a = -a are consequences of the axioms, but doesn't show how they are derived. There was a similar question asked before, but I'm not sure about the accepted answer.

https://en.wikipedia.org/wiki/Field_(mathematics)#Elementary_consequences_of_the_definition

Also, it would seem logical that if a ≠ b, and if c ≠ 0, then c·a ≠ c·b, a uniqueness property that should hold true for a finite field (unordered) that I'm wondering if it can be derived from the axioms (perhaps something like induction?) .

• These are in fact standard easy arguments (your guess in the last sentence is correct). I'm sure someone will step up and provide them. Here's a hint for the first one: consider $(0 + 0) \times a$. I'm curious: what prompts your curiousity? A course you're taking? – Ethan Bolker Apr 22 '17 at 0:12
• @EthanBolker - a question came up at another forum. An "addition" table was given for a finite field with 4 numbers (0,1,2,3), where the "addition" turns out to be exclusive or, and the problem was asking to produce the multiplication table, based on the axioms rather than knowledge of GF(4). That prompted me to wonder how the "consequences" as noted in the wiki article are derived, and what operations are considered acceptable as derivations for field math. – rcgldr Apr 22 '17 at 0:42
• @EthanBolker - I'm aware that algebra works with finite finite fields (I've worked with RS ECC), but never thought about the derivations based on the axioms. For that other forum question, after taking into account 0·a = a·0 = 0 and 1·a = a·1 = a, only 4 products in the multiplication table (2·2, 2·3, 3·2, 3·3) needed to be determined, so this could be done by trial and error, where an axiom would fail if the wrong set of 4 products was chosen. Noting that every number has a multiplicative inverse, meant that two of those products = 1 => 2·3 = 3·2 = 1, which simplified the problem. – rcgldr Apr 22 '17 at 1:06

For $0 \times a$: $$0 \times a = (0+0)\times a = 0\times a + 0 \times a .$$ Now whatever $0 \times a$ is, it has an additive inverse, so you can subtract it from each side of that equation to conclude that $0 \times a = 0$.

For $(-1) \times a$: $$0 \times a = (1 + (-1)) \times a = 1 \times a + (-1) \times a = a + (-1) \times a$$ but $a$ has a unique additive inverse $-a$.

If $c \ne 0$ then it has a multiplicative inverse $d$. Then $$ca = cb \implies ca - cb = c(a-b) = 0 \implies dc(a-b) = 0 \implies a-b = 0 \implies a = b.$$

• Shouldn't the third part somewhere include a - b ≠ 0 or perhaps a - b = e ≠ 0 ? – rcgldr Apr 22 '17 at 0:26
• @rcgldr No. I proved your claim by reasoning with the contrapositive form. If $a \ne b$ then the last entry in my chain of correct implications is false, so the first entry must be false. – Ethan Bolker Apr 22 '17 at 0:30
• OK, thanks for the explanation. I was also considering let a-b = e ≠ 0, then ca - cb = ce ≠ 0, since c ≠ 0 and e ≠ 0. – rcgldr Apr 22 '17 at 0:35

I presume that by "-a" you mean the additive inverse of a so that you want to prove that -1 times a is the additive inverse of a. Again, that follows from the distributive law. (1+ (-1))a= 0a= 0. But, by the distributive law, (1+ (-1)))a= 1a+ (-1)a= a+ (-1)a= 0 also. "a+ (-1)a= 0" is precisely what is meant by "additive inverse of a".

The statement "if a ≠ b, and if c ≠ 0, then c·a ≠ c·b" is most easily proved by proving the "contrapositive". For any statement "if p then q", the contrapositive is the statement "if not q then not p" and it is easy to show in general that a statement is true if and only if its contrapositive is true.

The contrapositive of "if a ≠ b, and if c ≠ 0, then c·a ≠ c·b" is "if ca= cb then a= b". To prove that add the additive inverse of (subtract) cb from both sides: ca- cb= 0. By the distributive law, c(a- b)= 0. Since c is not 0 we must have a- b= 0 so a= b.

• Should the contrapositive be stated as $$\text{If ca=cb the either a=b or c=0}?$$ – Juniven Apr 22 '17 at 0:37

$(-1)a=-a$ means $a+(-1)a=0$ as additive inverses are unique.

$a+(-1)a=1*a+(-1)a$ (existence of multiplicative identity.

$=(1+(-1))a$ (distributive property)

$=0*a$. Remains to show $0*a=0$ for all $a$

$0*a= (0+0)*a$ (Associativity and definition of additive property.

$=0*a+0*a$.

$0a=0a+0a$

$0a+(-0a)=0a+0a+(-0a)$ (existence of additive inverse and acknowledgement that addition is a closed binary opperation)

$0=0a+0=0a$(additive inverse and implied associtivity.)

So $a+(-1)a=0$. So $(-1)a=-a$ (and $a= -(-1 (a))$.)


This document presents the close relationship between Spearman correlation and Pearson correlation. Namely, that Spearman correlation is just a Pearson correlation on $$rank$$ed $$x$$ and $$y$$. It is an appendix to the post “Common statistical tests as linear models”.

TL;DR: Below, I argue that this approximation is good enough when the sample size is 12 or greater and virtually perfect when the sample size is 20 or greater.

# 1 Simple example

Here, I start by creating some mildly correlated data. Note that the distribution of the data does not matter since everything is ranked, so the results are identical for non-normal data.

data = MASS::mvrnorm(50, mu=c(2, 2), Sigma=cbind(c(1, 0.4), c(0.4, 1)))
x = data[,1]
y = data[,2]
plot(rank(y) ~ rank(x))

Now let’s test it
# Three ways of running the same model
spearman = cor.test(x, y, method='spearman')
pearson = cor.test(rank(x), rank(y), method='pearson')  # On ranks
linear = summary(lm(rank(y) ~ rank(x)))  # Linear is identical to pearson; just showing.

# Present the results
print('coefficient rho is exactly identical:')
rbind(spearman=spearman$estimate, ranked_pearson=pearson$estimate,
ranked_linear=linear$coefficients[2]) print('p is approximately identical:') rbind(spearman=spearman$p.value,
ranked_pearson=pearson$p.value, ranked_linear=linear$coefficients[8])
## [1] "coefficient rho is exactly identical:"
##                      rho
## spearman       0.1848259
## ranked_pearson 0.1848259
## ranked_linear  0.1848259
## [1] "p is approximately identical:"
##                     [,1]
## spearman       0.1982118
## ranked_pearson 0.1988046
## ranked_linear  0.1988046

The correlation coefficients are exact. And the p-values are quite close! But maybe we were just lucky? Let’s put it to the test:

# 2 Simulate for different N and r

The code below does the following:

1. Generate some correlated data for all combinations of the correlation coefficient ($$r = 0, 0.5, 0.95$$) and sample sizes 1 $$N = 6, 8, 10, ... 20, 30, 50, 80$$.
2. Compute coefficients and p-values for Spearman correlation test and Pearson correlation on the ranked values, many times for each combination.
3. Calculate the number of errors.
library(tidyverse)

# Settings for data simulation
PERMUTATIONS = 1:200
Ns = c(seq(from=6, to=20, by=2), 30, 50, 80)  # Sample sizes to model
rs = c(0, 0.5, 0.95)  # Correlation coefficients to model

# Begin
D = expand.grid(set=PERMUTATIONS, r=rs, N=Ns) %>%  # Combinations of N and r
mutate(
# Use the parameters to generate correlated data in each row
data = map2(N, r, function(N, r) MASS::mvrnorm(N, mu=c(0, 4), Sigma=cbind(c(1, r), c(r, 1)))),

# Tests
pearson_raw = map(data, ~cor.test(rank(.x[, 1]), rank(.x[, 2]), method='pearson')),
spearman_raw = map(data, ~cor.test(.x[, 1], .x[, 2], method = 'spearman')),

# Tidy it up
pearson = map(pearson_raw, broom::tidy),
spearman = map(spearman_raw, broom::tidy),
) %>%

# Get estimates "out" of the tidied results.
unnest(pearson, spearman, .sep='_') %>%
select(-data, -pearson_raw, -spearman_raw)

# If you want to do a permutation-based Spearman to better handle ties and
# overcome issues on how to compute p (computationally heavy).
# Corresponds very closely to R's spearman p-values
#mutate(p.perm = map(data, ~ coin::pvalue(coin::spearman_test(.x[,1] ~ .x[,2], distribution='approximate')))) %>%
#unnest(p.perm)

head(D)

### 2.0.1 Inspect correlation coefficients

Just as we saw in the simple example above, correlation coefficients (estimate and estimate1) are identical per definition. We can see that by calculating the difference and see that it is always zero:

summary(D$estimate - D$estimate1)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
##

Yup. The minimum difference is zero. The maximum is zero. All of them are identicalNo need to look further into that.

### 2.0.2 Inspet p-values

Before getting started on p-values, note that there are several ways of calculating them for Spearman correlations. This is the relevant section from the documentation of cor.test:

For Spearman’s test, p-values are computed using algorithm AS 89 for n < 1290 and exact = TRUE, otherwise via the asymptotic t approximation. Note that these are ‘exact’ for n < 10, and use an Edgeworth series approximation for larger sample sizes (the cutoff has been changed from the original paper).

You can get beyond all of that by doing permutation-based tests at the cost of computational power (see outcommented code above). However, there is almost virtually perfect correspondence, so I just rely on R’s more computationally efficient p-values.

Below, I plot the difference between Spearman-derived p-values and ranked-Pearson-derived p-values. I have chosen not to plot as a function or r in the name of simplicity because it makes no difference whatsoever, but feel free to try it out yourself.

library(patchwork)

# A straight-up comparison of the p-values
p_relative = ggplot(D, aes(x=spearman_p.value, y=pearson_p.value, color=N)) +
geom_line() +
geom_vline(xintercept=0.05, lty=2) +
geom_hline(yintercept=0.05, lty=2) +

labs(title='Absolute relation', x = 'Spearman p-value', y = 'Pearson p-value') +
#coord_cartesian(xlim=c(0, 0.10), ylim=c(0, 0.11)) +
theme_gray(13) +
guides(color=FALSE)

# Looking at the difference (error) between p-values
p_error_all = ggplot(D, aes(x=spearman_p.value, y=pearson_p.value-spearman_p.value, color=factor(N))) +
geom_line() +
geom_vline(xintercept=0.05, lty=2) +

labs(title='Error', x='Spearman p-value', y='Difference') +
coord_cartesian(ylim=c(-0.02, 0.005)) +
theme_gray(13) +
guides(color=FALSE)

# Same, but zoomed in around p=0.05
p_error_zoom = ggplot(D, aes(x=spearman_p.value, y=pearson_p.value-spearman_p.value, color=factor(N))) +
geom_line() +
geom_vline(xintercept=0.05, lty=2) +

labs(title='Error zoomed', x='Spearman p-value', y='Difference') +
coord_cartesian(ylim=c(-0.02, 0.005), xlim=c(0, 0.10)) +
theme_gray(13)

# Show it. Patchwork is your friend!
p_relative + p_error_all + p_error_zoom

Figure: The jagget lines in the leftmost two panels are due to R computing “exact” p-values for Spearman when N < 10. For samples larger than N>=10, an approximation is used which gives a smoother relationship.

It is clear that larger N gives a closer correspondence. For small p, the Pearson-derived p-values are lower than Spearman, i.e. more liberal. Importantly, when N=12, this effect is at most 0.5% in the “significance-region” which I deem acceptable enough for the present purposes. When N=20, the error is at most 0.2% for the whole curve and there is no difference when N=50 or N=100.

# 3 Conclusion

The correlation coefficients are identical.

The p-value is exact enough when N > 10 and virtualy perfect when N > 20. If you are doing correlation studies on sample sizes lower than this, you have larger inferential problems than those posed by the slight differences between Spearman and Pearson.


# Negative roots when simplifying surds

#### NeedingWD40

##### New member
Something that puzzles me:

The square roots of 100 are + or -10, yes? So why when one simplifies surds are we able to ignore negative roots?

e.g. ⎷300 -⎷48 =⎷100x3 -⎷16x3 = 10⎷3 - 4⎷3 = 6⎷3

what about the negative root of 100, and the negative root of 16?

Thanks in advance for your help.

#### Dr.Peterson

##### Elite Member
Something that puzzles me:

The square roots of 100 are + or -10, yes? So why when one simplifies surds are we able to ignore negative roots?

e.g. ⎷300 -⎷48 =⎷100x3 -⎷16x3 = 10⎷3 - 4⎷3 = 6⎷3

what about the negative root of 100, and the negative root of 16?

Thanks in advance for your help.
When we write the radical sign, it stands for the square root function, which means it has only one value. We call this the principal value, namely the non-negative root. The symbol does not represent both roots, but just one.

So although 100 has two square roots, ⎷100 is just 10. And that is what you are given when you simplify a surd, so you ignore the negative roots.

Have you noticed that if you chose always to keep both roots, an expression like yours would have more and more values? To take a simpler example, ⎷9 -⎷4 would have to be (±3) - (±2), with no requirement that the two signs are correlated, so its values would be 3 - 2 = 1, 3 - -2 = 5, -3 - 2 = -5, and -3 - -2 = -1. More radicals would make it even more confusing! That's why we insist on a square root function.

#### Jomo

##### Elite Member
Something that puzzles me:

The square roots of 100 are + or -10, yes? So why when one simplifies surds are we able to ignore negative roots?

e.g. ⎷300 -⎷48 =⎷100x3 -⎷16x3 = 10⎷3 - 4⎷3 = 6⎷3

what about the negative root of 100, and the negative root of 16?

Thanks in advance for your help.
By definition, $$\displaystyle \sqrt{x}$$ is > 0 providing x > 0. If x<0, then $$\displaystyle \sqrt{x}$$ has no real solution.

#### NeedingWD40

##### New member
When we write the radical sign, it stands for the square root function, which means it has only one value. We call this the principal value, namely the non-negative root. The symbol does not represent both roots, but just one.

So although 100 has two square roots, ⎷100 is just 10. And that is what you are given when you simplify a surd, so you ignore the negative roots.

Have you noticed that if you chose always to keep both roots, an expression like yours would have more and more values? To take a simpler example, ⎷9 -⎷4 would have to be (±3) - (±2), with no requirement that the two signs are correlated, so its values would be 3 - 2 = 1, 3 - -2 = 5, -3 - 2 = -5, and -3 - -2 = -1. More radicals would make it even more confusing! That's why we insist on a square root function.
Ah! That's why the formula for solving a quadratic equation explicitly says + or -⎷(b2-4ac), otherwise only the positive root would be found. Thank you very much, I'm clearer now

#### JeffM

##### Elite Member
Something that puzzles me:

The square roots of 100 are + or -10, yes? So why when one simplifies surds are we able to ignore negative roots?

e.g. ⎷300 -⎷48 =⎷100x3 -⎷16x3 = 10⎷3 - 4⎷3 = 6⎷3

what about the negative root of 100, and the negative root of 16?

Thanks in advance for your help.
You are absolutely correct that

$$\displaystyle (-\ 10)^2 = 100 \text { and } (+\ 10)^2 = 100.$$

But the technical definition of the square root function of a non-negative real number is:

$$\displaystyle \text {If } a \ge 0, \text { then } \sqrt{a} \ge 0 \text { and } \sqrt{a} * \sqrt{a} = a.$$

That is why if we get an equation that looks like

$$\displaystyle x^2 = 100 \implies x = \pm \sqrt{100} \implies x = 10 \text { or } x = -\ 10.$$

That plus or minus sign is there for a reason.

#### NeedingWD40

##### New member
By definition, $$\displaystyle \sqrt{x}$$ is > 0 providing x > 0. If x<0, then $$\displaystyle \sqrt{x}$$ has no real solution.
Thank you, I'm clearer now.

#### NeedingWD40

##### New member
You are absolutely correct that

$$\displaystyle (-\ 10)^2 = 100 \text { and } (+\ 10)^2 = 100.$$

But the technical definition of the square root function of a non-negative real number is:

$$\displaystyle \text {If } a \ge 0, \text { then } \sqrt{a} \ge 0 \text { and } \sqrt{a} * \sqrt{a} = a.$$

That is why if we get an equation that looks like

$$\displaystyle x^2 = 100 \implies x = \pm \sqrt{100} \implies x = 10 \text { or } x = -\ 10.$$

That plus or minus sign is there for a reason.
So, I'm thinking, if a problem includes the square root function, but no plus or minus, you assume the primary root, however, if you create a square root when solving a problem you need to be aware of the negative root in the context of the question....

#### tkhunny

##### Moderator
Staff member
So, I'm thinking, if a problem includes the square root function, but no plus or minus, you assume the primary root, however, if you create a square root when solving a problem you need to be aware of the negative root in the context of the question....
"problem includes ... no plus or minus" - Not a fan.

What is $$\displaystyle \sqrt{4}$$? It is 2

I do not ever recall seeing such a simplification problem as: What is $$\displaystyle \pm\sqrt{4}$$? It is +/- 2.

Let the context be your guide.

Don't make up things that don't exist. $$\displaystyle \sqrt{4} = 2$$ and that is NOT the same as "+/-2".
Don't miss things that do exist. $$\displaystyle Solve\;x^{2} = 4$$. $$\displaystyle x = +2\;or\;x = -2$$ and that is the same as "x = +/-2".

Also, keep your eyes on that "+/-" symbol. It means several different things in different contexts.

#### Otis

##### Senior Member
I'm not sure whether you've yet learned anything about solving algebraic equations or the concept of a function (you posted on the Pre-Algebra board), but here's my comments, anyways.

When you are given a surd, it always represents the positive root (only is exception is √0).

When you are solving an equation, sometimes you need to introduce surds (like the algebraic step of taking the square root of each side of an equation). In this case, you need to consider both the positive and negative root. Otherwise, you might miss solution(s).

As Jeff wrote, we express a positive root as √n, and when we need to express the negative root we write -√n. Cheers :cool:

#### JeffM

##### Elite Member
So, I'm thinking, if a problem includes the square root function, but no plus or minus, you assume the primary root, however, if you create a square root when solving a problem you need to be aware of the negative root in the context of the question....
Please read the other responses to this post because they are good and will give you different perspectives.

$$\displaystyle \sqrt{a} \ge 0$$ by definition if we are dealing with real numbers.

So, we do not assume that the square root is the primary root (meaning a non-negative number), we know that it is a non-negative number by definition.

However, if the square root of a is positive, we know that the square of the square root's additive inverse is also a. In notation,

$$\displaystyle \sqrt{a} * \sqrt{a} = a = (-\ \sqrt{a}) * (-\ \sqrt{a}).$$

So if we have an equation like

$$\displaystyle x^2 = c \ge 0$$,

we have no way to know from the equation itself whether

$$\displaystyle x = \sqrt{c} \text { or } x = - \ \sqrt{c}.$$

It may be that either solution works or that only one of the two works, but that must be decided based on information not contained in the equation itself.

Uncertainty about which answer applies does not mean that there is any uncertainty about what the surd means. To clarify where the uncertainty lies we say

$$\displaystyle x = \pm \sqrt{c}.$$

The sign of the surd is certain. It is the sign of x that is uncertain.

Last edited:

#### pka

##### Elite Member
Something that puzzles me:
The square roots of 100 are + or -10, yes? So why when one simplifies surds are we able to ignore negative roots?
e.g. ⎷300 -⎷48 =⎷100x3 -⎷16x3 = 10⎷3 - 4⎷3 = 6⎷3
what about the negative root of 100, and the negative root of 16?
I have had issues with this my whole active research & teaching life.
I had problems with students in my complex variables classes with the use of the radical: $$\displaystyle \sqrt{z}$$.
Now I belong to a school of analyst who think that the symbol $$\displaystyle \sqrt~$$ must be applied only to a non-negative real number.
So that means $$\displaystyle \sqrt{-1}$$ has no meaning whatsoever. So what is $$\displaystyle \bf{i}~?$$
It is so simple, $$\displaystyle \bf{i}$$ is a solution of the equation of the equation $$\displaystyle z^2+1=0$$
Now clearly the number $$\displaystyle 100$$ has two square roots, $$\displaystyle \pm 10$$.
BUT $$\displaystyle \sqrt{100}=10$$ that is one number. Moreover, $$\displaystyle -\sqrt{100}=-10$$ that is one number.
Thus it is absolutely incorrect to write that $$\displaystyle \sqrt{100}=\pm 10$$

#### JeffM

##### Elite Member
I have had issues with this my whole active research & teaching life.
I had problems with students in my complex variables classes with the use of the radical: $$\displaystyle \sqrt{z}$$.
Now I belong to a school of analyst who think that the symbol $$\displaystyle \sqrt~$$ must be applied only to a non-negative real number.
So that means $$\displaystyle \sqrt{-1}$$ has no meaning whatsoever. So what is $$\displaystyle \bf{i}~?$$
It is so simple, $$\displaystyle \bf{i}$$ is a solution of the equation of the equation $$\displaystyle z^2+1=0$$
Now clearly the number $$\displaystyle 100$$ has two square roots, $$\displaystyle \pm 10$$.
BUT $$\displaystyle \sqrt{100}=10$$ that is one number. Moreover, $$\displaystyle -\sqrt{100}=-10$$ that is one number.
Thus it is absolutely incorrect to write that $$\displaystyle \sqrt{100}=\pm 10$$

I like this convention that the surd can only be applied to non-negative real numbers. It would avoid a lot of confusion.

Unfortunately, conventions are necessarily extra-individual. And the surd is frequently used in a broader sense.

Anyway, I do like your school's approach. But is there not an inconsistency when it comes to

$$\displaystyle \sqrt[3]{-\ 8}$$

or is that prohibited as well. If so, is there a convenient notation to deal with it?

#### pka

##### Elite Member
Anyway, I do like your school's approach. But is there not an inconsistency when it comes to
$$\displaystyle \sqrt[3]{-\ 8}$$
or is that prohibited as well. If so, is there a convenient notation to deal with it?
No, because the any real number has a real root if the index is odd. Example $$\displaystyle \large\sqrt[5]{-32}=-2$$

#### JeffM

##### Elite Member
No, because the any real number has a real root if the index is odd. Example $$\displaystyle \large\sqrt[5]{-32}=-2$$
Ahh so your school's rule is

$$\displaystyle \sqrt[a]{b} \text { is defined only if (1) } a \in \mathbb Z \text {, (2) } a \ge 1 \text {, (3) } \ b \in \mathbb R,$$

$$\displaystyle \text {and (4) } a \text { even } \implies b \ge 0.$$

#### pka

##### Elite Member
Ahh so your school's rule is
$$\displaystyle \sqrt[a]{b} \text { is defined only if (1) } a \in \mathbb Z \text {, (2) } a \ge 1 \text {, (3) } \ b \in \mathbb R,$$
$$\displaystyle \text {and (4) } a \text { even } \implies b \ge 0.$$
.
At meetings we would have new jokes about exponents. Prof Paterson has address this above.
The objection is this $$\displaystyle \sqrt{-16}=\pm 4\bf{i}$$. I dare say you can some such in many basic mathematics textbooks.
It is true that many in my tradition are hyper sensitive to vocabulary abuse.

There are four fourth roots of $$\displaystyle -16\bf{i}$$ but there is just one answer to $$\displaystyle \large\sqrt[4]{-16\bf{i}}=~?$$

#### lookagain

##### Senior Member
NeedingWD40 said:
e.g. ⎷300 -⎷48 =⎷100x3 -⎷16x3 = 10⎷3 - 4⎷3 = 6⎷3
The above is incorrect. You need grouping symbols, such as below. I used asterisks for the products.

e.g.

⎷300 - ⎷48 =

⎷(100*3) - ⎷(16*3) =

10⎷3 - 4⎷3 =

6⎷3

Last edited:

#### topsquark

##### Full Member
.
At meetings we would have new jokes about exponents. Prof Paterson has address this above.
The objection is this $$\displaystyle \sqrt{-16}=\pm 4\bf{i}$$. I dare say you can some such in many basic mathematics textbooks.
It is true that many in my tradition are hyper sensitive to vocabulary abuse.

There are four fourth roots of $$\displaystyle -16\bf{i}$$ but there is just one answer to $$\displaystyle \large\sqrt[4]{-16\bf{i}}=~?$$
When I see stuff like "i"s under a radical I go straight over to the exponential format.
$$\displaystyle \sqrt[4]{-16 i } = \sqrt[4]{16} \cdot \sqrt[4]{-i}$$

$$\displaystyle = 2 \cdot \left ( e^{ 3i \pi/2 + 2 i \pi n } \right ) ^{1/4}$$

$$\displaystyle = 2 \cdot e^{3i \pi /8 + i \pi n / 2}$$

$$\displaystyle = 2 \cdot e^{3i \pi / 8} \cdot i^n$$

$$\displaystyle = 2 (i ^n) \left ( cos ( 3 \pi /8 ) + i~sin (3 \pi / 8) \right )$$
where n is an integer.

Perhaps it's a bit messy but I don't have to think so hard about any negative signs.

-Dan

#### pka

##### Elite Member
When I see stuff like "i"s under a radical I go straight over to the exponential format.
$$\displaystyle \sqrt[4]{-16 i } = \sqrt[4]{16} \cdot \sqrt[4]{-i}$$
-Dan
I did a very poor job explaining our objections. No one would ever need to use $$\displaystyle \sqrt[4]{-16 i }.$$
On the other hand there could be a need to solve $$\displaystyle z^4+16\bf{i}=0$$
There are four roots of that equation. If $$\displaystyle \rho=2\exp\left(\frac{-\pi\bf{i}}{8}\right)$$ then $$\displaystyle \rho^4+16\bf{i}=0$$, in other words one solution.
Let $$\displaystyle \zeta=\exp\left(\frac{2\pi\bf{i}}{4}\right) \$$ now the collection $$\displaystyle \rho\cdot\zeta^k,~k=1,2,3$$ are the other three roots(solutions).

#### NeedingWD40

##### New member
Thanks all, one clear message I've got is I need to clarify my sometimes foggy thought processes!

So, to attempt to summarise:

⎷a is always positive.
The solution to an equation involving a square root may have two solutions:⎷a or -⎷a
The context of the question should clarify if the negative of the root gives a valid solution.

#### JeffM

##### Elite Member
Thanks all, one clear message I've got is I need to clarify my sometimes foggy thought processes!

So, to attempt to summarise:

⎷a is always positive.
The solution to an equation involving a square root may have two solutions:⎷a or -⎷a
The context of the question should clarify if the negative of the root gives a valid solution.
That is very good. I'd make one slight emendation to your third point. I'd phrase it as

The context of the equation should clarify if both solutions are valid or else which one of the two is valid.

I would not assume that the negative solution is the only one that may not apply. In some cases, what is classified as positive and negative is arbitrary. It is of course true that we usually classify in a way that ensures a valid answer is non-negative, but that is not guaranteed.

Very minor point. You seem to have it. Good job.


# Trick to find multiples mentally

We all know how to recognize numbers that are multiple of $2, 3, 4, 5$ (and other). Some other divisors are a bit more difficult to spot. I am thinking about $7$.

A few months ago, I heard a simple and elegant way to find multiples of $7$:

Cut the digits into pairs from the end, multiply the last group by $1$, the previous by $2$, the previous by $4$, then $8$, $16$ and so on. Add all the parts. If the resulting number is multiple of $7$, then the first one was too.

Example:

$21553$

Cut digits into pairs:

$2, 15, 53$

Multiply $53$ by $1, 15$ by $2, 2$ by $4$:

$8, 30, 53$

$8+30+53=91$

As $91$ is a multiple of $7$ ($13 \cdot 7$), then $21553$ is too.

This works because $100-2$ is a multiple of 7. Each hundreds, the last two digits are 2 less than a multiple of $7 (105 → 05 = 7 - 2, 112 → 12 = 14 - 2, \cdots)$

I figured out that if it works like that, maybe it would work if we consider $7=10-3$ and multiplying by $3$ each digit instead of $2$ each pair of digits.

Exemple with $91$:

$91$

$9, 1$

$9\cdot3, 1\cdot1$

$27, 1$

$28$

My question is: can you find a rule that works with any divisor? I can find one with divisors from $1$ to $19$ ($10-9$ to $10+9$), but I have problem with bigger numbers. For example, how can we find multiples of $23$?

-
You'll want to see this KCd blurb. –  Guess who it is. Dec 8 '11 at 15:49
I've seen the following method for 7: Take the last digit, double it, and subtract it from the first digits. So $21553$ leads to $2155-3*2=2149$. Next step is $214-2*9=196$. Next $19-2*6=7$. –  Thomas Andrews Dec 8 '11 at 16:04
How can we find divisors of 23? Very easily: they are just $\pm 1$ and $\pm 23$. Presumably, you want to recognize multiples of 23? –  Arturo Magidin Dec 8 '11 at 16:11
For $23$, here is something fairly simple, for numbers that have $3$ or $4$ digits, say $abcd$. Take the number $ab$, add to it $3$ times the number $cd$. Call the result $t$. Then $abcd$ is a multiple of $23$ iff $t$ is. One could probably tweak this to be even simpler to execute. –  André Nicolas Dec 8 '11 at 16:16
@ArturoMagidin Yes, I meant multiples. It's corrected. –  Oltarus Dec 9 '11 at 8:46

One needn't memorize motley exotic divisibility tests. There is a universal test that is simpler and much easier recalled, viz. evaluate a radix polynomial in nested Horner form, using modular arithmetic. For example, consider evaluating a $3$ digit radix $10$ number modulo $7$. In Horner form $\rm\ d_2\ d_1\ d_0 \$ is $\rm\: (d_2\cdot 10 + d_1)\ 10 + d_0\ \equiv\ (d_2\cdot 3 + d_1)\ 3 + d_0\ (mod\ 7)\$ since $\rm\ 10\equiv 3\ (mod\ 7)\:.\:$ So we compute the remainder $\rm\ (mod\ 7)\$ as follows. Start with the leading digit then repeatedly apply the operation: multiply by $3$ then add the next digit, doing all of the arithmetic $\rm\:(mod\ 7)\:.\:$

For example, let's use this algorithm to reduce $\rm\ 43211\ \:(mod\ 7)\:.\:$ The algorithm consists of repeatedly replacing the first two leading digits $\rm\ d_n\ d_{n-1}\$ by $\rm\ d_n\cdot 3 + d_{n-1}\:\ (mod\ 7),\:$ namely

$\rm\qquad\phantom{\equiv} \color{red}{4\ 3}\ 2\ 1\ 1$

$\rm\qquad\equiv\phantom{4} \color{green}{1\ 2}\ 1\ 1\quad$ by $\rm\quad \color{red}4\cdot 3 + \color{red}3\ \equiv\ \color{green}1$

$\rm\qquad\equiv\phantom{4\ 3} \color{royalblue}{5\ 1}\ 1\quad$ by $\rm\quad \color{green}1\cdot 3 + \color{green}2\ \equiv\ \color{royalblue}5$

$\rm\qquad\equiv\phantom{4\ 3\ 5} \color{brown}{2\ 1}\quad$ by $\rm\quad \color{royalblue}5\cdot 3 + \color{royalblue}1\ \equiv\ \color{brown}2$

$\rm\qquad\equiv\phantom{4\ 3\ 5\ 2} 0\quad$ by $\rm\quad \color{brown}2\cdot 3 + \color{brown}1\ \equiv\ 0$

Hence $\rm\ 43211\equiv 0\:\ (mod\ 7)\:,\:$ indeed $\rm\ 43211 = 7\cdot 6173\:.\:$ Generally the modular arithmetic is simpler if one uses a balanced system of representatives, e.g. $\rm\: \pm\{0,1,2,3\}\ \:(mod\ 7)\:.$ Notice that for modulus $11$ or $9\:$ the above method reduces to the well-known divisibility tests by $11$ or $9\:$ (a.k.a. "casting out nines" for modulus $9\:$).

-

One approach is to find some higher multiple that makes it easy. For your example of $23$, note that $3*23+1=70$, so $a(70-1)$ will be a multiple of $23$. Now you have a single digit multiply followed by a subtraction. If you pick $a=301, a(70-1)=301(70-1)=301*70-301=2107-301=1806$, which is a multiple of $23$

-
I generally find this method much easier than the standard modular divisibility test. It does require that you memorize a starting list of multiples, but it's not so bad because you can just multiply them by powers of $10$. –  Qiaochu Yuan Dec 8 '11 at 18:28
@QiaochuYuan: It seemed the OP's request was to find multiples, not do a divisibility test. But this generalizes. In the spirit of the test for $7$, to test divisibility by $23$, you can take off the last digit, multiply the rest by $7$ and add the last digit. Keep going until you have only two digits left. You just have to find a multiple that ends in $1$ or $9$ to do this. –  Ross Millikan Dec 8 '11 at 18:40

In general, if you are doing things by dividing the number into groups of $k$ digits, you can think of it as looking at the number in base $10^k.$

More generally, if $B$ is a base, and $n$ is a number with no common factors with $B$, then you can always find an $m$ such that $Bm\equiv 1\pmod n$. Then $X\equiv 0\pmod n$ if and only if $Xm\equiv 0\pmod n$. But if $X=Bu+v$, then $Xm\equiv u+mv\pmod n$. So if we take the last digit, base B, multiply it by $m$ and add it to the other digits, the result is divisible by $n$ if and only if the original number was divisible by $n$.

In the case of $n=23$ base $B=10$, you get $m=7$, so you can take the last digit, multiply it by seven, and add it to the rest.

Or, if you use $B=100$, you get $m=3$, and you can take the lsat two digits, multiply by 3, and add to the other digits.

In general, you can always find a $k$ such that $10^k=1\pmod n$ if $n$ is not even or divisible by 5. Then if you take base $B=10^k$, you get $m=1$, and you can separate $m$ into groups of $k$ digits and add them. That's hardly useful when $k$ is large. For example, the smallest $k$ for $n=23$ is $k=22$, so this part only helps if your starting number was more than 23 digits long.

Usually, you want to find a relatively small pair $(m,k)$ so that $m10^k\equiv \pm 1\pmod n$. Then you take the last $k$ digits, multiplied by $m$, and add to or subtract from the other digits, depending on whether $+1$ or $-1$.

-

The following is a simple method to check divisibility by $7$ or $13$:

Cut the digits in pairs of 3 and calculate their alternating sum. If this is a multiple of 7 or 13, the original number is.

For example

$12345631241$ leads to

$$241-639+345-12=-65 \,.$$

Thus our number is a multiple of $13$, but not a multiple of $7$.

This works because $1001=7*11*13$ meaning that any number of the form $abcabc$ is a multiple of $7, 11$ and $13$...The trick also works for 11, but there is another simple trick.

Method 2 If the number is relatively prime to 10 (if it is not, you can make it), look for a multiple on $n$ which ends in 1.

Let say that this multiple is $a_1..a_k1$.

Then you simply pick the large number and subtract $a_1...a_k*$last digit from the remaining digits.

The trick works because $a_1...ak1$*last digit is always a multiple of $n$, and subtracting this from the original number you get a multiple of 10..Since $n$ is relatively prime to 10, you can erase the 0 at the end...

A simple such example, for $7$ the smalest such desires multiple is ... 21, which leads to the criteria posted by Thomas Andrews .

Also, for small numbers the following

-

I will try to explain a general rule using modular congruence. We can show that any integer in base $10$ can be written as $$z = a_0 + a_1 \times 10 + a_2 \times 10^2 + a_3 \times10^3 + \cdots + a_n \times 10^n$$

Lets say we have to find a divisibility rule of $7$,Hence for congruence modulo $7$ have,

$$10 \equiv 3, 10^2 \equiv 2, 10^3 \equiv -1, 10^4 \equiv -3, 10^5 \equiv -2, 10^6 \equiv 1,$$

The successive remainder then repeat. Thus our integer $z$ is divisible by $7$ iff if the remainder expression $$r= a_0 + 3a_1 +2a_2 -a_3-3a_4-2a_5+a_6+3a_7+\cdots$$ is divisible by $7$

To understand why the divisibility of $r$ indicate the divisibility of $z$, find $z-t$ which is given by :$$z-t = a_1 \times (10-3) + a_2 \times (10^2-2) + a_3 \times (10^3+1) + \cdots + a_6 \times (10^6-1) + \cdots$$ Since all this numbers $(10-3),(10^2-2),(10^3+1),\cdots$ are congruent to 0 modulo $7,z-t$ is also, and therefore $z$ leaves the same remainder on division by $7$ as $r$ does.

Using this approach we can derive divisibility of any integer.

-

This can be explained using Horner's method as below. z=a0 + a1 ×10 + a2 ×102 + a3 × 10^3+⋯+ an × 10^n

i.e. Z = ((((...(an × 10 + an-1)*10 + an-2 )*10 + an-3).....)*10 + a0

                           ( Shift-add representation of number)

Z % 7 =  ((((...(an × 3 + an-1)*3 + an-2 )*3 + an-3).....)*3 + a0  (Mod 7)

Now evaluate from left to right using modulo algebra.

To know more about shift- add representation of number and its application in interbase conversion and divisibility refer tinyurl.com/mlxk8pw .

                         Regard,


1. ## Help with reducing quadratic equation to final terms

$x^4 - 6x^2 + 8 = 0
$

$x^2=t$
$x^4=t^2$

a= 1, b = 6, c = 8

$\Rightarrow x = \frac {6 \pm \sqrt {6^2 - 4(1)(8)}}{2(1)}$

$\Rightarrow x = \frac {6 \pm \sqrt {4}}{2}$

$\Rightarrow x = \frac {3 \pm \sqrt {2}}{}$

How do I reduce this further to get the final answer or is my answer correct? I always get confused. Thanks in advance.

2. Originally Posted by lilrhino
$x^4 - 6x^2 + 8 = 0
$

$x^2=t$
$x^4=t^2$

a= 1, b = 6, c = 8

$\Rightarrow x = \frac {6 \pm \sqrt {6^2 - 4(1)(8)}}{2(1)}$

$\Rightarrow x = \frac {6 \pm \sqrt {4}}{2}$

$\Rightarrow x = \frac {3} \pm \frac { \sqrt {2}}{2}$

How do I reduce this further to get the final answer. I always get confused. Thanks in advance.
ok, so this is wrong, since b = -6 not 6. but we don't need the quadratic formula here

you replaced $x^2$ with $t$, so you should have

$t^2 - 6t + 8 = 0$ .........now factor

$\Rightarrow (t - 4)(t - 2) = 0$

$\Rightarrow t = 4 \mbox { or } t = 2$

But $t = x^2$

$\Rightarrow x^2 = 4 \mbox { or } x^2 = 2$

$\Rightarrow x = \pm 2 \mbox { or } x = \pm \sqrt {2}$

3. Hello, lilrhino!

You're making it unnecessarily complicated . . . and wrong.

$x^4 - 6x^2 + 8 \:= \:0$
Factor: . $(x^2 - 4)(x^2 - 2)\:=\:0$

And we have two equations to solve:

. . $x^2 - 4\:=\:0\quad\Rightarrow\quad x^2 \:=\:4\quad\Rightarrow\quad x \:=\: \pm2$

. . $x^2 - 2\:=\:0\quad\Rightarrow\quad x^2\:=\:2\quad\Rightarrow\quad x\:=\:\pm\sqrt{2}$

4. Originally Posted by Jhevon
ok, so this is wrong, since b = -6 not 6. but we don't need the quadratic formula here

you replaced $x^2$ with $t$, so you should have

$t^2 - 6t + 8 = 0$ .........now factor

$\Rightarrow (t - 4)(t - 2) = 0$

$\Rightarrow t = 4 \mbox { or } t = 2$

But $t = x^2$

$\Rightarrow x^2 = 4 \mbox { or } x^2 = 2$

$\Rightarrow x = \pm 2 \mbox { or } x = \pm \sqrt {2}$
Thanks Jhevon, I have to watch my signs. I was looking at an example in the book, so I thought I had to use the quadratic formula even though it could be factored as you demonstrated. Thanks again!

5. Originally Posted by lilrhino
Thanks Jhevon, I have to watch my signs. I was looking at an example in the book, so I thought I had to use the quadratic formula even though it could be factored as you demonstrated. Thanks again!
yeah, the quadratic formula can be overkill at times. However, i do agree with Soroban that we are making it more complicated than it should be. Generally it helps students to visualize what to do by replacing a squared term with a single variable, but i believe that's wasting writing space. ordinarily, i'd just factor as Soroban did and cut out the middle man, but i didn't really want to shock you. you'd be surprised at some of the things students are shocked with

6. Originally Posted by Soroban
Hello, lilrhino!

You're making it unnecessarily complicated . . . and wrong.

Factor: . $(x^2 - 4)(x^2 - 2)\:=\:0$

And we have two equations to solve:

. . $x^2 - 4\:=\:0\quad\Rightarrow\quad x^2 \:=\:4\quad\Rightarrow\quad x \:=\: \pm2$

. . $x^2 - 2\:=\:0\quad\Rightarrow\quad x^2\:=\:2\quad\Rightarrow\quad x\:=\:\pm\sqrt{2}$
I do that often unfortunately. Math is not my strongest subject, so I sometimes make things more complicated that necessary. Thanks for your response.

7. Originally Posted by Jhevon
yeah, the quadratic formula can be overkill at times. However, i do agree with Soroban that we are making it more complicated than it should be. Generally it helps students to visualize what to do by replacing a squared term with a single variable, but i believe that's wasting writing space. ordinarily, i'd just factor as Soroban did and cut out the middle man, but i didn't really want to shock you. you'd be surprised at some of the things students are shocked with
I'm not easily shocked, I may spend more time thinking about it though . It's easier the way Soroban demonstrated it actually, so I'm not shocked. Thanks...


# Tiling a square with rectangles

Consider the set of all the rectangles with dimensions $$2^a\times 2^b\,a,b\in \mathbb{Z}^{\ge 0}$$. We want to tile an $$n\times n$$ square by rectangles from this set (you can use a rectangle several times). What is the minimum number of rectangles we need?

If $$f(n)$$ is the sum of digits of $$n$$ in base $$2$$, I think we need at most $$f(n)^2$$ rectangles. I have an example for this number: write $$n=2^{a_1}+2^{a_2}+...2^{a_{f(n)}}$$ and split each side to segments with length $$2^{a_1},2^{a_2},...,2^{a_{f(n)}}$$ and consider $$f(n)^2$$ rectangles obtained this way.

On the other hand, you need at least $$f(n)$$ rectangles to tile a raw (or column) so I think you need $$f(n)^2$$ rectangles, but I can't prove it.

Any ideas?

• By $f(n)$ do you mean the sum of the bits in the binary representation of $n$? Apr 3, 2019 at 20:18
• @JohnWaylandBales yes f(n) is the least number such that $n=2^{a_1}+2^{a_2}+...2^{f(n)}$
– ali
Apr 3, 2019 at 20:23
• You mean $f(n)$ is the least number such that $n = 2^{a_1} + 2^{a_2} + \cdots + 2^{a_{f(n)}}$ right? Because $f(n)$ counts the number of terms, but it is not the highest exponent. Apr 4, 2019 at 3:51
• @antkam yes i edited the question
– ali
Apr 4, 2019 at 4:36

MAJOR UPDATE:

What I am about to show is not a proof for the minimum number of rectangles. However, in some cases I found the number of rectangles can be less than $$f(n)^2$$. The smallest $$N×N$$ grid that I have found that can have less than $$f(n)^2$$ rectangles is $$15×15$$, which is displayed below: $$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline 1&1&1&1&2&2&3&4&4&4&4&4&4&4&4\\ \hline 1&1&1&1&2&2&3&4&4&4&4&4&4&4&4\\ \hline 1&1&1&1&2&2&3&4&4&4&4&4&4&4&4\\ \hline 1&1&1&1&2&2&3&4&4&4&4&4&4&4&4\\ \hline 1&1&1&1&2&2&3&5&5&5&5&5&5&5&5\\ \hline 1&1&1&1&2&2&3&5&5&5&5&5&5&5&5\\ \hline 1&1&1&1&2&2&3&6&6&6&6&6&6&6&6\\ \hline 1&1&1&1&2&2&3&7&8&9&9&10&10&10&10\\ \hline 11&11&11&11&11&11&11&11&8&9&9&10&10&10&10\\ \hline 12&12&12&12&12&12&12&12&8&9&9&10&10&10&10\\ \hline 12&12&12&12&12&12&12&12&8&9&9&10&10&10&10\\ \hline 13&13&13&13&13&13&13&13&8&9&9&10&10&10&10\\ \hline 13&13&13&13&13&13&13&13&8&9&9&10&10&10&10\\ \hline 13&13&13&13&13&13&13&13&8&9&9&10&10&10&10\\ \hline 13&13&13&13&13&13&13&13&8&9&9&10&10&10&10\\ \hline \end{array}$$

The method used in the above $$15×15$$ square can be generalized not just to other squares but to rectangles as well. In order to make full use of this method, I will expand the op's method to rectangles. Let the length of a rectangle be equal to $$m$$ units and the width be $$n$$ units. Then the number of base-2 rectangles used to cover a $$m × n$$ rectangle by the op's method is $$f(m)f(n)$$. My method isn't fundamentally different from the op's method. It splits the $$m×n$$ rectangle into five sub-rectangles, then the op's method is applied to each of the five rectangles. In some cases the number of base-2 rectangles that covers the five sub-rectangles is less than the number of base-2 rectangles that cover original $$m$$×$$n$$ rectangle using the op's method. The five rectangles are arranged so that their are two pairs of rectangles that occupy the corners and one rectangle that is in the middle (not touching the perimeter). Each pair of rectangles are the same size and orientation but in opposite corners. (Top left and bottom right or Top right and bottom left.) The length and width of the five rectangles are constructed from two other unit lengths $$a$$ and $$b$$. $$a$$ is the smallest number such that $$m+a$$ is a power of two. $$b$$ is the smallest number such that $$n+b$$ is a power of two. This means that $$f(m+a)$$ and $$f(n+b)$$ are each one. The length and width of the two rectangles in the first pair are $$f\left(\frac{m+a}{2}\right)$$ and $$f\left(\frac{n-b}{2}\right)$$ respectively. The length and width of the two rectangles in the second pair are $$f\left(\frac{m-a}{2}\right)$$ and $$f\left(\frac{n+b}{2}\right)$$ respectively. The formula for the total number of base-2 rectangle used is $$2f\left(\frac{m+a}{2}\right) f\left(\frac{n-b}{2}\right)+2f\left(\frac{m-a}{2}\right) f\left(\frac{n+b}{2}\right)+f(a)f(b)$$.

Note that if a square with a length of $$n$$ units is of the form $$2^xy$$ where $$x,y\in\Bbb{N}|x\ge 1,y\ge 1$$ and $$y$$ is odd. Then the number of base-2 rectangles used for both the op's method and my method are the the same as the number of base-2 rectangles used for a square of length $$y$$ because each of the dimensions of the sub-rectangles can be multiplied by $$2^x$$. For example if we want to determine how many base-2 rectangles is rectangles are required to cover a $$30×30$$ square using my method. We just use the $$15×15$$ example near the top of this post and multiply the length and width of each base-2 rectangle by $$2$$. So this means the $$30×30$$ square requires the same number of base-2 rectangles as the $$15×15$$ square. So the problem can be simplified to just rectangles where $$m$$ and $$n$$ are odd.

A simple inequality can be made which would indicate which method uses less base-2 rectangles. Let $$N_l$$ be the number of ones in the number for length of the rectangle in binary and $$N_w$$ be the number of ones in the width in binary $$\bigl($$or more simply $$N_l=f(m)$$ and $$N_w=f(n)\bigr)$$. Also Let $$Z_l$$ be the number of zeros in the number for length of the rectangle in binary, $$Z_w$$ be the number of zeros in the width in binary. My method uses less rectangles than the op when $$2f\left(\frac{m+a}{2}\right) f\left(\frac{n-b}{2}\right)+2f\left(\frac{m-a}{2}\right) f\left(\frac{n+b}{2}\right)+f(a)f(b)\lt f(m)f(n)$$

$$f\left(\frac{m+a}{2}\right)=1$$ $$f\left(\frac{n+b}{2}\right)=1$$ $$f\left(\frac{m-a}{2}\right)=N_l-1$$ $$f\left(\frac{n-b}{2}\right)=N_w-1$$ $$f(m)=N_l$$ $$f(n)=N_w$$ $$f(a)=Z_l+1$$ $$f(b)=Z_w+1$$

With the above substitutions the inequality can be changed to:

$$2(N_l-1)+2(N_w-1)+(Z_l+1)(Z_w+1)\lt N_lN_w$$ $$2N_l+2N_w-4+(Z_l+1)(Z_w+1)\lt N_lN_w$$ $$(Z_l+1)(Z_w+1)\lt N_lN_w-2N_l-2N_w+4$$ $$(Z_l+1)(Z_w+1)\lt (N_l-2)(N_w-2)$$

In the specific case of the square (where the length equals the width) my method uses less base-2 rectangles than the op when the number ones in the binary representation of the length is at least four more than than the number of zeros. To get the maximum utility out of my method the inequality shouldn't only be applied to the entire length and width of the main square it should also be applied to components of the square. For example consider the square $$1927×1927$$. The binary representation of 1927 is 11110000111. There are three more ones than zeros in this number so my method would normally break even with the op, covering the square with 49 base-2 rectangles. There is a way to cover the square using less base-2 rectangles by spliting the square into four rectangles $$1920×1920$$, $$1920×7$$, $$7×1920$$, and $$7×7$$. Splitting this way doesn't change the net result of the op's method. The first three sub rectangles satisfies the inequality. If I use my method on the first three sub rectangles I use 13, 11, and 11 base-2 rectangles respectively. Using the op's method on the last sub rectangle then counting up all of the base-2 rectangles I can cover the $$1927×1927$$ square using 44 base-2 rectangles.(13+11+11+9) So if a combination of sub-strings in the binary value of the length and width satisfies the inequality like it did three times with the sub rectangles then my method will use less base-2 rectangles than the op's method. Finding the minimum number of base-2 rectangles for some squares will inevtably involve searching for the best way to split the square.

For large enough squares the worst digit combination where my method does no better than the op is a block of three ones and the rest are alternating zeros and ones. For example the square $$\require{enclose}\enclose{horizontalstrike}{343×343}$$, its binary representation is 101010111. This square requires 36 base-2 rectangles and is tied for most number of required base-2 rectangles amoung the nine digit squares. This means that a upper bound can be made for the minimum number of rectangles required. Let $$\enclose{horizontalstrike}{d_l}$$ be the number of digits in the binary representation of the length of the rectangle. ($$\enclose{horizontalstrike}{d_l=N_l+Z_l}$$) Let $$\enclose{horizontalstrike}{d_w}$$ be the number of digits in the binary representation of the width of the rectangle. ($$\enclose{horizontalstrike}{d_w=N_w+Z_w}$$) Then the upper bound is: $$\enclose{horizontalstrike}{\left(\left\lceil\frac{d_l}{2}\right\rceil+1\right)\left(\left\lceil\frac{d_w}{2}\right\rceil+1\right)}$$

I conjecture that the combination of my method and the op's method is the optimal way of minimizing the number of base-2 rectangles. The only way that someone might use be able to use less rectangles is to find a another way of spliting the square into sub-rectangles such that using the op's method on those sub-rectangles uses less base-2 rectangles than using my method and the op's method on the whole square.

Rob Pratt's(RP's) post shows that there is a third method for covering the $$n×n$$ square with less base-2 rectangles than my method or the op's method for some $$n×n$$ squares. In order to describe how many rectangles RP's method uses I will continue to use the the term $$b$$ from my method (where $$b$$ is the smallest number such that $$b+n$$ is a power of 2). I will also need a new sets of terms $$c_k$$ and $$s_k$$ where $$k\in\Bbb{N}|1\le k\le f(b)$$. $$c_1$$ is the value of left most ones digit of b in binary form. $$c_2$$ is the value of the second ones digit from the left of b in binary form. $$c_3$$ is the value of the third ones digit from the left of b in binary form. Etc. $$s_v=\sum_{j=1}^vc_v$$. For example if $$n=23$$ then $$b=9$$, $$c_1=8$$, $$c_2=1$$, $$s_1=8$$, $$s_2=9$$. RP's method uses $$2f\left(\frac{n+b}{2}\right)f\left(\frac{n-b}{2}\right)+f\left(\frac{n+b}{2}\right)f\left(\frac{n-b}{2}+s_k\right)+f\left(\frac{n-b}{2}\right)f\left(\frac{n+b}{2}-s_k\right)+f(b)f(b-s_k)$$

base-2 rectangles. Each $$f(•)f(•)$$ product contains the length and width of each of the sub-rectangles that covers the square inside the f function. RP's method has $$k$$ ways of covering the $$n×n$$ square one for each $$s$$ element. Obviously the particular $$s_k$$ element that uses the least number of base-2 rectangles according to the above formula is the one that is used for the minimum. A sufficient condition for when RP's method uses less base-2 rectangles than both my method and the op's method when the binary representation of $$n$$ has at least three more ones than zeros, the second digit to the left is a zero, and the spliting method that was mentioned for the $$1927×1927$$ square doesn't apply.

• absolutely brilliant!! IMHO well worth the bounty. :) Apr 11, 2019 at 2:02
• What does your method obtain for $n\in\{23,30,31\}$? Mar 5, 2020 at 22:36
• @Rob_Pratt 16,13, and 17 base-2 rectangles respectively Mar 5, 2020 at 23:30
• @RobPratt I realized that the way I explained it in my edited post it doesn't show how n=30 is 13 base-2 rectangles with my method. I will edit accordingly. Mar 5, 2020 at 23:37

For fixed $$n$$, you can solve this problem via integer linear programming as follows. Let $$R$$ be the set of rectangles. For $$i\in\{1,\dots,n\}, j\in\{1,\dots,n\}$$, let $$R_{i,j}\subset R$$ be the subset of rectangles that contain cell $$(i,j)$$. Let binary decision variable $$x_r$$ indicate whether rectangle $$r\in R$$ is used. The problem is to minimize $$\sum_r x_r$$ subject to: \begin{align} \sum_{r \in R_{i,j}} x_r &= 1 &&\text{for i\in\{1,\dots,n\}, j\in\{1,\dots,n\}} \\ x_r &\in \{0,1\} &&\text{for r \in R} \end{align}

Here are several optimal values that differ from $$f(n)^2$$: $$\begin{matrix} n &15 &23 &30 &31 &46 &47 &55 &59 &60 &61 &62 &63\\ \hline f(n)^2 &16 &16 &16 &25 &16 &25 &25 &25 &16 &25 &25 &36\\ \text{optimal} &13 &15 &13 &17 &15 &19 &20 &20 &13 &20 &17 &21\\ \end{matrix}$$

For example, here is an optimal solution for $$n=23$$ with $$15$$ rectangles (shown here with hex values $$0$$ through $$E$$ for compactness): $$\begin{matrix} 0&0&0&0&0&0&0&0&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ 5&5&5&5&5&5&5&5&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ 5&5&5&5&5&5&5&5&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ 5&5&5&5&5&5&5&5&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ 5&5&5&5&5&5&5&5&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ 3&3&3&3&3&3&3&3&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ 3&3&3&3&3&3&3&3&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ C&E&E&E&E&D&D&7&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ C&E&E&E&E&D&D&7&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ C&E&E&E&E&D&D&7&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ C&E&E&E&E&D&D&7&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ C&E&E&E&E&D&D&7&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ C&E&E&E&E&D&D&7&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ C&E&E&E&E&D&D&7&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ C&E&E&E&E&D&D&7&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ C&E&E&E&E&D&D&1&A&A&A&A&8&B&B&B&B&B&B&B&B&9&9\\ C&E&E&E&E&D&D&6&6&6&6&6&6&6&6&6&6&6&6&6&6&6&6\\ C&E&E&E&E&D&D&6&6&6&6&6&6&6&6&6&6&6&6&6&6&6&6\\ C&E&E&E&E&D&D&6&6&6&6&6&6&6&6&6&6&6&6&6&6&6&6\\ C&E&E&E&E&D&D&6&6&6&6&6&6&6&6&6&6&6&6&6&6&6&6\\ C&E&E&E&E&D&D&4&4&4&4&4&4&4&4&4&4&4&4&4&4&4&4\\ C&E&E&E&E&D&D&4&4&4&4&4&4&4&4&4&4&4&4&4&4&4&4\\ C&E&E&E&E&D&D&2&2&2&2&2&2&2&2&2&2&2&2&2&2&2&2\\ \end{matrix}$$

• I should have mentioned this earlier but good job finding this. As of when this comment being posted you are the only one who has helped me with this problem. If no one else posts am answer by the end of the bounty grace period you will receive the bounty. Also I have made a formula for your method in my most recent edit that I just made, you might want to take a look. Mar 8, 2020 at 23:13
• Thanks. I added a few more values $< f(n)^2$ just now. Mar 9, 2020 at 23:35

NOTE:This doesn't work, the induction hypothesis is too strong (and false).

Lets first consider a more general question, where we tile a rectangle $$R$$ by smaller rectangles, where all vertices are points in an (ambient) integer lattice.

We have a row of rectangles $$T_i$$ touching the bottom edge of $$R$$, and each of these has a top edge $$e_i$$. For each $$T_i$$ we define the number $$\lambda(T_i)$$ to be the minimal number of our tiling rectangles that intersect any column starting in $$T_i$$.

Our first claim is that for the total number of rectangles in $$R$$, denoted $$r(R)$$, we have $$\sum_i \lambda(T_i) \leq r(R)$$

Lets prove this by induction on the height of the rectangle $$R$$ (drawing a picture may help see whats happening). First, if the height is $$1$$, then we are done trivially.

So now for the inductive step, let $$R_0$$ have height $$n$$, and consider the edges $$e_i$$ that have minimal height, and define $$a$$ to be this height. This is to say, they border the $$a$$th row, if the first row is the bottom row of $$R_0$$. Say that we have $$k$$ minimal edges $$e_i$$ bordering this row.

We now consider the new rectangle $$R_0'$$ we obtain by chopping off the first $$a$$ rows of $$R$$. On one hand, this has strictly smaller height, so we have, by induction and our definition of $$k$$: $$\sum_i \lambda(T_i') \leq r(R_0)-k$$

But each rectangle on the bottom row of $$R_0'$$ is either one of the rectangles of $$R_0$$, chopped, but not removed, or a rectangle of $$R_0$$ lying above one of our minimal edges $$e_i$$.

Now note if our original $$T_i$$ is chopped but not removed, $$\lambda(T_i)=\lambda(T_i')$$, and if our original $$T_j$$ is removed (so top edge has minimal height), then $$\lambda(T_j')=\lambda(T_j)-1$$, where $$T_j'$$ is any of the rectangles lying directly over $$T_j$$. Thus, adding $$k$$ to both sides of our previous equality, we have:

$$\sum_i \lambda(T_i) \leq \sum_i \lambda(T_i')+k\leq r(R_0)$$

So we are done by induction.

So for your case, note that each column must have at least $$f(n)$$ rectangles in it, and note the bottom row has at least $$f(n)$$ rectangles. This follows since $$f(n)$$ is the minimal number of powers of two needed to express $$n$$. Thus, $$f(n)^2\leq r(R)$$ in your case.

• I think your original claim is false.I don't know how to send a picture in comment but you can easily draw $3\times 4$ counter examples(two horizontal dominos and two vertical dominos in first two rows and two $1\times1$square and a domino on the third row).the problem is on your induction step the rectangles above two removed rectangle may not be distinict.
– ali
Apr 5, 2019 at 9:41
• your last statement have counter example too.if each row intersect k rectangle and each column intersect k rectangle doesn't mean we need $k^2$ rectangle.
– ali
Apr 5, 2019 at 9:43
• True, I'll leave this up in case someone can make this approach work. Apr 5, 2019 at 10:03


# Finding a tricky composition of two piecewise functions

I have a question about finding the formula for a composition of two piecewise functions. The functions are defined as follows:

$$f(x) = \begin{cases} 2x+1, & \text{if x \le 0} \\ x^2, & \text{if x > 0} \end{cases}$$

$$g(x) = \begin{cases} -x, & \text{if x < 2} \\ 5, & \text{if x \ge 2} \end{cases}$$

My main question lies in how to approach finding the formula for the composition $g(f(x))$. I have seen a couple of other examples of questions like this online, but the domains of each piecewise function were the same, so the compositions weren't difficult to determine.

In this case, I have assumed that, in finding $g(f(x))$, one must consider only the domain of $f(x)$. Thus, I think it would make sense to test for individual cases: for example, I would try to find $g(f(x))$ when $x <= 0$. $g(f(x))$ when $x <= 0$ would thus be $-2x-1$, right? However, I feel like I'm missing something critical, because I'm just assuming that the condition $x < 2$ for $g(x)$ can just be treated as $x <= 0$ in this case. Sorry for my rambling, and many thanks for anyone who can help lead me to the solution.

The first branch point is associated with $f$, and happens at $x=0$. We note that: $$g\circ f(x) = \begin{cases} g(2x+1), & \text{if x≤ 0} \\ g(x^2), & \text{if x > 0} \end{cases}$$

Let's just consider each case separately. First, $g(2x+1)$ when $x≤0$. We remark that $x≤0\implies 2x≤0\implies 2x+1≤1$ so we see that $x≤0\implies g(2x+1)=-2x-1$.

Now consider $g(x^2)$ when $x>0$. We see that we hit a branch point at $x=\sqrt 2$. Specifically, $x<\sqrt 2\implies x^2<2$ so $x<\sqrt 2\implies g(x^2)=-x^2$. Of course $x≥\sqrt 2 \implies x^2≥2$ so $x≥\sqrt 2\implies g(x^2)= 5$. Combining all this we see that $$g\circ f(x) = \begin{cases} -2x-1, & \text{if x ≤ 0} \\ -x^2, & \text{if 0< x < \sqrt 2}\\ 5 & \text{if x≥\sqrt 2} \end{cases}$$

As a sanity check, we try the special case $x=1$. We note that $g\circ f(1)=g(f(1))=g(1)=-1$ as desired. I advise picking a few other special values just to check.

• Thanks for your answer. You say that "[w]e see that we hit a branch point at $x = \sqrt 2$," but how do we see this? I don't know how one would conclude this from any of the information given. – Linear Pedant Jul 22 '16 at 22:19
• @LinearPedant The branch point happens when the argument of $g$ reaches $2$. Because the argument is $x^2$, the branch point is at $x=\sqrt2$. – f'' Jul 22 '16 at 23:41
• @LinearPedant Did the reply from f'' help? It is exactly right. We know that $g(x)$ has a branch point at $x=2$, thus $g(x^2)$ has a branch point at $x^2=2$, a.k.a. $x=\sqrt 2$ (since we already know $x>0$ so we can ignore the negative square root). – lulu Jul 23 '16 at 1:45
• @lulu @f" Yes, thank you. This reply greatly helped me understand the problem better. – Linear Pedant Jul 23 '16 at 1:52

$$g(x) = \begin{cases} -x, & x < 2 \\ 5, & x \ge 2 \end{cases}$$

Therefore

$$g(f(x)) = \begin{cases} -f(x), & f(x) < 2 \\ 5, & f(x) \ge 2 \end{cases}$$

So now we need to know when $f(x) < 2$ and when $f(x) \ge 2$.

$$f(x) = \begin{cases} 2x + 1, & x \le 0 \\ x^2, & x > 0 \end{cases}$$

Let's look at one piece of $f$ at a time. On the first piece, $2x + 1 \ge 2$ means $x \ge 1/2$. But this is impossible because $x \le 0$ on the first piece. Also on the first piece, $2x + 1 < 2$ means $x < 1/2$. Well, on the first piece we have $x \le 0 < 1/2$, therefore $f(x) < 2$ on the entire first piece, i.e., $f(x) < 2$ if $x \le 0$.

On the second piece, $x^2 \ge 2$ means $x \ge \sqrt{2}$ or $x \le -\sqrt{2}$. On the second piece we always have $x > 0$, therefore we have $x^2 \ge 2$ when $x \ge \sqrt{2}$. Also on the second piece, $x^2 < 2$ means $-\sqrt{2} < x < \sqrt{2}$. And since on the second piece we always have $x > 0$, then we must have $0 < x < \sqrt{2}$ in order to have $x^2 < 2$.

Putting it all together so far, we have the following:

$$f(x) \ge 2 \text{ if and only if } x \ge \sqrt{2}$$

$$f(x) < 2 \text{ if and only if } x \le 0 \text{ or } 0 < x < \sqrt{2}$$

Notice that this last one can be simplified but we need to keep them separate. Why is this? We'll see as we continue. Recall:

$$g(f(x)) = \begin{cases} -f(x), & f(x) < 2 \\ 5, & f(x) \ge 2 \end{cases}$$

Therefore:

$$g(f(x)) = \begin{cases} -f(x), & x \le 0 \text{ or } 0 < x < \sqrt{2} \\ 5, & x \ge \sqrt{2} \end{cases}$$

Separating the conditions gives us:

$$g(f(x)) = \begin{cases} -f(x), & x \le 0\\ -f(x), & 0 < x < \sqrt{2} \\ 5, & x \ge \sqrt{2} \end{cases}$$

And we need to do this because $f(x)$ itself is different for $x \le 0$ and $0 < x < \sqrt{2}$. Finally, we end up with:

$$h(x) := g(f(x)) = \begin{cases} -(2x+1), & x \le 0 \\ -x^2, & 0 < x < \sqrt{2} \\ 5, & x \ge \sqrt{2} \end{cases}$$

• Ack, someone beat me to it. I guess that's what happens when I let myself get distracted. – tilper Jul 22 '16 at 15:59

You're correct about the value of $g(f(x))$ when $x\leq 0$; since $f(x)$ will be at most $2\cdot0+1=1$, $g$ is only going to evaluate $f(x)$ according to the definition for $x<2$. Testing for cases here is a good approach, and you've just resolved the $x\leq0$ case. When $x>0$, consider the values of $f(x)$: when will they be less than $2$, and when will they be greater? This will determine where $g(f(x))$ takes on its values.

$$g(x) = \begin{cases} -x, & \text{if x < 2} \\ 5, & \text{if x \ge 2} \end{cases}$$ In finding $g(f(x))$ with $g$ defined as above, one should remember that $$g(f(x)) = \begin{cases} -x, & \text{if f(x) < 2} \\ 5, & \text{if f(x) \ge 2} \end{cases}$$ with $f(x)$, not $x$, being the thing about which one asks when it is $<2$ and when it is $\ge 2$.

(The calculus text by Salas and Hille had a problem like this that used to get assigned in every academic term by vast numbers of professors who taught the course to thousands of students. And every time an hour would be spent answering the students questions about it, which in every case said something like "I didn't even know where to start", and that's an hour that could have been spent on something that would help the students understand calculus. It was an even worse waste of time and effort than what usually goes on in the kind of math courses where the students don't want to learn math.)


# Can every proof by contradiction also be shown without contradiction?

Are there some proofs that can only be shown by contradiction or can everything that can be shown by contradiction also be shown without contradiction? What are the advantages/disadvantages of proving by contradiction?

As an aside, how is proving by contradiction viewed in general by ‘advanced’ mathematicians. Is it a bit of an ‘easy way out’ when it comes to trying to show something or is it perfectly fine? I ask because one of our tutors said something to that effect and said that he isn’t fond of proof by contradiction.

## Answer

To determine what can and cannot be proved by contradiction, we have to formalize a notion of proof. As a piece of notation, we let $\bot$ represent an identically false proposition. Then $\lnot A$, the negation of $A$, is equivalent to $A \to \bot$, and we take the latter to be the definition of the former in terms of $\bot$.

There are two key logical principles that express different parts of what we call “proof by contradiction”:

1. The principle of explosion: for any statement $A$, we can take “$\bot$ implies $A$” as an axiom. This is also called ex falso quodlibet.

2. The law of the excluded middle: for any statement $A$, we can take “$A$ or $\lnot A$” as an axiom.

In proof theory, there are three well known systems:

• Minimal logic has neither of the two principles above, but it has basic proof rules for manipulating logical connectives (other than negation) and quantifiers. This system corresponds most closely to “direct proof”, because it does not let us leverage a negation for any purpose.

• Intuitionistic logic includes minimal logic and the principle of explosion

• Classical logic includes intuitionistic logic and the law of the excluded middle

It is known that there are statements that are provable in intuitionistic logic but not in minimal logic, and there are statements that are provable in classical logic that are not provable in intuitionistic logic. In this sense, the principle of explosion allows us to prove things that would not be provable without it, and the law of the excluded middle allows us to prove things we could not prove even with the principle of explosion. So there are statements that are provable by contradiction that are not provable directly.

The scheme “If $A$ implies a contradiction, then $\lnot A$ must hold” is true even in intuitionistic logic, because $\lnot A$ is just an abbreviation for $A \to \bot$, and so that scheme just says “if $A \to \bot$ then $A \to \bot$”. But in intuitionistic logic, if we prove $\lnot A \to \bot$, this only shows that $\lnot \lnot A$ holds. The extra strength in classical logic is that the law of the excluded middle shows that $\lnot \lnot A$ implies $A$, which means that in classical logic if we can prove $\lnot A$ implies a contradiction then we know that $A$ holds. In other words: even in intuitionistic logic, if a statement implies a contradiction then the negation of the statement is true, but in classical logic we also have that if the negation of a statement implies a contradiction then the original statement is true, and the latter is not provable in intuitionistic logic, and in particular is not provable directly.

Attribution
Source : Link , Question Author : sonicboom , Answer Author : Carl Mummert


# Understanding and writing limit proofs

I got this question :

Consider $a_n,\:b_{n\:}$ sequences such that for every n , $0\le a_n\le \:b_{n\:}$.

Let $\lim _{n\to \infty }\left(\frac{b_n}{a_n}\right)\:=\:1$, and $a_n$ is a bounded sequence.

Prove that $\left(a_n-b_n\right)_{n=1}^{\infty \:}\:\rightarrow \:0$.

I tried little bit by myself to understand what is given:

By definition of limit, we see that : $\forall\epsilon, \exists n_{0} \forall n> n_{0}\Rightarrow |\frac{b_{n}}{a_{n}}- 1|< \varepsilon$, and also that exist some $M$ that for every $n$, $-M<a_n<M$.

Now i'm looking for that right?: i need to prove that $\forall\epsilon, \exists n_{0} \forall n> n_{0}\Rightarrow |a_{n}-b_{n}- 0|< \varepsilon$

Here is where i struggle: i choose some $\epsilon$. What i need to find? an $N2$ that for every $n>N2, |(a_{n}-b_{n})- 0|< \varepsilon$ ?

How to start?

Hint: use may the fact that $|a_n-b_n|=|a_n||\frac{b_n}{a_n}-1|$ in your proof .

• Yes, but what i actually need to write to do it formal? – user2637293 Sep 18 '14 at 1:09
• I would recommend you to fill in the gap yourself. but if not interested you can look at the proof provided by @Ishfaaq down here. – BigM Sep 18 '14 at 1:12

Hints

First of all notice that $\left|{\dfrac{b_n}{a_n} - 1}\right| = \dfrac{|(b_n - a_n) - 0|}{|a_n|}$

Secondly note that $|a_n| \le M \;\; \forall n \in \Bbb N$

Now let $\epsilon \gt 0$ be arbitrary and think what you can do with a quantity like $$\dfrac{\epsilon}{M} \gt 0$$

UPDATE: Just extract what the definition of $\lim \dfrac{a_n}{b_n}$ means. It says for every $\epsilon \gt 0$ there is $n_0$ such that the difference between $\dfrac{a_n}{b_n}$ and the limit ($= 1$ ) is less than $\epsilon$. So all you need to write in your paper would be to let $\epsilon$ be arbitrary. Then since $\lim \dfrac{a_n}{b_n}$ is given, choose $\dfrac{\epsilon}{M}$ as your arbitrary positive quantity. Then as per the definition of the limit there exists $n_0 \in \Bbb N$ such that $$n \ge n_0 \implies \left|{\dfrac{b_n}{a_n} - 1}\right| = \dfrac{|(b_n - a_n) - 0|}{|a_n|} \lt \dfrac{\epsilon}{M} \implies | (b_n - a_n ) - 0 | \lt |a_n | \cdot \dfrac{\epsilon}{M} \le \epsilon$$

And you're done!

• This is the problem. i don't know what to write after i choose arbitrary epsilon. i need to someone write the answer like in a test. it's not the question of HOW to do this, its about WHAT to write to do it formally, tnx! – user2637293 Sep 18 '14 at 1:15
• @user2637293: I edited the answer. Comment if you need more help. – Ishfaaq Sep 18 '14 at 6:58


Let $A$ and $B$ are two $n \times n$ Hermitian matrices . Suppose $A-B$ is positive semidefinite.

(a) Show that $\lambda_k(A) \geq \lambda_k(B)$ for $k=1,2,\dots ,n,$ where $\lambda_i(A)$ and $\lambda_i(B)$ are eigenvalues of matrices $A$ and $B$ with $\lambda_i (A)\geq \lambda_{i+1}(A)$ and $\lambda_i (B)\geq \lambda_{i+1}(B)$.

(b) Find a counter example to prove the converse of (a) is not true. i.e.,

Find two Hermitian matrices $A$ and $B$ such that $\lambda_k(A) \geq \lambda_k(B)$ but $A-B$ is not positive semidefinite.

My Trial:

(a) Since $A-B$ is positve semi-definite, $x^*(A-B)x \geq 0 \Rightarrow x^{*}Ax \geq x^{*}Bx.$

Also, as $A$ and $B$ are two Hermitian matrices, by spectral decomposition, we have $x^{*}Ax = x^{*}UD_{A}U^{*}x=x^{*}D_{A} x$ and $x^{*}Bx = x^{*}UD_{B}U^{*}x=x^{*}D_{B} x$.

It implies $D_{A} \geq D_{B}$ and so for each of theire eigenvalues.

(b) Let $A=\begin{pmatrix} 1 &0 \\ 0 &1 \end{pmatrix}$ and $B=\begin{pmatrix} 2 &0 \\ 0 &0 \end{pmatrix}$.

Please help to check if I make mistake in my presentation because I am very new to this topic so I do not know if am correct or not. Thank you very much in advance.

• How $U^*x$ becomes $x$? And the $U$ matrix is not the same for $A$ and for $B$. – enzotib May 7 '14 at 10:03
• Is it given that $A,B$ have the same eigenvectors? – Samrat Mukhopadhyay May 7 '14 at 10:21
• @enzotib Since $U$ is unitary, norm of $U*x$ = norm of $x$. Can I say this? – nam May 7 '14 at 10:22
• @Samrat The question does not give that $A$ and $B$ have the same eigenvectors. – nam May 7 '14 at 10:23
• @Pavel So, am I correct on the whole? – nam May 7 '14 at 10:24

To prove (a), you need to use the variational characterisation of eigenvalues (the min-max principle, Courant-Fischer theorem). For a Hermitian $A$ and the specified ordering of eigenvalues, $$\tag{*} \lambda_k(A)=\max_{S:\dim S=k}\min_{x\in S\setminus\{0\}}\frac{x^*Ax}{x^*x}=\min_{S:\dim S=n-k+1}\max_{x\in S\setminus\{0\}}\frac{x^*Ax}{x^*x}.$$ Let $S_*$ be the subspace of the dimension $n-k+1$ for which the minimum on the right-hand side of ($*$) is attained, that is, $$\lambda_k(A)=\max_{x\in S_*\setminus\{0\}}\frac{x^*Ax}{x^*x}.$$ Then (using $x^*Ax\geq x^*Bx$ for all $x$) $$\lambda_k(A)=\max_{x\in S_*\setminus\{0\}}\frac{x^*Ax}{x^*x} \geq\max_{x\in S_*\setminus\{0\}}\frac{x^*Bx}{x^*x} \geq\min_{S:\dim S=n-k+1}\max_{x\in S\setminus\{0\}}\frac{x^*Bx}{x^*x}=\lambda_k(B).$$

I was also thinking whether there would be an alternative proof which does not involve the variational characterisation and can thus be considered as more elementary. This is motivated by the proof here.

By exactly the same technique as in that proof, one can show the following:

Let $A,B\in\mathbb{C}^{n\times n}$ be Hermitian and let $i_{+,0}(A)$ and $i_+(A)$ denote the number of non-negative and positive eigenvalues, respectively, of $A$. Then $$i_{+,0}(A+B)\leq i_+(A)+i_{+,0}(B).$$

Now replace in the statement above $A$ and $B$, respectively, by $A-\lambda_k(A)I$ and $\lambda_k(A)I-B$. We get $$\tag{1} i_{+,0}(A-B)\leq i_+(A-\lambda_k(A)I)+i_{+,0}(\lambda_k(A)I-B).$$ Since our $A$ and $B$ are such that $A-B$ is positive semi-definite and hence all its eigenvalues are non-negative, we have $i_{+,0}(A-B)=n$. Note that adding a multiple of the identity to $A$ just shifts the eigenvalues, so the $k$th eigenvalue of $A-\lambda_k(A)I$ is zero and thus there is at most $k-1$ positive eigenvalues of $A-\lambda_k(A)I$, that is, $i_+(A-\lambda_k(A)I)\leq k-1$. Putting this stuff to (1) gives $$n\leq k-1+i_{+,0}(\lambda_k(A)I-B)\quad \Rightarrow \quad n-k+1\leq i_{+,0}(\lambda_k(A)I-B).$$ Since there is at least $n-k+1$ non-negative eigenvalues of $\lambda_k(A)I-B$, there is at most $k-1$ negative eigenvalues. Equivalently, the matrix $B-\lambda_k(A)I$ has a most $k-1$ positive eigenvalues and hence the $k$th eigenvalue of $B-\lambda_k(A)I$ is non-positive. But this eigenvalue is nothing but $\lambda_k(B)-\lambda_k(A)$ and hence $$\lambda_k(B)-\lambda_k(A)\leq 0\quad\Rightarrow\quad\lambda_k(A)\geq\lambda(B).$$

• Very thanks to your hint. I just had some readings on Rayleigh quotients but I get lost and do not know how to apply your hint on this question (it seems to pick up second large eigenvalue from some subspaces of smaller dimension). Would you mind giving me lines of answers. Thank you so much. – nam May 7 '14 at 10:45
• @nam It should be quite complete now :) – Algebraic Pavel May 7 '14 at 10:49
• Thank you for your illustration for the theorem. I know what I missed and I am looking for other examples to check. Thank you so much. – nam May 7 '14 at 11:02
• You are welcome. – Algebraic Pavel May 7 '14 at 11:02
• Just for fun, I've tried to make a proof which does not use the min-max principle. However, it's a bit longer. – Algebraic Pavel May 7 '14 at 13:03


Where do these Mobius transformations map the coordinate half-planes?

They are

$$\frac{z-1}{z+1}, \frac{z+1}{z-1},\frac{z-i}{z+i},\frac{z+i}{z-i}.$$

All four look virtually identical, so I would like to know how to best distinguish between them.

For example, the first mapping, I notice that $1$ maps to $0$, $-1$ maps to $\infty$.

But since $1$ and $-1$ are symmetric points w.r.t. the imaginary axis, the Mobius transformations always maps symmetric points to symmetric points. So, the images, $0$ and $\infty$, being symmetric points, must mean that they are symmetric with respect to a circle. So the real axis gets mapped to some circle centered at the origin.

How do I tell how big this circle is? And since the real axis separates the upper and lower half planes, either the upper or lower plane gets mapped inside the circle, but how can I tell? I tried a sample point: I see that $i$ gets mapped to $i$, but is this telling me that the mapping is "fixing" the UHP? I doubt it, because it's mapping stuff to some open disk.

Edit: Also, with the first mapping, there doesn't seem to be a nice symmetry argument to exploit for the imaginary axis.

Thanks.

• The first mapping does not map the real line to a circle centered at the origin: The image of $1$ under that map is the origin itself. Jul 1 '15 at 2:47
• It does, however, map the imaginary axis to the unit circle, as each imaginary $z$ is equidistant from $1$ and $-1$, and hence $\left\vert\frac{z - 1}{z + 1}\right\vert = 1$. Jul 1 '15 at 2:51
• Hmm...can I think of it this way, @Travis? I realize my mistake, I think. So, since 1 and -1 are symmetric w.r.t. the imaginary axis, and their images must also be symmetric w.r.t. the image of the imaginary axis (not the real axis), which I don't know yet. But 1 and -1 map to 0 and infinity, which are symmetric w.r.t. the image, which implies that the imaginary axis must get mapped to a circle. Is this valid? If so, how do I know how big the circle is? Your equidistant argument I have never heard of...thanks, Jul 1 '15 at 3:00
• I've expanded my comments into a proper answer below, and I think it answers all these questions, but feel free to comment there if not. Jul 1 '15 at 3:16

The first mapping, $$f : z \mapsto \frac{z - 1}{z + 1},$$ does not map the real axis to a circle centered at the origin: Indeed, we have $f(1) = 0$.

On the other hand, any imaginary number $iy$ is, by symmetry, equidistant from $-1$ and $1$, and so $$|f(iy)| = \left\vert\frac{iy - 1}{iy + 1}\right\vert = \frac{|iy - 1|}{|iy + 1|} =1,$$ that is, $f$ maps the imaginary axis to the unit circle (bijectively, if we include $f(\infty) = 1$). Using the above fact that $f(1) = 0$ (and continuity) gives that $f$ maps that right half-plane to the inside of the unit circle, and so the left half-plane to the outside.

As for the real axis, note that if $x$ is real, then so is $f(x) = \frac{x - 1}{x + 1}$ (unless $x = -1$, in which case we have $f(-1) = \infty$), so $f$ maps the real axis (including $\infty$) to itself.

The other three maps can be analyzed similarly, or we can use the fact that they are precisely the compositions of $f$ with the rotation maps $z \mapsto i^k z$, $k = 1, 2, 3$.

• Hi @Travis, if this mapping fixes the real line, must it also fix the upper and lower half planes, too? I'm thinking of the connectedness argument - a continuous function maps connected sets to connected sets. Or, does the fact that f is not continuous at -1 make this connectedness argument invalid? Also, your getting the norm = 1, because of iy being equidistant from -1 and 1, but isn't iy really equidistant from 1+iy and -1+iy? I might not be understanding the meaning of equidistant, but I feel like we should just be extending iy out in a straight line, along the real axis.... Jul 1 '15 at 3:26
• actually, I think got the equidistant argument, if I just use the modulus definition. and I'm pretty sure the connectedness argument is not valid. but feel free to comment, if you have something more to say. Thanks so much for your help, @Travis. Jul 1 '15 at 3:35
• Like you say, it fixes the upper and lower half-planes (as sets, not pointwise). Also, we can think of Mobius transformations as continuous maps from the Riemann sphere, $\Bbb C \cup \{\infty\}$ to itself, so there's no issue with continuity here. Jul 1 '15 at 3:37
• Sure, for any region $A$ we can ask what the region $f(A)$ is, and it happens that our particular map $f$ maps deveral familiar regions to other familiar regions. Note that this lets us conclude, e.g., that the first quadrant is mapped the the upper half of the unit disk. Jul 1 '15 at 3:55
• You're welcome, I'm glad you found it useful, and you too. Jul 1 '15 at 4:57

I'll do the first one for you, and you can do the rest. Consider the unit circle $z=e^{i\theta}$. Then $$\frac{z-1}{z+1}=\frac{e^{i\theta}-1}{e^{i\theta}+1}=\frac{1-e^{-i\theta}+e^{i\theta}-1}{1+e^{i\theta}+e^{-i\theta}+1}=\frac{2i\sin(\theta)}{2+2\cos(\theta)}=i\frac{\sin\theta}{1+\cos\theta}=i\tan\frac{\theta}{2}.$$ Thus, the unit circle gets mapped to the imaginary axis. The inside of the unit circle gets mapped to one side of the imaginary axis, while the outside gets mapped to the other. To tell which side gets mapped where, check $z=0$. This maps to $-1$. Thus, the inside of the unit circle gets mapped to the left side of the imaginary axis, while the outside of the nit circle gets mapped to the right. This method should work for all of your functions.

• Hi @AlexS, nice argument - and based on Travis's comment above, then this mapping also maps the imaginary axis back to the unit circle. So, it is its own inverse, sort of? And, how do we know the imaginary axis gets mapped to a unit circle? Thanks, Jul 1 '15 at 3:09
• It is not quite its own inverse: $$f(f(z))=\frac{\frac{z-1}{z+1}-1}{\frac{z-1}{z+1}+1}=\frac{-2}{2z}=-1/z.$$ But, if $z$ is on the unit circle, so is $-1/z$. To show directly that $f$ takes the imaginary axis to the unit circle, multiply $f(iy)$ by its complex conjugate, and check that the result is 1. Jul 1 '15 at 3:17
• ah, nice suggestion about multiplication by complex conjugates to check the norm -- thanks so much @AlexS. Jul 1 '15 at 4:17

Every Mobius map can be broken down into a composition of scaling maps, translations and inversions. Just do a standard long division of polynomials, i.e., fo $\frac {az+b}{cz+d}$ , do along division $(az+b)| (cz+d)$ and you will get this decomposition. Then you can apply these operations of inversion, scaling and translation, rotation in the right order to get your map. Maybe you could study more between games too!!

• ok, got it. Thanks @Gary. I will try to, but with Kyrie and Kevin injured, and me playing 48minutes per game, it's been hard to keep up with complex analysis :-( Jul 1 '15 at 4:21


# calculus

Evaluate integral of e^x^(1/2) / x^(1/2)
I've looked at the answer but I don't understand what people do in their steps.

When I substitute x^(1/2) for u, I get:

2du = 1/x^(1/2) dx

But what do you do with the 1/x^(1/2) dx? It just disappears in the solutions I've seen people give.

1. 👍 0
2. 👎 0
3. 👁 104
1. yes. the substitution is correct:
let u = x^(1/2)
thus du = 1/[2(x^(1/2))] dx, or
dx = 2(x^(1/2)) du, or
dx = 2u du
substituting these to original integral,
integral of [e^x^(1/2) / x^(1/2)] dx
integral of [(e^u) / u] * (2u) du
the u's will cancel out:
integral of [2*e^u] du
we can readily integrate this to
2*e^u + C
substituting back the value of u,
2*e^(x^(1/2)) + C

hope this helps~ :)

1. 👍 0
2. 👎 0
posted by Jai
2. This helped alot :)

1. 👍 0
2. 👎 0

## Similar Questions

1. ### calculus

a) Let f(z) = z^2 and γ(t) = 1 + it^3, t ∈ [0,1]. i) Write out the contour integral ∫γ f(z)dz as an integral with respect to t. You do not need to evaluate this integral. ii) Evaluate the integral ∫0,1+i z^2dz iii) What is

asked by jack on March 5, 2016
2. ### Calculus

1. Express the given integral as the limit of a Riemann sum but do not evaluate: integral[0 to 3]((x^3 - 6x)dx) 2.Use the Fundamental Theorem to evaluate integral[0 to 3]((x^3 - 6x)dx).(Your answer must include the

asked by Ernie on May 7, 2018
3. ### Math

Let A denote the portion of the curve y = sqrt(x) that is between the lines x = 1 and x = 4. 1) Set up, don't evaluate, 2 integrals, one in the variable x and one in the variable y, for the length of A. My Work: for x:

asked by Daisy on January 25, 2017
4. ### Math

1. Evaluate the indefinite integral ([6x^2 + 12x^(3/2) +4x+9]/sqrt x)dx. Answer = + C 2. Evaluate the indefinite integral (12sin x+4tan x)dx. Answer = + C 3. Evaluate the indefinite integral. (x^7)e^(x^8)dx. Answer = + C Thank you

asked by Matt on September 19, 2011
5. ### calculus

Suppose R is the rectangle 1

asked by Bryant on March 18, 2012
6. ### Math. Need help

1.if g=5 and k=1, evaluate 9+gk a. 13 b. 14 c. 15 d. 16 2.Evaluate a-b^2/b if a = 10 and b = 2. a. 1 b. 3 c. 8 d. 32 3.Evaluate 3xy^2 - y if x = 2 and y = 5. a. 26 b. 30 c. 120 d. 145 4.Evaluate 2d + d^2/3 if d = 6. a. 4 b. 12 c.

asked by A. on August 28, 2016
7. ### statistics

I don't understand how to answer this: relate the three measures of central tendency to a normal distribution- I looked it up online and it said that the mean, median, and mode are all the same but i don't understand how that can

asked by katie on February 12, 2009
8. ### Calculus

Evaluate integral, don't use calculator. I'll use the S as the sign for integral S 4 sin (theta)d(theta) from pi/2 to pi I know the answer is 4 because I enter it in on my calculator, but don't know the steps to get to the answer.

asked by Janet on May 9, 2011
9. ### calculus

8). Part 1 of 2: In the solid the base is a circle x^2+y^2=16 and the cross-section perpendicular to the y-axis is a square. Set up a definite integral expressing the volume of the solid. Answer choices: integral from -4 to 4 of

asked by Sally on April 14, 2013
10. ### Calculus

Hello. I would appreciate it if someone could check my answers. I'm sorry it is so long. 1.) Let R denote the region between the curves y=x^-1 and y=x^-2 over the interval 1

asked by Shay on September 3, 2005

More Similar Questions


# Determine remainder when dividing polynomials if we know other remainders

If 3 is the remainder when dividing $P(x)$ with $(x-3)$, and $5$ is the remainder when dividing $P(x)$ with $(x-4)$, what is the remainder when dividing $P(x)$ with $(x-3)(x-4)$?

I'm completely puzzled by this, I'm not sure where to start...

Any hint would be much appreciated.

Hint: $$P(x)=Q(x)(x-3)(x-4)+Ax+B$$ $$P(3)=3$$ $$P(4)=5$$

Can you solve for $A$ and $B$?

Let $$P(x)=(x-4)(x-3)Q(x)+ax+b$$

Now you have $$3a+b=3$$ and $$4a+b=5$$ So solve for $a$ and $b$.

• Thank you!! I only accepted the other answer because he posted first. – kok_nikol Nov 26 '16 at 18:27
• @kok_nikol You haven't accepted anything: you just upvoted one answer. You can upvote all the answers you find informative, good and helpful . Then you will accept only one: the best for you, which is not necessarily the first one, of course. – DonAntonio Nov 26 '16 at 18:34
• @DonAntonio Sorry, you can't accept an answer right away, you have to wait 10 minutes. And I can't upvote yet because I don't have enough reputation. – kok_nikol Nov 27 '16 at 0:26
• @kok_nikol Perhaps, yet you did not accepted anything when I first commented. I thought posters can upvote answers to their questions. – DonAntonio Nov 27 '16 at 0:30
• @DonAntonio Yes, when you commented I hadn't yet accepted anything. As for the voting Votes cast by those with less than 15 reputation are recorded, but do not change the publicly displayed post score. – kok_nikol Nov 27 '16 at 0:37

If you carry out the division you should be able to show that $$P(x)=Q(x)(x-3)(x-4)+R(x)$$

Two questions: What is the maximum degree of the remainder $R(x)$? Can you see how to use the remainder theorem for the cases $x=3,4$ that you know already so that $Q(x)$ becomes irrelevant?

Euclidean division by $(x-3)(x-4)$ can be written as $$P(x)=(x-3)(x-4)Q(x)+R(x),\quad\deg R(x)\le1.$$ Instead of writing $R(x)$ with the standard basis $\{1,x\}$, use the basis $\{x-3,x-4\}$. Thus $$P(x)=(x-3)(x-4)Q(x)+A(x-3)+B(x-4)$$ Setting $x=3$, you get right away $B=-P(3)=-3$. Similarly $A=P(4)=5$, and finally $$R(x)=5(x-3)-3(x-4)=2x-3.$$

This formula can be generalised: the remainder when dividing a polynomial $P(x)$ by $(x-a)(x-b),\enspace a\ne b$, is $$R(x)=\frac1{b-a}\Bigl[P(b)(x--a)-P(a)(x-b)\Bigr]=\frac1{b-a}\begin{vmatrix}x-a&x-b\\[1ex]P(a)&P(b)\end{vmatrix}.$$

Notice $\ p\, = 3+(x\!-\!3)q\,\$ by $\,\ p(3) = 3$

$5 = p(4) = 3+q(4)\,\Rightarrow\, {\color{#c00}2}\, =\, q(\color{#0a0}4) \,$

Hence $\, p = 3 + (x\!-\!3)(\underbrace{\color{#c00}2+(x\!-\!\color{#0a0}4)r}_{\Large q}\!)\, =\, 2x\!-\!3\, +\, (x\!-\!3)(x\!-\!4)r$


# $\mathbb{R^n}$ Second Countable

Lemma: Let U be an open subset of $$\mathbb{R}^n$$ and $$x \in U$$ then there exists a ball with rational center $$(\mathbb{Q}^n$$ and rational radius containing x, that is contained in U.

What I want to show: Every open subset of $$\mathbb{R}^n$$ is the union of some some collection of elements in the following set:

$$\{$$ Open balls with rational center and rational radius $$\}$$

Let U be an arbitrary open set in $$\mathbb{R}^n$$. By the above lemma, for each $$x \in U$$ you can find a rational center $$x'$$ and rational radius $$r'>0$$ so that $$x \in B(x',r')$$ $$\subset U$$. Hence the following collection:

$$B'=$$ $$\{$$ B(a,r) : a $$\in \mathbb{Q}^n$$ , $$r>0$$ , B(a,r) $$\subset U$$ $$\}$$ gives us

$$U=$$ $$\bigcup_{B \in B'}$$ $$B$$

is the proof correct?

As this is part of the proof of showing that $$\mathbb{R}^n$$ is second countable, it follows that since every metric space is Hausdorff, $$\mathbb{R}^n$$ is Hausdorff, and trivially, it is locally euclidean of dimension n (take the identity map, which is a homeomorphism) and so $$\mathbb{R}^n$$ is a n- dimensional topological manifold, am I correct?

• I assume that you require manifolds to be second countable? Then yes and yes. – freakish Aug 5 at 18:39
• @freakish Yup, a topological space M is a n - dimensional manifold if (1) It is second countable (2) It is Hausdorff (3) Every point in M has a neighborhood homeomorphic to an open subset of $\mathbb{R}^n$ – topologicalmagician Aug 5 at 18:40
• @topologicalmagician You prove correctly that $\mathbb{R}^n$ is second countable. Also, since $\mathbb{R}^n$ is Hausdorff and locally euclidean for the reasons you said above, it is a topological manifold too. – bing-nagata-smirnov Aug 5 at 19:03

Yes, your proofs that $$\mathbb R^n$$ is cecond countable and that $$\mathbb R^n$$ is an $$n$$- dimensional topological manifold are correct.