# Prove a function is continuos

#### ianchenmu

##### Member
Prove that
$f(x,y)=\left\{\begin{matrix} e^{-1/|x-y|},x\neq y\\ 0,x=y \end{matrix}\right.$
is continuous on $\mathbb{R}^2$.

Can I conclude since $e^{-1/t}$ is continuous then for $x$,$y\in \mathbb{R}^2$, $x\neq y$,$e^{-1/|x-y|}$ is continuous on $\mathbb{R}^2$ since here $t=|x-y|$? And how to prove it when $x=y$, using the $\delta- \varepsilon$ way? Thank you a lot!

#### girdav

##### Member
Re: Prove a function is continuous

We have to show that the function is continuous at each point of $\Bbb R^2$. Your argument threats the case of points of the form $(x,y),x\neq y$. So now, fix $x_0$. We have to show that $\lim_{(x,y)\to (x_0,x_0)}=0$.

Hint: use $e^{-u}\leqslant \frac 1{1+u}$ for each non-negative real number $u$.

Last edited:

#### Fernando Revilla

##### Well-known member
MHB Math Helper
Re: Prove a function is continuous

Hint: use $e^{-u}\leqslant \frac 1{1+u}$ for each real number $u$.
Surely is a typo, I suppose you meant for each real number $u$ close to $0$.

Another way: denoting $\Delta=\{(t,t):t\in\mathbb{R}\}$ (diagonal of $\mathbb{R}^2$), in a neighborhood $V$ of $(0,0)$ we have:

$\displaystyle\lim_{(x,y)\to (x_0,x_0)\;,\;(x,y)\in V-\Delta}f(x,y)=\displaystyle\lim_{(x,y)\to (x_0,x_0)\;,\;(x,y)\in V-\Delta}e^{-\dfrac{1}{|x-y|}}=e^{-\infty}=0$

$\displaystyle\lim_{(x,y)\to (x_0,x_0)\;,\;(x,y)\in V\cap\Delta}f(x,y)=\displaystyle\lim_{(x,y)\to (x_0,x_0)\;,\;(x,y)\in V\cap\Delta}0=0$

In a finite partition of $V$ all the limits coincide, so $\displaystyle\lim_{(x,y)\to (x_0,x_0)}f(x,y)=0$ .

#### ianchenmu

##### Member
The definition of continuity for a function is {lim x-> a} f(x) = f(a). For this function we only need to worry about when |x - y| -> 0, since the exponential function is continuous everywhere. So in other words we need to show that
{lim |x - y|-> 0} f(x,y) = f(x,x) = f(y,y)
Now f(x,x) = f(y,y) = 0 so we just need to show that
{lim |x - y| -> 0} f(x,y) = 0.
Since |x - y| > 0, -1/|x - y| -> infinity so f(x,y) -> 0 and we're done.

Is a proof like this correct?


# How to solve an equation with $x^4$?

Today, I had this question on a Maths test about Algebra. This was the equation I had to solve:

$$(1-x)(x-5)^3=x-1$$

I worked away the brackets and subtracted $x-1$ from both sides and was left with this:

$$-x^4+16x^3-90x^2+199x-124=0$$

Problem is, I haven't a clue how to solve this? First thing I tried was replacing $x^2$ with another variable like $u$ but that got me no further. Dividing the whole equation by $x^2$ (as is suggested by a lot of sites on this matter) also did not get me any further. I then tried something incredibly ludicrous;

$$(ax+b)(cx^3+dx^2+ex+f)=0$$ \left\{ \begin{aligned} ac &= -1 \\ ad + bc &= 16 \\ ae + bd &= -90 \\ af + be &= 199 \\ bf &= -124 \end{aligned} \right.

which got even worse when there were 3 brackets;

$$(ax+b)(gx+h)(ix^2+jx+k)=0$$ \left\{ \begin{aligned} ac &= agi &&= -1 \\ ad + bc &= agj + ahi + bgi &&= 16 \\ ae + bd &= agk + ahj + bgj + bhi &&= -90 \\ af + be &= ahk + bgk + bhj &&= 199 \\ bf &= bhk &&= -124 \end{aligned} \right.

only to be left with no result.

When using Wolfram Alpha on this question, it performs a rather strange step I don't understand:

$$-x^4+16x^3-90x^2+199x-124=0$$ $$\downarrow$$ $$-((x-4)(x-1)(x^2-11x+31))=0$$

Could somebody explain how to properly tackle this problem? And if possible, also show me how to get the non-real answers for it. Thanks.

• You needn't expand: first, observe that $x=1$ is a solution. Now, looking for other solutions (at most 3 other possible ones), you can divide each side by $x-1$ and obtain the equation $(x-5)^3=-1$ to solve (for values $x\neq 1$). – Clement C. May 20 '15 at 13:13

Euh... I think you overcomplicated things here...

$(1-x)(x-5)^3=x-1$ is equivalent to $(1-x)[(x-5)^3+1]=0$

Either $x=1$ or $(x-5)^3=-1$...

• I feel so stupid, that I didn't see this in the first place... Thanks for the great answer! – Martijn May 20 '15 at 13:19
• @TitoPiezasIII Done :) – Martijn May 20 '15 at 13:27
• You can also use the identity $a^3+1=(a+1)(...)$ – k1.M May 20 '15 at 14:00

There are already answers on how to get the real solutions, so I will only show you the non-real solutions.

You have obtained that $(x-5)^3=-1$. Expanding and simplifying we get: $$x^3-15x^2+75x-124=0$$ However, we know that $x=4$ is a solution so we can say that $(x-4)(ax^2+bx+c)=0$. You can equate coefficients or use polynomial division, but as you have already found with Wolfram Alpha, $(ax^2+bx+c)=(x^2-11x+31)$.

To solve, complete the square: $$x^2-11x+31=0$$ $$(x-\frac{11}{2})^2-\frac{121}{4}+31=0$$ $$(x-\frac{11}{2})^2=-\frac{3}{4}$$ $$x-\frac{11}{2}=\sqrt{-\frac{3}{4}}= \pm \frac{\sqrt{3}}{2}i$$ $$x=\frac {11 \pm i \sqrt{3}}{2}$$

• Awesome work! I have two kinds of mathematics in the Netherlands and the other kind is about imaginary numbers a.t.m. so this will definitely come in handy, I'm sure! – Martijn May 20 '15 at 13:45
• In general the solutions to $(x-a)^m=-1$ are $x_n=e^{i\pi\frac{2n-1}{m}}+a$, for $n=1,2,\dots m$. – Kwin van der Veen May 21 '15 at 4:10

From the beginning: $$(1-x)(x-5)^3=x-1\\ (1-x)(x-5)^3+1-x=0\\ (1-x)(x-5)^3+(1-x)=0\\ (1-x)[(x-5)^3+1]=0\\$$

This implies $1-x=0$ or $(x-5)^3=-1$. I believe you can solve these.

• Isn't there a sign that got flipped between the first and second line? – Clement C. May 20 '15 at 13:18
• Sorry fixed it. @ClementC. – KittyL May 20 '15 at 13:24

I know that the problem has already been answered but I want to show you a more general method, let's suppose that you don't se how to rewrite the equation:

$-x^4+16x^3-90x^2+199x-124=0$

Or I prefer to write:

$x^4-16x^3+90x^2-199x+124=0$

You can use something called the Ruffini rule: search for integers divisor (both positive and negative) of the constant term and then set $x$ equals to the them and see if one of them is a solution. Starting from one we have:

$1-16+90-199+124=0$

So $x=1$ is a solution, now via Ruffini's rule, which can be seen here, we can rewrite the equation as:

$(x-1)(x^3-15x^2+75x-124)=0$

Now you have $3$ options to end this exercise:

$1.)$ Note that the second factor is a perfect cube;

$2.)$ Use Ruffini's rule again;

$3.)$ Use the general formula for third degree equation (which I'd not advise you to if your interested only in real solution).

This is a more general method so I hope this will help you in the future!


# Numbers of distinct values obtained by inserting $+ - \times \div ()$ in $\underbrace{2\quad2 \quad2 \quad2\quad…\quad 2}_{n \text{ times}}$

This question is inspired by How many values of $2^{2^{2^{.^{.^{.^{2}}}}}}$ depending on parenthesis? (By the way, I sincerely hope this kind of questions can receive more attention)

Insert $$+ - \times \div ()$$ in $$\underbrace{2\quad2 \quad2 \quad2\quad...\quad 2}_{n \text{ times}}$$ Denote the number of distinct values which can be obtained in this way by $$D(n)$$. Is there a general formula (or recurrence relation at least) for $$D(n)$$?

This is basically the $$+ - \times \div ()$$ version of @barakmanos question. It seems this question is easier than the power tower version. Or maybe not?

For $$n=1$$ , there is only $$2$$ values $$-2,2$$;

For $$n=2$$, there are $$5$$ values $$-4,-1,0,1,4$$;

For $$n=3$$, there are $$13$$ values $$-8,-6,-3,-2,-1,-\frac{1}{2},0,\frac{1}{2},1,2,3,6,8$$;

And for $$n=4$$ I'm reluctant to calculate with bare hands. (See @DanUznanki answer for what follows)

Any idea is appreciated. Sorry if this is a duplicate.

Edit: My research shows that the version with distinct generic variables $$a_1,a_2,...,a_n$$ is solved. See A182173 for your reference.

• Based on what you say for $n=1$ and $n=2$, we may use $-$ as either the negation operator on one number or the subtraction operator between two numbers? – alex.jordan Mar 24 at 4:53
• @alex.jordan Yes. – YuiTo Cheng Mar 24 at 4:59
• OK. I wonder if you did not allow $-$ to be unary negation, if then the sequence might appear in OEIS. We have an overloaded symbol, $-$. And one of its meanings happens to be unary negation. It's odd that we don't have a symbol for unary inversion. Well, we do: $^{-1}$. But it does not reuse the division symbol $\div$. All this is just to say that it feels like the special double meaning for $-$, that is not mirrored with $\div$, could break what would otherwise be a simpler pattern in the sequence of counts. – alex.jordan Mar 24 at 5:22
• @alex.jordan If we disallow $-$ to be a unary operator, will things become vastly simpler? – YuiTo Cheng Mar 24 at 5:27
• @alex.jordan: I once toyed with the idea of "$/$" for unary inversion. (So, "$/2$" means "$1/2$" in the same way that "$-2$" means "$0-2$".) I still kinda like it. :) – Blue Mar 25 at 20:09

It looks like we can do this "inductively": for the calculations of size $$n$$, we can take values from the list for $$1 \le k\le n/2$$, and values from the list for $$(n-k)$$, and operate on them using the 64 different operation orders.

Fortunately, it's really only 10 classes of operation, because many are duplicates:

• There's four values we can get from addition and subtraction: $$a+b$$, $$-(a+b)$$, $$a-b$$, and $$b-a$$.
• There's only two we can get from multiplication: $$ab$$ and $$-ab$$.
• There are four cases we can get from division: $$a/b$$, $$b/a$$, $$-a/b$$, and $$-b/a$$.

Also conveniently we only need to try the nonnegative entries in previous lists.

So for $$n=4$$, we have:

$$-16$$, $$-12$$, $$-10$$, $$-8$$, $$-6$$, $$-5$$, $$-4$$, $$-3$$, $$-5/2$$, $$-2$$, $$-3/2$$, $$-1$$, $$-2/3$$, $$-1/2$$, $$-1/3$$, $$-1/4$$, $$0$$, $$1/4$$, $$1/3$$, $$1/2$$, $$2/3$$, $$1$$, $$3/2$$, $$2$$, $$5/2$$, $$3$$, $$4$$, $$5$$, $$6$$, $$8$$, $$10$$, $$12$$, $$16$$

Which is $$33$$ entries.

I've written a short script which finds them all, and told it to run up to $$n=10$$, which gave the following sizes: $$2,5,13,33,77,185,441,1051,2523,6083$$. Apparently this sequence is not in OEIS, nor is the positive-values-only version! I am very surprised.

• You mentioned we can do it 'inductively', do you have any clue about the recurrence relation? – YuiTo Cheng Feb 20 at 13:31
• Not sure, but it isn't gonna be nice. actually, different values for $x$ instead of $2$ give different results: using $3$ gives $2$, $7$, $21$, $67$, using $4$ gives $2$,$7$,$21$,$77$. Interestingly, when I tried to get the generic result, where $x$ is just $x$, I did get some partial results in the realm of restricted lattice walks: $2$,$7$,$23$,$91$, $347$ appears twice in OEIS. – Dan Uznanski Feb 20 at 13:39
• Interesting. What about the original power tower question? Compared with this, which one do you think is simpler? – YuiTo Cheng Feb 20 at 13:45
• for generic $x$ in the power tower thing you're getting exactly the Catalan Numbers. But assuming you have the ability to actually store the numbers you're working with, the power tower version is vastly simpler: there's only one operator instead of 10. – Dan Uznanski Feb 20 at 13:46
• What about the case without bracket? – YuiTo Cheng Feb 21 at 3:36

@Dan Uznanski if your results are correct i noticed that the ratio between the successive terms of the sum is almost constant. So maybe:

Conjecture

$$D(n) \sim K^n$$

Where:

$$2.3\leq K \leq 2.5$$


# Homomorphism between cyclic groups

I have some confusion in relation to the following question.

Let $\langle x\rangle = G$, $\langle y\rangle = H$ be finite cyclic groups of order $n$ and $m$ respectively. Let $f:G \mapsto H$ be the mapping sending $x^i$ to $y^i.$ Determine the conditions on $n$ and $m$ so that $f$ is going to be a homomorphism.

In verifying the definition of a homomorphism I seem to be able to conclude that there is no restriction on $n$ and $m$ since the definition of a homomorphic mapping is satisfied:

$$f(x^i x^j) = f(x^{i+j}) = y^{i+j} = y^i y^j = f(x^i)(x^j) \; \; \; (1)$$

But assuming n < m we would get

$$f(x^n) = f(e) = e \not = y^n$$

So clearly we have to have $m | n.$

My questions are

1. Are there any other conditions on $n$ and $m$ besides $m |n.$

2. What exactly did I do wrong in $(1)$ ?

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For the (1), What would happen if $i+j \ge m$? It always helps to define the map this way: $x^{i(n)} \mapsto x^{i(m)}$. Note that the map may not even be well-defined if $m \nmid n$. –  user21436 Mar 16 '12 at 9:50
I'll write up a more detailed answer in an hour if it's still unanswered. This has its direct connection to the universal property of the quotient maps. –  user21436 Mar 16 '12 at 9:52
I see that there are many contradictions if $m \not | n.$ My main concern is that $(1)$ seems like such a innocent algebraic manipulation that I would never suspect that something is wrong with it.I would like to hear an "aha" explanation on why one cannot apply the definition of a homomorphism on $f$ so blindlessly. –  Jernej Mar 16 '12 at 10:02
–  lhf Mar 16 '12 at 10:43
Well, either you use a homomorphism $\mathbb Z \rightarrow G$ (and talk about lifting that homomorphism) or you must face the fact that the Element $x^i$ does NOT determine the number $i$. –  Blah Mar 16 '12 at 10:50

The discussion about "well-defined" is perhaps a bit obscure. Here's the problem:

Remember that in general, an element of $G$ may have many different "names." For example, if $n=10$, then the element $x^{11}$ is equal to $x$; in fact, $x$ has infinitely many different "names" as a power of $x$: $$\cdots = x^{-9} = x^1 = x^{11} = x^{21} = \cdots$$

The problem is that the definition of $f$ as given depends on the name of the element! That is, if we are furiously working and somebody hands us a power of $x$, say, $x^{3781}$, we are supposed to, unthinkingly, map it to $y^{3781}$. The problem is that $x^{3781}$ is the same element as $x$, which we are supposed to send to $y^1$. That means that unless $y^1=y^{3781}$, what we have is not really a function: because the same input, $x$ (who, when being teased by bullies is called "$x^{3781}$") may be sent to $y^1$ or to $y^{3781}$, depending on what "name" we just heard for it.

Checking that the value of the function is the same regardless of what name we are using for an element, even though the function is defined in terms of the name, is called "checking that the function is 'well-defined.'"

An example of a function that is not well-defined would be one in which the input is an integer, and the output is the number of symbols used to express that integer. For example, $f(3)$ would be $1$ (because 3 is only one symbol), but $f(4-1)$ would be $3$ (because we are using 4, -, and 1). This is not well defined as a function of the integers, because $3$ is the same as $4-1$, but $f$ assigns it two different outputs.

So in order for the expression given to actually define a function, we need the following to be true: $$\text{if }x^i=x^j,\text{ then }y^i = f(x^i)\text{ is equal to }y^j=f(x^j).$$ Now, $x^i=x^j$ if and only if $i\equiv j\pmod{n}$; and $y^i=y^j$ if and only if $i\equiv j\pmod{m}$. Therefore, we need: $$\text{if }i\equiv j\pmod{m},\text{ then }i\equiv j\pmod{n}.$$ In other words: every number that is a multiple of $m$ must be a multiple of $n$.

This is equivalent to $n|m$.

In fact, you noticed that in what you wrote, because $x^n=e$, so we need $f(x^n)$ to be the same as $f(x^0)$.

Once we know that $n|m$, then $f$ is well defined. Once it is well-defined, we can start verifying that it is indeed a homomorphism (it is). Technically, it's incorrect to start working to see if it is a homomorphism before you even know whether or not it is a function.

Note that the condition actually works if we allow $G$ or $H$ to be infinite cyclic groups, if we call infinite cyclic groups "groups of order $0$". Then $i\equiv j\pmod{0}$ means $i=j$, every number divides $0$, but the only multiple of $0$ is $0$. So if $G$ is infinite cyclic then the value of $n$ does not matter; if $H$ is infinite cyclic then we must have $G$ infinite cyclic.

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I don't agree with your take on well-defined. The rule $x^i\mapsto y^i$ is well-defined in the set $\{x^0, x^1, \ldots, x^{n-1}\}$ because every element in this set has a single name. The problem is that this rule may not be a homomorphism because there is no guarantee that say $x^{2n-3}$ goes to $f(x)^{2n-3}=y^{2n-3}$. –  lhf Mar 16 '12 at 16:47
@lhf: I agree that if we were specifying $0\leq i\lt n$, then the rule would yield a well-defined function. However, I don't see any such specification in the statement of the problem. It just says "sending $x^i$ to $y^i$", with no restrictions on $i$. Of course, checking that it is a homomorphism is important: I note it later in the post, I don't ignore it. –  Arturo Magidin Mar 16 '12 at 16:58
Right, I assumed that $0\leq i\lt n$ was a natural implicit assumption. –  lhf Mar 16 '12 at 17:04
@lhf: In that case, there is a different problem with the definition, which is that it would only make sense if $n\leq m$ (otherwise, $y^i$ for $i\geq m$ would be an "invalid name"). So we would end up with the conclusion that the map only makes sense and is a homomorphism when $m=n$; or else we need to treat the exponent differently in the domain (restricted to a special representation for the elements) but not the codomain, which seems somewhat perverse. –  Arturo Magidin Mar 16 '12 at 18:21
I don't think $y^i$ is an invalid name. Just raise $y$ to $i$. The exponent is well defined, isn't it? –  lhf Mar 16 '12 at 21:27

You may want to check first if the map is well defined, ie., if $x^i=x^j$ does it follow that $f(x^i)=f(x^j)$?

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+1 Why would anyone downvote this? Ok, it would be better to phrase it: "if $x^i=x^j$ does it follow that $y^i=y^j$?" It is somewhat dangerous to talk about values of $f$ before you know that $f$ is well defined. –  Jyrki Lahtonen Mar 16 '12 at 14:07
it is perfectly acceptable to ask whether or not a definition of f makes f a function. –  David Wheeler Mar 16 '12 at 15:22
@David, you're right, of course. I was just trying to guess, why there was a downvote for a little while. –  Jyrki Lahtonen Mar 16 '12 at 15:44

I am not sure if this completely explains why the naive calculations in $(1)$ are not valid. Here is my go at it:

I'll list down several points of concern.

• Firstly your map $x^i \mapsto y^i$ is very naive that it misses some essential details.

For convenience of notation, I prefer to look at this as $\varphi(\bar i)=\bar i$ if you'll allow me to do so. This is also naive but to some extent captures what is happening.

So, an equivalence class $\bmod n$ is mapped to the same equivalence class $\bmod m$ under $\varphi$. This bit of information is missed in $x^i \mapsto y^i$. This explains my comment under your question that you will want to write your maps as $x^{i(n)} \mapsto y^{i(m)}$. This notation makes things more transparent.

That you have dicovered that $m \mid n$, it can be obtained through the universal property of the quotients, which I will write about later.

The quotient map $G \to G/H$ factors over $G \to G/K$ if and only if $K \subseteq H$. That is there is a map from $G/H$ to $G/K$ if and only if $K \subseteq H$.

Proof: Please Go through Dummit and Foote's description in their Abstract Algebra in the section $\S3.3$ Isomorphism Theorems.

Now consider your cyclic groups as $\Bbb Z/n\Bbb Z$ and $\Bbb Z/m\Bbb Z$ and observe that above condition will tell you, $m\Bbb Z \subseteq n \Bbb Z$ whic happens if and only if...

As you had earlier in your comments observed, the map won't even be well defined if $m \nmid n$.

I'll leave it to you as a challenge to use this blooper to prove that two groups of same order are isomorphic, which really is not the case. Hint: Cyclic groups exist for any given order.

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Will the down voter explain why? –  user21436 Mar 16 '12 at 14:01
Why do you say that $y^{i+j}$ might not be an element of $H$? Of course it is an element of $H$! A group is a set closed under the group operation, so if $y$ is in there, so are all its powers. Also, are you sure that $i(m)$ is a standard notation for the remainder? Color my old-fashioned, but I think that binary mod should stay in programming classes. This last bit is just a pet-peeve of mine :-) –  Jyrki Lahtonen Mar 16 '12 at 14:04
Well, $y^{i+j}$ is an element of the group. For example, if $H$ is a subgroup of the multiplicative group of complex numbers generated by the imaginary unit $i$, you seem to be saying that $i^5$ is not an element of $H$. Well, you are wrong: $$i^5=i^{4+1}=i^4\cdot i=1\cdot i=i.$$ Looks like it is in there! The same holds for all cyclic groups. I just picked a concrete one to make you realize your error. –  Jyrki Lahtonen Mar 16 '12 at 14:16
No. I'm not prone to making a mistake. I check that my mappings are well-defined. If (granted, a big if in some cases) $y^i=y^j$ whenever $x^i=x^j$, then the mapping $x^i\mapsto y^i$ is trivially a homomorphism: $$f(x^i*x^j)=f(x^{i+j})=y^{i+j}=y^i*y^j=f(x^i)f(x^j)$$ AFTER I have checked that it is well-defined, then the above always makes sense, because in a cyclic group the power of an element is defined in such a way that we always have $y^{i+j}=y^i*y^j$ and so forth. –  Jyrki Lahtonen Mar 16 '12 at 14:31
as a matter of fact, it IS true that $(i+j)$ (mod n) = $i$ (mod n) + $j$ (mod n). the problem is that $(i+j)$ (mod n) may not equal $i+j$ (mod m). –  David Wheeler Mar 16 '12 at 15:32

The rule $x^i\mapsto y^i$ gives a perfectly well-defined function in the set $\{x^0, x^1, \ldots, x^{n-1}\}$. However, for that rule to be a homomorphism we need that $x^{i+j} \mapsto y^{i+j}$ for all $i$ and $j$. To find where $x^{i+j}$ goes we need to reduce $i+j \bmod n$ to a $k$ in $\{0,1,\ldots,n-1\}$ and we need that $k \equiv i+j \bmod m$. In other words, we need $k \equiv k' \bmod n \implies k \equiv k' \bmod m$ and this can only happen when $m\mid n$.

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THEOREM:

Define $f:\langle x \rangle \to \langle y \rangle$, where the orders of $\langle x \rangle$ and $\langle y \rangle$ are n and m respectively, by $f(x) = y^u$. Then f is a well-defined homomorphism if and only if $\frac{m}{\gcd(m,n)} \mid u$.

PROOF: Suppose that the mapping $f:\langle x \rangle \to \langle y \rangle$ is a well-defined homomorphism.

Let $f(x) = y^u$ for some non negative integer u.

We must also have $1 = f(x^n) = y^{un}$

Then $m \mid un$. Let $g = \gcd(m, n)$. Then $\frac m g \mid u \frac n g$. It follows that $\frac m g \mid u$.

The converse is pretty straightforward.

Suppose we define $f:\langle x \rangle \to \langle y \rangle$ by $f(x) = y^u$ where $\frac{m}{\gcd(m,n)} \mid u$

The only real problem with this definition is "Is it well defined?" Since the order of $\langle x \rangle$ is $n$:

\begin{align*} x^a = x^b &\Rightarrow a \equiv b \mod n \cr &\Rightarrow ua \equiv ub \mod {un} \cr &\Rightarrow ua \equiv ub \mod {\frac{mn}{\gcd(m,n)}} \cr &\Rightarrow ua \equiv ub \mod {m} \cr &\Rightarrow y^{ua} = y^{ub}\cr &\Rightarrow f(x^a) = f(x^b) \end{align*}

Linearity is pretty obvious, so $f$ is a well-defined homomorphism.

So, if you want $f(x^i) = y^i$, then you want $f(x) = y$. You must therefore have $\frac{m}{\gcd(m,n)} \mid 1$. Which implies $m \mid n$.

Conversely, if $m \mid n$ and $f(x) = y$, it follows that $f(x^i) = y^i$.

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# induction and proper use of the inductive step

Suppose I have a conjecture of the form $\forall x \in \mathbb{N}^{+}$ $f(x) > g(x)$.

Without explicitly giving the functions $f$ and $g$ I would like to prove this conjecture using induction.

Specifically I have shown $f(1) > g(1)$. My inductive assumption is now that $f(x - 1) > g(x - 1)$ and I'm trying to show $f(x) > g(x)$.

I haven't been able to do this, but I've noticed that I can if I assume $f(y) > g(y)$ $\forall y \in \mathbb{N}^{+}$ such that $y < x$ instead of my inductive argument I can prove $f(x) > g(x)$.

My question is, is this ok or am I using circular logic? Typically when we use induction we use $x - 1$ to prove the $x$ case after an initial base case. For my problem it doesn't appear possible to prove $f(x) > g(x)$ without assuming that for all values less than $x$ the conjecture holds ($f$ and $g$ are recursive). I suspect that it is ok, but it sounds extremely circular and I haven't been able to convince myself that it isn't.

Note also that I've included proof verification as a tag because although I have not included the explicit proof, I have included a proof approach that I'm interested in validating.

• That's fine. That's called strong induction. It might be clearer if worded it slightly differently. You know $f(1) > g(1)$ and you have proven that if $f(y)> g(y)$ for all $y \le x-1< x$ then $f(x) > f(x)$. Hence... induction. Is that clearer? – fleablood Oct 14 '17 at 17:13

So you just want to make the stronger assumption that some statement $P(y)$ holds for all $y < x$ instead of only for $y = x-1$?

This is totally fine. It is called strong induction and is equivalent to simple induction.

• Ok, that's what I thought. I was just a bit unsure because I couldn't remember doing something like this in any course I've taken. Thank you! – HXSP1947 Oct 7 '17 at 1:54

"Typically when we use induction we use x−1 to prove the x case after an initial base case. For my problem it doesn't appear possible to prove .. the proposition.. without assuming that for all values less than x the conjecture holds"

Think about this. If being true for all $y \le x-1 < x \implies$ being true for $x$, then being true for $y < x$ and being true for $x$ implies being true for all $y < x+1$. So true for $x+1$. And so on.

In principal, this is the exact same reasoning as the reasoning for induction. (if it's true for one case, it's true for the next, and via infinite countable reptitions it must be true for all).

The only difference is that whereas in "weak" induction we only make use of the fact that the proposition is true of $1$ and for $x-1$ and therefore true for $x$ we don't need to and we don't make use of the fact that it must also be true for $1,.......,x-2$ as well as just $x-1$.

But we know they must be true so we can make use of them if we need to.

This is called "strong" induction. It is equivalent to weak induction. The only difference is we might need to make use of earlier values than just the $x-1$ case.

This can be particularly useful with recursive definitions.

Say you know that $a_1 = something$ and $a_n =$ something to do with $a_{n-1}$ if $n$ is odd and something to do with $a_{\frac n2}$ and $a_{n-1}$ if $n$ is even. And you want to prove something about $a_n$. Well, it is not just enough to assume that the proposition is true of $a_{n-1}$. You also have to assume it is true of $a_{\frac n2}$. Can you assume that? Of course you can. To get to $a_{n-1}$ you had to have "passed through" $a_{\frac n2}$ "along the way". So why not use it if you have it?


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# In the figure above if l1||l2, what is the value of x?

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15 Mar 2018, 23:19
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In the figure above if $$l_1||l_2$$, what is the value of x?

A. 70
B. 100
C. 110
D. 150
E. 170

Attachment:

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In the figure above if l1||l2, what is the value of x?  [#permalink]

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15 Mar 2018, 23:19
Bunuel wrote:

In the figure above if $$l_1||l_2$$, what is the value of x?

A. 70
B. 100
C. 110
D. 150
E. 170

Attachment:
2018-03-16_1010.png

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Re: In the figure above if l1||l2, what is the value of x?  [#permalink]

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15 Mar 2018, 23:53
I got A - 70°, you can complete the top triangle as 30°, 40° and 110° because l1 and l2 are parallel, x = 180° - 110° which is 70°
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Re: In the figure above if l1||l2, what is the value of x?  [#permalink]

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16 Mar 2018, 08:35
Bunuel wrote:
In the figure above if $$l_1||l_2$$, what is the value of x?

A. 70
B. 100
C. 110
D. 150
E. 170

Attachment:

2018-03-16_1010ed.png [ 10.8 KiB | Viewed 1021 times ]

A couple of other ways to approach . . .

$$l_1$$ and $$l_2$$ are cut by a transversal

Alternate interior angles are congruent.
Lower right angle = $$40°$$
Other angle of lower triangle, given = $$30°$$

1) $$x$$ = exterior angle
Exterior angle $$x$$ = sum of opposite interior angles of lower triangle
Exterior angle $$x$$ $$= (30° + 40°) =$$ $$70°$$

OR

2) Straight line: Third angle of lower ∆ + $$x = 180°$$
Interior angles of a triangle sum to 180°
Lower ∆'s third angle = $$110°$$
($$(180° - 30° - 40°) = 110°$$ = third angle))

Third angle lies on a straight line with $$x$$
$$110° + x° = 180°$$
$$x = 70°$$

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Re: In the figure above if l1||l2, what is the value of x?  [#permalink]

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18 Mar 2018, 10:40
Bunuel wrote:

In the figure above if $$l_1||l_2$$, what is the value of x?

A. 70
B. 100
C. 110
D. 150
E. 170

Attachment:
The attachment 2018-03-16_1010.png is no longer available

Drawing a additional parallel line l1 and l2 and using property of equal alternate angle between 2 parallel line

we get x= 30 +40
x=70
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Re: In the figure above if l1||l2, what is the value of x?  [#permalink]

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20 Oct 2019, 10:06
Bunuel wrote:

In the figure above if $$l_1||l_2$$, what is the value of x?

A. 70
B. 100
C. 110
D. 150
E. 170

Attachment:
2018-03-16_1010.png

Draw a line through the vertex of angle x that is parallel to both l1 and l2. We will see that angle x is the sum of the two given angles of 40 degrees and 30 degrees. Thus angle x = 70 degrees.

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Re: In the figure above if l1||l2, what is the value of x?  [#permalink]

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20 Oct 2019, 10:11
Bunuel wrote:

In the figure above if $$l_1||l_2$$, what is the value of x?

A. 70
B. 100
C. 110
D. 150
E. 170

Attachment:
2018-03-16_1010.png

x = 180 - (180 - 40 - 30) = 70

IMO A
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Re: In the figure above if l1||l2, what is the value of x?   [#permalink] 20 Oct 2019, 10:11
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# In the xy-plane, the straight-line graphs of the three equations above

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In the xy-plane, the straight-line graphs of the three equations above  [#permalink]

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Updated on: 13 Nov 2017, 21:24
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y = ax - 5
y = x + 6
y = 3x + b
In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?

(1) a = 2
(2) r = 17

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Originally posted by Bunuel on 23 Jul 2015, 10:47.
Last edited by Bunuel on 13 Nov 2017, 21:24, edited 4 times in total.
Edited the question.
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In the xy-plane, the straight-line graphs of the three equations above  [#permalink]

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26 Oct 2015, 10:02
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Re: In the xy-plane, the straight-line graphs of the three equations above  [#permalink]

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23 Jul 2015, 11:30
12
8
mcelroytutoring wrote:
DS 136 from OFG 2016 (new question)

y = ax - 5
y = x + 6
y = 3x + b

In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?

1) a = 2

2) r = 17

Solution provided by : mcelroytutoring

Let's start by substituting the point (p,r) into all equations in place of (x,y) which will make step #2 a bit easier to comprehend but is not necessary to solve the question. Then, let's consider the number of variables left in each equation.

#1: r = ap - 5 (3 variables R,A,P)
#2: r = p + 6 (2 variables R,P)
#3: r = 3p + b (3 variables R,B,P)

1) a = 2 allows us to reduce equation #1 to the variables r and p, which are the same two variables as equation #2. Thus we have simultaneous equations. As soon as we verify that the equations are different, we know that we can solve for both variables. Once we know r and p, we can substitute in equation #3 to solve for b. Sufficient.

2) r = 17 allows us to do the same thing, more or less. It reduces equation #2 to only one variable, allowing us to solve for p. Once we have p (and r), we can use equation #3 to solve for b. Sufficient.
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##### General Discussion
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Re: In the xy-plane, the straight-line graphs of the three equations above  [#permalink]

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24 Sep 2015, 01:53
5
6
if each of the 3 equations contains points (p,r) this means that they intersect in that point
1. a=2
Find the intercept Intercept for three simultaneous equations
y=2x-5
y=x+6
y=3x+b
Let's use the first 2 equations: plug y=x+6 in the secod equation
x+6=2x-5 -> x=11, y=17 we can use the values to calulate b in the 3rd equation
17=33+b -> b=-16 SUFFICIENT

2. Here we have directly the value for Y, let's plug it in the 2nd equation
y=x+6 -> 17=x+6 -> x=11, y=17; We can plug these values in the 3rd equation and find b as we did above
17=33+b -> b=-16 SUFFICIENT

Answer (D) Most important point is here to catch the hint about intersection of 3 lines at one point
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Re: In the xy-plane, the straight-line graphs of the three equations above  [#permalink]

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26 Oct 2015, 10:50
4
I think it's D.

Keeping point (p,r) in all the equations we get :

p = ar -5 -----(1)
p = r + 6 ------(2)
p = 3r + b ------(3)

Now consider (1) if a = 2 from (1) and (2) we get
r = 11 , p=17 and putting in (3) we can get b.

Similarly for (2) we can get the values for r and p and hence can get the value for b.

So both statements individually are correct to answer the question.
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Re: In the xy-plane, the straight-line graphs of the three equations above  [#permalink]

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26 Oct 2015, 19:14
2
m2k wrote:
I think it's D.

Keeping point (p,r) in all the equations we get :

p = ar -5 -----(1)
p = r + 6 ------(2)
p = 3r + b ------(3)

Now consider (1) if a = 2 from (1) and (2) we get
r = 11 , p=17 and putting in (3) we can get b.

Similarly for (2) we can get the values for r and p and hence can get the value for b.

So both statements individually are correct to answer the question.

Approach if right but the values you derived are wrong. According to me r = 17 and that is what stmt b also states.
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Re: In the xy-plane, the straight-line graphs of the three equations above  [#permalink]

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10 Jun 2016, 11:23
3
y = ax - 5 ... eq 1
y = x + 6 ... eq 2
y = 3x + b ...eq 3

Total of 4 variables are present.

Statement 1 : a = 2

Insert in eq 1

We have y = 2x -5 and y = x+6

Solving we get x = 11 and y = 17

Substitute in eq 3 and we get value of b

Statement 2: r=17

This means the y co- ordinate is 17
Substitute in eq 2 we get x as 11

Again can find value of b from equation 3

Hence D
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Re: In the xy-plane, the straight-line graphs of the three equations above  [#permalink]

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18 Jan 2017, 15:18
we used information from both 1 and 2 then how can the answer be D... should it not be C... some one kindly clarify.......... clearly am a zero in ds and that too a big one
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Re: In the xy-plane, the straight-line graphs of the three equations above  [#permalink]

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18 Jan 2017, 20:18
1
y = ax - 5---------1
y = x + 6---------2
y = 3x + b--------3
In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?

(1) a = 2
(2) r = 17

All three line intersect each other at common point (p,r).
1. given a = 2
putting in equation 1 .= y=2x-5
equating 1(after replacing value of a) and 2 we will get value of (p,r)
putting (p,r) in equation 3 we will get value for b---suff..

2 given r = 17.
putting in equation 2 we will get value of x.i'e p.
Now as we know common point of intersection ,putting p,r in equation 3 , we will get value of b
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Re: In the xy-plane, the straight-line graphs of the three equations above  [#permalink]

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22 Jan 2017, 15:51
thanks sobby for your response... highly appreciated...
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Re: In the xy-plane, the straight-line graphs of the three equations above  [#permalink]

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16 Mar 2017, 08:58
1
(eq1) $$y = ax - 5$$
(eq2) $$y = x + 6$$
(eq3) $$y = 3x + b$$

In the xy-plane, the straight-line graphs of the three equations above each contain the point (p, r). If a and b are constants, what is the value of b?

1) $$a = 2$$
2) $$r = 17$$

Solution:

1) $$a = 2$$
- Putting the value of a in eq1, we get: $$y = 2x - 5$$
- At this point you can solve for (x, y), plug (x, y) in (eq3) and solve for (b) [though this approach might take few seconds]
(alternatively, faster method)
- you can skip solving for (x, y) and deduce that given 3 equations and 3 unknowns (since a is given in statement 1) we can solve for all of them (including b), since the lines are have different slopes i.e. different lines. Hence, we can get single value of 'b', proving the condition SUFFICIENT.
NOTE: 3 equations and 3 unknowns does not ALWAYS mean that we can find 3 unknown. We have to make sure that 2 of them or all of them are not the same line.

2) $$r = 17$$
- Since, point (p, r) lie on all the line, we can plugin the point in above equation
$$r = ap - 5 => 17 = ap - 5$$
$$r = p + 6 => 17 = p + 6$$
$$r = 3p + b => 17 = 3p + b$$
- Again, we do not need to solve for all the variables and just recognize that the above equations will lead to single value of b. Hence, SUFFICIENT.

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Re: In the xy-plane, the straight-line graphs of the three equations above  [#permalink]

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30 Apr 2017, 23:51
y = ax - 5
y = x + 6
y = 3x + b

In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?

1) a = 2

2) r = 17

My 2 cents.

It is important to realize from the Question stem that the 3 equations intersect as (p,r).

For 1), as we know a =2, we can equate the first and second equation to get the value of x, and then use that value of x to find value of y and the find value of b.

For 2), similarly, use r = 17 (which is value of y) to find value of x using the second equation. And then plug it back to the third equation.

So D.
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Re: In the xy-plane, the straight-line graphs of the three equations above  [#permalink]

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10 Sep 2017, 19:04
1
Would it be correct to simply say that we have 4 variables with 3 equations so eliminating any one variable gets us to three equations and three variables and is therefore sufficient? Is that logic sound?
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Re: In the xy-plane, the straight-line graphs of the three equations above  [#permalink]

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13 Nov 2017, 20:21
2
what is the significance of the the line "In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r)",

i solved the problem, but wud have did the same even if they didnt provide line above . as question has 3 equations with 4 variables
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Re: In the xy-plane, the straight-line graphs of the three equations above  [#permalink]

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13 Nov 2017, 22:31
2
Cheryn wrote:
what is the significance of the the line "In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r)",

i solved the problem, but wud have did the same even if they didnt provide line above . as question has 3 equations with 4 variables

The highlighted statement in effect says that all these 3 lines meet each other at one point and so there is a single value of (x,y) that satisfies these 3 equations. It is only because of this highlighted statement you can solve this set of equations for a unique value of x,y, a and b.

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Re: In the xy-plane, the straight-line graphs of the three equations above  [#permalink]

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11 Dec 2017, 14:28
3
1
Hi All,

We're given the equations for 3 lines (and those equations are based on 4 unknowns: 2 variables and the 2 'constants' A and B):

Y = (A)(X) - 5
Y = X + 6
Y = 3X + B

We're told that the three lines all cross at one point on a graph (p,r). We're asked for the value of B. While this question looks complex, it's actually built around a 'system' math "shortcut" - meaning that since we have 3 unique equations and 4 unknowns, we just need one more unique equation (with one or more of those unknowns) and we can solve for ALL of the unknowns:

1) A =2

With this information, we now have a 4th equation, so we CAN solve for B.
Fact 1 is SUFFICIENT

2) R = 17

This information tell us the x co-ordinate where all three lines will meet, so it's the equivalent of having X=17 to work with. This 4th equation also allows us to solve for B.
Fact 2 is SUFFICIENT

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Re: In the xy-plane, the straight-line graphs of the three equations above  [#permalink]

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02 Jan 2018, 11:00
1
1
Bunuel wrote:
y = ax - 5
y = x + 6
y = 3x + b
In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?

(1) a = 2
(2) r = 17

We can begin by substituting p and r for x and y, respectively, in the three given equations.

1) r = ap – 5

2) r = p + 6

3) r = 3p + b

Statement One Alone:

a = 2

We can substitute 2 for a in the equation r = ap – 5. Thus, we have:

r = 2p – 5

Next we can set equations 1 and 2 equal to each other.

2p – 5 = p + 6

p = 11

Since p = 11, we see that r = 11 + 6 = 17

Finally, we can substitute 11 for p and 17 for r in equation 3. This gives us:

17 = 3(11) + b

17 = 33 + b

-16 = b

Statement one alone is sufficient to answer the question.

Statement Two Alone:

r = 17

We can substitute r into all three equations and we have:

1) 17 = ap – 5

2) 17 = p + 6

3) 17 = 3p + b

We see that p = 11. Now we can substitute 11 for p in equation 3 to determine a value for b.

17 = 3(11) + b

-16 = b

Statement two alone is also sufficient to answer the question.

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Re: In the xy-plane, the straight-line graphs of the three equations above  [#permalink]

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20 Jul 2019, 23:56
Video solution for the same

https://gmatquantum.com/official-guides ... ial-guide/
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Re: In the xy-plane, the straight-line graphs of the three equations above  [#permalink]

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17 Aug 2019, 23:34
Bunuel wrote:
y = ax - 5
y = x + 6
y = 3x + b
In the xy-plane, the straight-line graphs of the three equations above each contain the point (p,r). If a and b are constants, what is the value of b?

(1) a = 2
(2) r = 17

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Re: In the xy-plane, the straight-line graphs of the three equations above   [#permalink] 17 Aug 2019, 23:34
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# Resolving $\sec{x}\left(\sin^3x + \sin x \cos^2x\right)=\tan{x}$

Going steadily through my book, I found this exercise to resolve

$$\sec{x}\left(\sin^3x + \sin x \cos^2x\right)=\tan{x}$$

Here's how I resolve it ($LHS$) and again bear with me as I truly reverting to a feeling of vulnerability, like a child actually

As $\sec x$ is equal to $\frac{1}{\cos x}$

$$\frac{(\sin^3x+\sin x\cos^2x)}{\cos x}$$

I'm factorizing one $\sin x$

$$\frac{\sin x(\sin^2x+\cos^2x)}{\cos x} = \frac{\sin x(1)}{\cos x} = \tan x$$

That seems to work otherwise I completly messed this up

Reading the book's solution, I have something different...

\begin{align*} LHS&=\frac{\sin^3x}{\cos x}+ \sin x \cos x \\[4pt] &=\frac{\sin x}{\cos x}-\frac{\sin x\cos^2x}{\cos x}+\sin x\cos x\\[4pt] &= \tan x\end{align*}

What did I miss?

• They messed this up. – Yves Daoust May 13 '16 at 10:00
• Shorter: multiplying both members by $\cot(x)$ yields $\sin^2(x)+\cos^2(x)=1$. – Yves Daoust May 13 '16 at 10:04
• Hi @YvesDaoust you see this is where dogma starts. My way is better than your way and at the end, you are disgusting a generation of people but for what purpose...? – Andy K May 13 '16 at 10:15
• There is absolutely no intent to disgust anyone. You seem to miss that my first remark was an compliment to you. There is beauty in making things short. – Yves Daoust May 13 '16 at 10:22
• @YvesDaoust I understood correctly what you said. I was more empathetic about these 2 chaps who went up to create a complicate solution for what it was. They could have done something simplier and also mentionned that, there were other ways (pluralist views) to solve that problem. As we said J'avais bien compris :) – Andy K May 13 '16 at 10:24

So the book does about the same thing as you but in a different order (your solution is hence correct too). By the trigonometric one we get $\sin^2 x= (1-\cos^2 x)$. Thus $$\frac{\sin^3 x}{\cos x}=\frac{\sin x (\sin^2 x)}{\cos x}= \frac{\sin x(1-\cos^2 x)}{\cos x} = \frac{\sin x-\sin x\cos^2 x}{\cos x}= \frac{\sin x}{\cos x} -\frac{\sin x\cos ^2 x}{\cos x}$$ Thus $$LHS =\frac{\sin^3 x}{\cos x}+\sin x\cos x= \frac{\sin x}{\cos x} -\frac{\sin x\cos ^2 x}{\cos x} +\sin x\cos x = \frac{\sin x}{\cos x}=\tan x$$


# Given $S_n = \sum \dots$ and $a_n = \sum \dots$ prove that $a_n = S_n + {1\over n\cdot n!}$

I'm trying to solve the following problem:

Let: \begin{align} S_n &= 2 + {1\over2!} + {1\over3!} + {1\over 4!} + \dots + {1\over n!} \\ a_n &= 3 - {1\over 1\cdot2\cdot2!} - {1\over 2\cdot3\cdot3!} - \dots - {1\over (n-1)\cdot n\cdot n!} \end{align} Prove that: $$a_n = S_n + {1\over n\cdot n!}$$

My thought on this are:

Manipulating the sums here would become pretty hard so I decided to use another approach. I've tried to look what the first terms of each sum are going to be in order to observe a pattern:

\begin{align} S_1 &= 2 \\ S_2 &= 2 + {1\over 2!} = S_1 + {1\over 2!} \\ S_3 &= 2 + {1\over 2!} + {1\over 3!} = S_2 + {1\over 3!}\\ &\vdots \\ S_{n+1} &= S_n + {1\over (n+1)!} \end{align}

A similar thing is done to $$a_n$$:

$$a_{n+1} = a_n - {1\over n\cdot (n+1) \cdot (n+1)!}$$

So potentially if we could find closed forms of both $$S_n$$ and $$a_n$$ it would become easier to reason about them.

By the way, I've observed both of the sums approach $$e$$ at infinity but from different sides. $$S_n$$ is starting from $$2$$ adding more terms as $$n$$ grows, while $$a_n$$ starts from 3 and decreases the sum as $$n$$ grows. So that will be our initial conditions for both recurrences:

$$S_1 = 2 \\ a_1 = 3$$

The problem just reduced to finding closed forms of the recurrences. Both of them are non-homogenous and here is where I got stuck. Solving for homogenous part is easy since roots of both characteristic equations $$\lambda_{a_n} = \lambda_{S_n} = 1$$.

I couldn't find a particular solution for them. I've tried yet another approach expressing $$S_{n+1}$$ and $$S_{n+2}$$ trying to get rid of the $$1\over (n+1)!$$ or at least change its form so the guess for particular solution is more obvious.

My questions are:

1. Is it even a valid approach to solve the problem?
2. If so then how could I find a closed form for the recurrences? Especially i'm interested in finding a particular solution. (Yet a complete flow is very appreciated and encourages learning by example)
3. Not sure how to phrase that better, but does there exist a table of "guesses" for particular solution of the recurrences in some form of $$x_{n+1} = x_n + F(n)$$. Like if the "free term" is a constant say $$F(n) = 2$$ then the guess for particular solution is some constant $$B$$.

Please excuse me if there is some vagueness or inaccuracy in the terminology, English is not my native language and I have almost no background in Maths.

• It seems unlikely to find a simple closed form, because, as you noticed, the limit converges to $e$. – user600464 Oct 5 '18 at 18:59
• @user600464 is that because $e$ is irrational and that is why no "simple" formula for an irrational number may be obtained? – roman Oct 5 '18 at 19:02
• There is actually a simple formula, the limit $\frac{n^n}{(n- 1)^n}$. However, to compare to this sums, we would need many terms – user600464 Oct 5 '18 at 19:34

$$a_2 - S_2 = 3- 1/4 - (1 + 1/1! + 1/2!)$$ [I think there is some typo in your expression of $$S_n$$] $$= 1/4 = 1/(2 \cdot 2!)]$$. Suppose the equation holds for $$n$$. Then for $$n+1$$, \begin{align*} a_{n+1} - S_{n+1} &= a_n - S_n - \frac 1 {n(n+1) \cdot (n+1)!} - \frac 1{(n+1)!} \\ &= \frac 1 {n!n }-\frac 1 {n(n+1) \cdot (n+1)!} - \frac 1{(n+1)!} \\ &= \frac 1 {(n+1)! (n+1)} \left( \frac {(n+1)^2}n -\frac 1n -(n+1) \right)\\ &= \frac 1 {(n+1)! (n+1)}. \end{align*} Thus the equation holds for all $$n \geqslant 2$$ by induction principle.
\begin{align} \color{red}{a_{n}-S_{n}} &=\left(3-\sum_{k=2}^{n}\frac{1}{(n-1)\cdot n\cdot n!}\right)-\left(2+\sum_{k=2}^{n}\frac{1}{n!}\right) \\[2mm] &=1-\sum_{k=2}^{n}\frac{1}{n!}\left(1+\frac{1}{(n-1)\cdot n}\right) =1-\sum_{k=2}^{n}\frac{1}{n!}\left(1+\frac{1}{n-1}-\frac{1}{n}\right) \\[2mm] &=1-\sum_{k=2}^{n}\frac{1}{n!}\left(\frac{n}{n-1}-\frac{1}{n}\right) =1-\sum_{k=2}^{n}\left(\frac{1}{(n-1)\cdot (n-1)!}-\frac{1}{n\cdot n!}\right) \\[2mm] &=1-\frac{1}{1\cdot1!}+\frac{1}{2\cdot2!}-\frac{1}{2\cdot2!}+\frac{1}{3\cdot3!}-\cdots+\frac{1}{n\cdot n!}=\color{red}{\frac{1}{n\cdot n!}}\quad\left\{\text{telescoping}\right\} \end{align}


# In proof writing, is it mathematically sound to prove uniqueness before proving existence?

As stated in the title, I'd like to find out is whether or not it is always mathematically sound to prove the uniqueness of something before proving the existence of said something.

I am still relatively new to proofs and, in this light, I was intrigued by an observation that some authors choose to prove uniqueness before existence. I understand that some do this because, allegedly, "it is easier to prove the uniqueness part first". And in an exam setting, it's better to tackle the easier parts of a task.

However, as I see it, if done this way, aren't we essentially proving a proposition about something that, at that point, we have not yet guaranteed exists? (Thus, might not even exist at all). I understand that in most cases the existence is proved immediately after the proof of uniqueness but, generally, doesn't uniqueness depend on existence?

My example will be derived from page $$40$$ of The Real Analysis Lifesaver: All the Tools You Need to Understand Proofs by Raffi Grinberg:

Theorem 5.8. (Existence of Roots in $$\mathbb{R}$$)

Every positive real number has a unique positive $$n$$th root, for any $$n \in \mathbb{N}$$.

In symbols:

$$∀x ∈ R$$ with $$x > 0, ∀n ∈ N, ∃y ∈ R$$ unique, such that $$y > 0$$ and $$y^n = x.$$

Note that even-numbered roots $$(\sqrt{x},\sqrt[4]x,\sqrt[6]x, etc.)$$ signify two numbers in $$\mathbb{R}$$, namely, $$+y$$ and $$−y$$. The theorem asserts that there is one and only one positive real root.

Proof.

The uniqueness of $$y$$ is the easiest part of the proof, so we’ll start there. For any two different positive real numbers, the fact that they are different means one must be greater than the other. If there were two positive real roots $$y_1$$ and $$y_2$$ such that $$y_1^n = x$$ and $$y_2^n = x$$ we would have $$0 < y_1 < y_2$$. But then $$0 < y_1^n < y_n^2$$ , meaning $$0 < x < x$$, which is impossible. Thus only one positive real root can exist.

To prove that $$\sqrt[n]x$$ exists in $$\mathbb{R}$$, let’s first figure out our game plan and then write it up formally $$...$$

While the proof is totally clear, I'd really like feedback on this format. As an aspiring mathematician, is this a format I can adapt and use in my proofs or perhaps it is not good practice in general? What if, say, I have a conjecture that there exist unique integers possessing some property and this conjecture is something new (i.e. hasn't been proven yet). I try proving they exist but run out of luck. Would it be worth the effort of trying to prove the uniqueness of such integers not knowing they exist?

• Yeah, that's perfectly fine. It doesn't matter: you're proving two things: 1) There is at least one of these; and 2) there aren't two different ones. There's no problem doing them whatever way round you like. There are a fair few results of the form "there are at most $n$ things with property X", and this is just the special case where $n = 1$. – user3482749 Jan 4 at 17:57
• Sure. Naturally, if you show uniqueness first what you have actually shown at that point is that there is at most one object with the desired property. – Andrés E. Caicedo Jan 4 at 17:57
• Even in the case where authors first prove uniqueness, they usually mean, under the assumption that there exists such an object, it is unique. – Dietrich Burde Jan 4 at 18:01
• I prefer existence before uniqueness, but that is just a preference and not a strong recommendation. – David G. Stork Jan 4 at 18:02
• An explanation: I prefer existence before uniqueness because in general existence is simpler. Imagine the awkward case where you prove uniqueness first, then find you cannot prove existence! THAT is weird. – David G. Stork Jan 4 at 18:16

I can see the unease: if what you're proving unique doesn't exist, might that invalidate whatever manipulations you've done with it?

This is similar to the situation with proof by contradiction, where you don't just reason about something that might not be true: you reason about it expecting it to be untrue, but making all the logical steps anyway.

In either case, you end up showing "If $$A$$ then $$B$$"—whether it's "If $$x$$ exists then it's unique" or "If $$x$$ exists then it has contradictory properties". In one case we go on to apply $$B$$ to $$x$$ once we know it exists, and in the other we use the impossibility of $$B$$ to disprove the existence of $$x$$.

A separate thought: proofs are presented as a sequence of steps and inferences because as humans, we process written information sequentially. But what's behind a mathematical theorem is more like a network of theorems and logical steps, some dependent on others, which together imply the theorem. We can't take them all in simultaneously, so we arrange them in a suitable order to let us see the logic most clearly. It's a matter of convenience.

If $$X$$ and $$Y$$ are both true and together imply $$Z$$, it doesn't really matter which order the various parts of that are proved in—all that matters from a validity point of view is that they're all proved and brought together.

• Thanks! Your first paragraph alone asks my question even better than I could in my long post! – E.Nole Jan 7 at 2:56

This is a perfectly reasonable argument. The author even tells you he's doing it this way. He then uses the argument to allow the reader to become comfortable with the assertion and the notation.

There are other reasons for starting with uniqueness. Often if you know there's just one of something then it might be easier to search for it.

When faced with a problem with an easy part and a hard part the argument for starting with the easy part is that it's a good warmup. That's what I usually do, and what I like to see in things I read.

The argument for starting with the hard part is that if you fail there you haven't wasted time on the easy part. For example, proving that there's at most one positive rational number whose square is $$2$$ is not going to tell you very much about the rational square root of $$2$$ that does not exist.

Starting with the hard part, the bottleneck, is often the preferred strategy when writing software.

PS The author is wrong to say that when $$x >0$$, $$\sqrt{x}$$ signifies two two numbers. $$4$$ does have two real square roots, but the mathematical convention is that only $$2 = \sqrt{4}$$.

• Nice typo near the end: "two two" (Shame to fix it really!) – timtfj Jan 4 at 19:15
• @timtfj I will leave it thanks. – Ethan Bolker Jan 4 at 19:33


# Ideal Generated by a Finite Number of Polynomials?

Background

I have been confused about a particular definition in the textbook for my abstract algebra class, Ideals, Varieties, and Algorithms by Cox, Little, and O'Shea. It is frustrating because I feel that I have a partial grasp on what the definition is trying to say, but when it comes down to it I simply find my confused. So, instead of suffering by myself for any longer with this definition, I have decided to turn to you lovely folks to help me understand this seemingly simple concept.

What I Understand

First off, I understand (at least in the context of this book) what an ideal is. The definition my book gives for an ideal is

Definition. A subset $$I\subseteq k[x_1,..., x_n]$$ is an ideal if it satisfies:

(i) $$0\in I$$.

(ii) If $$f,g\in I$$, then $$f+g\in I$$.

(iii) If $$f\in I$$ and $$h\in k[x_1,...,x_n]$$, then $$hf\in I$$.

I find this to be a simple, easy to understand definition. My problem, however, arises a few lines later when they define an ideal generated by a finite number of polynomials.

What I Don't Understand

And now, I give you the definition that has been causing me an incredible amount of confusion and frustration.

Definition. Let $$f_1,...,f_s$$ be polynomials in $$k[x_1,...,x_n]$$. Then we set $$\langle f_1,...,f_s \rangle=\Big\lbrace \sum_{i=1}^s h_if_i \ | \ h_1,...,h_s\in k[x_1,...,x_n] \Big\rbrace.$$

I know what you are thinking. How does he not understand this? I wish I knew the answer to that question, but in the meantime, can someone please help me visualize what this set looks like? I understand that $$\langle f_1,...,f_s \rangle$$ is an ideal, but I don't understand its structure, if that makes sense. In other words, I can't visualize this definition in a way that makes sense to me. The authors do make a slightly helpful note, saying that "we can think of $$\langle f_1,...,f_s \rangle$$ as consisting of all 'polynomial consequences' of the equations $$f_1=f_2=...=f_s=0$$."

To elaborate a little more on my confusion, what I'm asking for is a less "compact" definition. When I read this definition, for whatever reason the only thing I can come up with is $$f_1h_1+f_2h_2+...+f_sh_s.$$ But that doesn't make sense, because $$\langle f_1,...,f_s \rangle$$ is supposed to generate a set, not just a single polynomial.

As always, thank you all for your time. If you find this to be a stupid or silly question, then I'm sorry to have disappointed you -- I'm a slow learner, and I get hung up on stupid things sometimes.

Oh, and Happy Halloween!

• Suggestion: Try looking at one or two specific examples, such as: (1) In $k[x,y]$, look at $\langle x^2, y^2 \rangle$. This is equal to the set $\{ x^2 h_1 + y^2 h_2 \mid h_1, h_2 \in k[x,y]\}$. Some elements include $x^2 \cdot 1 + y^2 \cdot 0$, $x^2 \cdot 0 + y^2 \cdot 1$,... $x^2 \cdot (x y^2 - 3 xy) + y^2 \cdot (2x^3 - 5)$, etc. (2) Look at examples in $\mathbb{Z}$, e.g., $\langle 6,4 \rangle = \{ 6 h_1 + 4 h_2 \mid h_1, h_2 \in \mathbb{Z} \}$. Can you prove this is equal to the set of even numbers? – Zach Teitler Nov 1 '17 at 4:55
• Makes alot more sense when you put it like that! Thank you @ZachTeitler! – Thy Art is Math Nov 1 '17 at 5:00
• One more suggestion. Make sure you do Exercise 1.4.2, or at least try to do it. The result in Exercise 1.4.2 gets used constantly, over and over, throughout the rest of the book, including in a lot of exercises. It's the main way to show that two ideals are equal, or that one ideal is contained in another. So it is extremely useful. Also it's good practice with the definition of ideal, and proving a subset relationship. (There are a lot of other good exercises, I just want to highlight that particular one. I made sure to assign it in the class I'm teaching from IVA this semester.) – Zach Teitler Nov 1 '17 at 5:49
• Read the Rings chapter in Algebra by Hungerford. He describes ideals in full generality very well (non-commutative and commutative). – user494247 Nov 11 '17 at 9:57

Would it help to see the see the set with explicit polynomials in place of the $f_1, f_2, \dots, f_n$?

For example, lets look at an explicit example when $n = 2$, so an ideal generated by 2 polynomials. Also, we'll work in a polynomial ring in two variables over $k$, i.e. $k[x,y]$.

Here is the ideal generated by the polynomials $x^2 - 1$, $yx+x$.$$\langle \, x^2 - 1, \, yx+x \, \rangle = \{ \, f\cdot(x^2 - 1) + g\cdot(yx + x) \, | \, f,g \in k[x,y] \, \}.$$

So the elements of the set are any polynomial that can be written in the form $f \cdot (x^2 - 1) + g \cdot (yx+x)$. But we can choose $f,g$ to be $\textit{any}$ polynomial we want. For example, we know $x^2 - 1$ itself is in that set because we can choose $f = 1, g = 0$. We also know that the polynomial $x^3-x+yx ^2 + x^2$ is in the set, because we can choose $f = x, g = x$.

If you are familiar with linear algebra you can maybe, in a way, draw a connection between an ideal generated by polynomials and the span of a set of vectors. You can think of it as, the set of all things that can be made from the objects defining it.

• Also, is your class an Algebraic Geometry class or an Abstract Algebra class? I think IVA is a strange book to teach general algebra out of. – Prince M Nov 1 '17 at 7:38
• Sadly I was absolutely terrible at linear algebra but seeing the explicit polynomials made this make so much more sense, so thank you for the help. The class is labeled rather generally as "abstract algebra," so you are probably right that the course should be called "algebraic geometry". – Thy Art is Math Nov 1 '17 at 15:30
• Just a note that the authors of IVA do write in the preface: "For instance, the book could serve as a basis of a second course in undergraduate abstract algebra, but we think that it just as easily could provide a credible alternative to the first course." – J W Jan 3 at 15:16

Perhaps a good way to understand this (or to understand anything, for that matter) is to look at examples.

The definition of ideal makes sense in any ring, so let’s look at $\Bbb Z$ first. Every ideal, you soon persuade yourself, is a set $d\Bbb Z\subset\Bbb Z$, that is, just the set of all multiples of a given number $d$. What if you try to take two numbers and look at $\langle d_1,d_2\rangle\subset\Bbb Z$? Please do this with specific numbers. Like what about $d_1=8$ and $d_2=6$? You want to describe the totality of all numbers writable in the form $8m+6n$. You rapidly see that this is an ideal in the sense of the definition, and you see that the set is equal to $2\Bbb Z$. This is basic number theory.

Now do the same thing with a ring of polynomials, but in just one variable, $R=\Bbb Q[x]$. One proves (you prove it, with Euclidean division of polynomials) that every ideal of $R$ is of form $fR$, where $f$ is a well-chosen polynomial. In fact, for a nonzero ideal $I$, you take $f$ to be a nonzero element of $I$ of least degree. So, what is $\langle f_1,f_2\rangle$, when $f_1$ and $f_2$ are two polynomials in $x$ that are given? Just as with numbers, it’s the ideal $gR$ where $g$ is the greatest common divisor of $f$ and $g$. Convince yourself that $\langle x^3-1,x^2-2x+1\rangle$ is the set of all multiples of $x-1$.

Things are no longer so simple when you have polynomials in more than one indeterminate. But at least you get some insight by looking at the simplest cases, and I hope you see that the set of all possible $h_1f_1+\cdots+h_nf_n$ is an ideal.

• @ThyArtisMath Note that the ring of polynomials in one variable is treated very explicitly and thoroughly in section 1.5 of the textbook Ideals, Varieties, and Algorithms. So, by all means feel free to think about that example and take a shot at proving some things about it, but don't feel like you have to work it out on your own. The textbook goes over all that. – Zach Teitler Nov 1 '17 at 5:40

You were close.

The ideal $(f_1,...,f_s)$ of $k[x_1,..., x_n]$ is the set of all polynomials which can be expressed as $h_1f_1+\cdots +h_sf_s$, for some $h_1,...,h_s \in k[x_1,..., x_n]$.

In other words, "linear combinations" of $f_1,...,f_s$ where the "linear coefficients" are arbitrary elements $h_1,...,h_s$ of $k[x_1,..., x_n]$.

Regarding the author's point about common zeros . . .

If $\bar{k}$ is an algebraically closed extension of $k$, and if there is some $a = (a_1,...,a_n) \in \left(\bar{k}\right)^n$ such that $f_1(a) = \cdots = f_s(a) = 0$, then $a$ is automatically a zero of any polynomial of the form $h_1f_1+\cdots +h_sf_s$, hence $a$ is a common zero for all members of the ideal $(f_1,...,f_s)$.

In particular, if you've identified a common zero $a \in \left(\bar{k}\right)^n$ for $f_1,...,f_s$, then if $g \in k[x_1,..., x_n]$ is such that $g(a) \ne 0$, it follows that $g$ is not in the ideal $(f_1,...,f_s)$.

An important, nontrivial result is a partial converse: If $g \in k[x_1,...,x_n]$ is such that $g,g^2,g^3,...$ are not in the ideal $(f_1,...,f_s)$, there is some $a \in \left(\bar{k}\right)^n$ such that $a$ is a common zero of $f_1,...,f_n$, but $a$ is not a zero of $g$.


# Solving this inequality

Question:

Solve: $$\frac{5x-6}{x+6}<1$$

My attempt:

$$\frac{5x-6-x-6}{x+6}<0$$ $$\Rightarrow \frac{4x-12}{x+6}<0$$ $$\Rightarrow \frac{x-3}{x+6}<0$$ $$\Rightarrow (x-3)(x+6) < 0$$

Thus, it implies that the answer is $-6 < x < 3$. However, in my textbook, it is given as wrong answer.

Can please someone tell what is the correct procedure for this.

Thanks.

If you're really unsure when something like that happens, you can always graph your rational function. But consider that in $\ \frac{x-3}{x+6} \$ , there is a vertical asymptote at $\ x \ = \ -6 \$ and an $\ x-$ intercept at $\ x = 3 \$ . So the real numbers are divided into intervals by these points, $\ x \ < \ -6 \ , \ -6 \ < \ x \ < \ 3 \ ,$ and $\ x \ > \ 3 \$ . Also, your rational function has a horizontal asymptote of $\ y \ = \ 1 \$ .

So the function is positive for "a lot" of the real numbers to start with. We can also use what we know about working with signed numbers. For large negative numbers, both the numerator and denominator are negative, and for large positive numbers, they are both positive. So for $\ x \ < \ -6 \$ and $\ x \ > \ 3 \$ , we can conclude that the ratio is positive. It will only be in the interval $\ -6 \ < \ x \ < \ 3 \$ that the numerator is negative, while the denominator is positive, so the ratio is negative in this interval.

So you are correct. Large introductory course textbook answers can be wrong anywhere from about $\ \frac{1}{3}$ % to $\ 2$ % of the time, depending upon how carefully the submitted answers (of multiple authorship) have been combed through...

• Thanks for the detailed method! And yes, you are right, the textbook is indeed large, so the author might have missed it. – Gaurang Tandon May 18 '14 at 19:47
• The authors (probably of just about all of the omnibus texts, and not just in math) generally hire teaching assistants to work out answers, which are then collected into the backs of these books. They do not check what can be two to three thousand problem answers for accuracy... – colormegone May 18 '14 at 19:54

Hint: There are two cases, either $x < -6$ or $x > -6$. You can multiply both sides of the inequality by $x + 6$ and depending on which case you are considering, either it will reverse the inequality or leave it intact. Then you can rewrite the inequality with $x$ on one side and a number on the other side by adding/subtracting/dividing.
• I am sorry but I do not understand you. The denominator has $(x+6)$, not $x-6$. – Gaurang Tandon May 18 '14 at 18:40
• @GaurangTandon Sorry, I should have written $-6$. The point is that there are two cases, either $x+6$ is positive or negative and it will flip the inequality or not depending on what case you're in. – user2566092 May 18 '14 at 18:59