Equivalence class problem

**sarahh** · July 10th 2006, 03:38 PM

...

**BubbleBrain_103** · July 10th 2006, 09:03 PM

c)

the only such I could find was [t] (obviously, since t^2 - 5 is a multiple of t^2 - 5). Here's my reasoning for suspecting there are no others:
suppose [a+bt]^2 = [5] for some a,b in Q. then we have
(a+bt)^2 - 5 = q(t^2 - 5) for some polynomial q, and rewriting:

a^2 + 2abt + (b^2)(t^2) = qt^2 + 5(q-1)

since the poly on the left is of second degree, so is the poly on the right... hence q must be of degree zero (a number), lest we have a t^3 term. Further, suppose that a is non-zero. Since the right has no t term, we must have 2ab = 0, and hence b = 0, and thus q = 0. then we are left with a^2 = -5, and there is certainly no such a. Thus a must be zero. so we are left with
b^2 t^2 = qt^2 + 5(q-1)

but now 5(q-1) must vanish, and so q = 1, and b=1. This was our original choice.

f)

a zero divisor is by definition a non-zero element which can be multiplied with another non-zero element to yield zero. if f and g are such that

fg = q(t^2 - 5)

for some q, then since t^2 - 5 is irreducible, it must be a factor of either f or g, making one of their equivalence classes equal to [0] = [t^2-5]. So there are no zero divisors.

g)

The equivalence class [a+bt] has an inverse precisely so long as a and b are not both zero. I'll leave out the calculation, but given any a and b not both zero, let

c = -a/(5b^2 - a^2) d = b/(5b^2 - a^2)

and you can show that (a + bt)(c + dt) is in the equivalence class of 1.

Hope that helps, don't crucify me if there are some typos, I did this late at night. Look closely at c), if your teacher says there's another one there very well may be, but I'd like to know where my reasoning is flawed. take care

**BubbleBrain_103** · July 10th 2006, 09:37 PM

oh, duh... for part c), you still need q = 1, but then b could be -1. So [-t] is the other possibility.

I should also note that a lot of the general questions fall easily from the fact that t^2 - 5 is irreducible. This implies that the ideal generated by t^2 - 5 is maximal in Q[t], and so the quotient ring Q[t]/(t^2 - 5) (the one we're looking at) must be a field. This gives you inverses, no non-zero divisors, and the fact that if a^2 = b^2, then a = b or a = -b, which might help you with h).

**ThePerfectHacker** · July 11th 2006, 09:22 AM

Originally Posted by sarahh

In Q[t], define the equivalence relation ~ as f(t) ~ g(t) precisely when f(t) - g(t) is a multiple of t^2 - 5. We define the addition and multiplication of equivalence classes as [f(t)] + [g(t)] = [f(t) + g(t)] and [f(t)] * [g(t)] = [f(t) * g(t)]
(Assume: ~ is an equivalence relation, Addition/Mutliplication of equivalence classes is well-defined, and every equivalence class contains exactly one element of the form a + bt, where a, b in Q)

d) Find a, b in Q such that [a + bt]^2 + [-2][a + bt] = [19] (Two possible answers)

We have,
$[a+bt]^2=[a^2+2abt+b^2t^2]$
and,
$[-2][a+bt]=[-2a-2bt]$
Thus,
$[b^2t^2+2abt-2bt+a^2-2a]=[19]$
Now these two equivalence classes are equal iff,
$b^2t^2-2abt-2bt+a^2-2a$ ~ $19$
Iff,
$b^2t^2+2abt-2bt+a^2-2a-19=k(t^2-5)$
Thus,
$b^2t^2+2abt-2bt+a^2-2a-19=kt^2-5k$
Organize as,
$(b^2)t^2+(2ab-2b)t+(a^2-2a-19)=kt^2-5k$
Since, $\mathbb{Q}$ is an infinite field, the two poynomials can only be equal when,
$\left\{ \begin{array} {c} b^2=k\\ 2ab-2b=0\\ a^2-2a-19=-5k$
The rational solutions are,
$(a,b)=(1,2),(1,-2)$
(there is another solution but it is not rational)
[Should I show how to solve these equations?]

**ThePerfectHacker** · July 11th 2006, 11:33 AM

Originally Posted by sarahh

Yes please. Thank you ThePerfectHacker. If it wouldn't be too much trouble could you do part c) and e) --h) like that? Your solution is very streamlined. If I could see the others worked out like this, I'd be well on my way!

First thing is that do you understand how I reached,
$<br /> \left\{ \begin{array} {c} b^2=k\\ 2ab-2b=0\\ a^2-2a-19=-5k<br />$ ?

If you do, good; I will countinue with the solution.
In the second equation we have,
$2ab-2b=0$ thus,
$2ab=2b$ divide by 2, $ab=b$ . Now this is when it gets tricky.

If $b\not = 0$ then you can divide by $b$ to get, $a=1$ , then in the third equation,
$1^2-2(1)-19=-5k$ , thus, $k=4$ .
Then in the first equation $b^2=4$ thus, $b=\pm 2$ . Now note that,
$a,b,k\in \mathbb{Q}$ so these solutions exist. Thus, $(a,b)=(1,2),(1,-2)$ .

If $b=0$ then you cannot divide by $b$ . Rather procede to the first equation $0^2=k$ thus, $k=0$ then it in the third equation,
$a^2-2a-19=0$ since the discrimant is $4+4(19)=80$ which is not a square the solutions are irrational. Thus, what we have $b,k\in \mathbb{Q}$ and $a\not \in \mathbb{Q}$ . Thus, there are no solutions for $b=0$ .
----
Should I re-explain BubbleBrains solutions or do you understand them (he did explain them well).

**ThePerfectHacker** · July 11th 2006, 06:26 PM

Originally Posted by sarahh

Thanks BubbleBrain for the help!! I'm pretty sure I got c) and e) as well. Just to clarify, so there are no zero divisors in f)? Woah...trick question! Also, could you do the calculation for part g) (Which equiv. classes have multiplicative inverses?)

f) is an interesting question.
----
In a ring $<\mathcal{R},+,\cdot >$
Definition: The elements $a,b\in \mathcal{R}$ are called zero-divisors whenever,
$ab=0$ where 0 is the identity element in the group $<\mathcal{R},+>$ and $a,b\not = 0$ .

Now we note that the equivalence class, $[0]$ is an idenity element because $[f(t)]+[0]=[0]+[f(t)]=[f(t)]$ .
Given a non-zero element $[f(t)]$ we need to find another non-zero element $[g(t)]$ such as, $[f(t)][g(t)]=[0]$
Thus, by definition of these equivalence classes
$f(t)g(t)=k(t^2-5)$
Now, note that the polynomial $t^2-5$ divides
$f(t)g(t)$ . Also note the most important feauture (as Bubblebrain said; I am not trying to steal his solution) that it is irreducible over $\mathbb{Q}$ (use Einstein Criterion, or note that $t^2-5=0$ has no rational solutions). Now, the fundamental property of an irreducible polynomial is that if two polynoms are divisible by it then one of them is divisible by it. Thus either $f(t)$ or $g(t)$ is divisible by $t^2-5$ but if one of is divisible by $t^2-5$ means that $f(t) \mbox{ or }g(t)=k(t^2-5)$ for some $k\in \mathbb{Q}$ then by definition equivalence classes $[f(t)]\mbox{ or }[g(t)]=[t^2-5]$
But, $[t^2-5]=[0]$ therefore, $[f(t)]\mbox{ or }[g(t)]=[0]$ ---->A zero element, which means there cannot be any because zero divisiors cannot be zero.

**BubbleBrain_103** · July 11th 2006, 09:56 PM

sarahh,

this is in regards to h). As it turns out, [6] has no square roots. What you have so far is good. So let's do some detective work.

we want to show that there do not exist any a, b, k, where a and b are rational and k is in Q[t], such that your equations hold. So suppose they do:

b^2 = k
2ab = 0
a^2 - 6 = -5k

Notice first that if k = 0, we would be left with a^2 - 6 = 0, which is impossible (no rational squared can equal 6). So k is non-zero, and b must also be non-zero. But then a must be zero, since 2ab = 0. And that leaves us with -6 = -5k, so k must be equal to 6/5. But then we have b^2 = 6/5... impossible again, since b is rational. And we are out of options.

Thus these equations can never hold, so [a+bt]^2 is never [6] for any [a+bt].

I'll try to answer the other part of your question some other time, or maybe hacker can help... I'm currently typing with one hand after slicing the other with a can opener. But no worries. And this sort of reasoning is similar to what is going on with g)... just follow the equations, narrow down the possibilities on what c and d could be. take care

**BubbleBrain_103** · July 12th 2006, 01:13 PM

Hi sarahh,

I assume your asking for help on g). The question asks, "which equivalence classes have inverses?" As I said, all equivalence classes have inverses except for [0].

Given a non-zero equivalence class [a+bt] (at least one of a or b not zero), its inverse [c + dt] is given by the formula

c = -a/(5b^2 - a^2) d = b/(5b^2 - a^2)

we can show this just by brute force calculation:

(a+bt)(c+dt) - 1 = (a+bt)(-a/(5b^2 - a^2) + bt/(5b^2 -a^2)) - 1

= (a+bt)(-a + bt)/(5b^2 - a^2) - 1

= (-a^2 + (ab-ba)t + b^2 t^2)/(5b^2 - a^2) - 1

= (-a^2 + b^2 t^2)/(5b^2 - a^2) - 1

= (-a^2 + b^2 t^2)/(5b^2 - a^2) - (5b^2 - a^2)/(5b^2 - a^2)

= (-a^2 + b^2 t^2 - 5b^2 + a^2)/(5b^2 - a^2)

= (b^2 t^2 - 5b^2)/(5b^2 - a^2)

= (b^2/(5b^2 - a^2)) * (t^2 - 5)

This last quantity is a constant (b^2/(5b^2 - a^2)) times t^2 - 5, which shows that (a+bt)(c+dt) is in the same equivalence class as 1. This is what it means for [(c+dt)] to be the inverse of [(a+bt)]. [1] acts as a multiplicative identity in this structure, and when you multiply [(a+bt)] and [(c+dt)] together, you get [1].

We've made a tacit assumption however, namely that our formulas for c and d are defined... this will be the case so long as the denominator (5b^2 - a^2) is not zero, which will happen when 5b^2 = a^2. But this can only happen when b and a are both zero. Why? If one is zero, then clearly the other must be zero... so suppose they are both non-zero. Then we are left with

(a/b)^2 = 5

But this is impossible, since a and b are both rational, thus a/b is rational.. but no rational squared can equal 5.

This just shows that your formulas for c and d will work for any a and b you choose, so long as they are not both zero.

Now, at this point, the problem is done, you've shown all that needs to be shown. But you are probably wondering where I came up with these formulas, and your teacher might also. We are looking for c and d such that

(a+bt)(c+dt) - 1 = k(t^2 - 5)

for some polynomial k. Rewriting:

ac + (bc + ad)t + bd t^2 = k t^2 - 5k + 1

From this we can infer that:

1) k = bd
2) bc + ad = 0
3) ac = 1 - 5k

substituting 1) in 3) and solving for c:

ac = 1 - 5bd

c = (1- 5bd)/a

Substituting this value for c back into 2) :

b(1-5bd)/a + ad = 0

This equation is linear in d, and can be solved as:

d = b/(5b^2 - a^2)

Finally substituting this back into c = (1-5bd)/a :

c = -a/(5b^2 - a^2)

and there you have it.

**BubbleBrain_103** · July 12th 2006, 08:24 PM

for h):

first you show that k can't be zero, so b can't be zero. This shows that a must be zero, since 2ab = 0. Then you get -6 = -5k, so k = 6/5. This leaves you with b^2 = 6/5. But this is impossible, since b is rational. So no, there isn't any class that you can square to get [6].

It is important to understand that these equations CAN hold for arbitrary real numbers... it's the fact that a,b and k are constrained to be rational that makes it impossible.

Also, you can forget about what I said in the beginning... I said it might help you, turns out it doesn't. I only said it because, if you were to have found one square root of [6], the other would be its negative, and you could be sure at that point that you'd found all of them.

for g):

yes, the equivalence class [c+dt] is the multiplicative inverse of the equivalence class [a+bt].

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