How many rational members
- Thread starter Loki123
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Because you and your teacher are thinking along different lines.
You say k must be divisible by six because you are thinking of k as the exponent [imath]\sqrt[3]{2}[/imath]. But really [imath](\sqrt[3]{2})^k[/imath] gives a rational number whenever k is divisible by 3. It is the necessity (thinking your way) that [imath](\sqrt{3})^{(20 - k)}[/imath] be rational that leads you to k must also be divisible by 2. So one of your terms, where k = 6, gives some binomial times 2^2 * 3^7
Your teacher says (k - 2) must be divisible by 6. Why? Your teacher is thinking of k as the exponent of [imath]\sqrt{3}.[/imath]
And [imath](\sqrt{3})^k[/imath] is rational whenever k is divisible by 2. It is the necessity (thinking your teacher’s way) that [imath](\sqrt[3]{2})^{(20-k)}[/imath] be rational that leads your teacher to (20 - k) be divisible by 3. So (20 - k} must be 18, 15, 12, 9, 6, 3, or 0. But 15, 9, and 3 are impossible because k must be even. Thus we get k = 2, 8, 14, or 20.
Now let’s see what we get when k = 14. That gives a term of some binomial * 3^7 * 2^3.
We get the same powers of the radicals but in a different order.
That leaves the binomials to deal with. Your k values are 0, 6, 1, and 18. Your teacher’s k values are 20, 14, 8, and 2. In short your teacher’s values are the difference between 20 and your values. But
[math]\dbinom{20}{k} = \dbinom{20}{20 - k}.[/math]
You say k must be divisible by six because you are thinking of k as the exponent [imath]\sqrt[3]{2}[/imath]. But really [imath](\sqrt[3]{2})^k[/imath] gives a rational number whenever k is divisible by 3. It is the necessity (thinking your way) that [imath](\sqrt{3})^{(20 - k)}[/imath] be rational that leads you to k must also be divisible by 2. So one of your terms, where k = 6, gives some binomial times 2^2 * 3^7
Your teacher says (k - 2) must be divisible by 6. Why? Your teacher is thinking of k as the exponent of [imath]\sqrt{3}.[/imath]
And [imath](\sqrt{3})^k[/imath] is rational whenever k is divisible by 2. It is the necessity (thinking your teacher’s way) that [imath](\sqrt[3]{2})^{(20-k)}[/imath] be rational that leads your teacher to (20 - k) be divisible by 3. So (20 - k} must be 18, 15, 12, 9, 6, 3, or 0. But 15, 9, and 3 are impossible because k must be even. Thus we get k = 2, 8, 14, or 20.
Now let’s see what we get when k = 14. That gives a term of some binomial * 3^7 * 2^3.
We get the same powers of the radicals but in a different order.
That leaves the binomials to deal with. Your k values are 0, 6, 1, and 18. Your teacher’s k values are 20, 14, 8, and 2. In short your teacher’s values are the difference between 20 and your values. But
[math]\dbinom{20}{k} = \dbinom{20}{20 - k}.[/math]
I don't think we understand each other.
The formula I know is:
my teacher puts y where x is and x where y is. so it's y^(n-k)x^k. This is what results in me getting a different answer.
Dr.Peterson
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Exactly what I would have said.
So, are you saying you have answered your own question, or is there still an issue?
[imath]\text{ }\\ (x+y)^n=(y+x)^n\\ \text{ }\\ \sum \limits_{k=0}^{n} \binom{n}{k}x^{n-k}y^k= \sum \limits_{k=0}^{n} \binom{n}{k}y^{n-k}x^k\\ \text{ }\\ (3^{\tfrac{1}{2}}+2^{\tfrac{1}{3}})^{20}=(2^{\tfrac{1}{3}}+3^{\tfrac{1}{2}})^{20}\\ \text{ }\\ (3^{\tfrac{1}{2}}+2^{\tfrac{1}{3}})^{20}=\sum \limits_{k=0}^{20} \binom{20}{k}3^{\tfrac{20-k}{2}}2^{\tfrac{k}{3}}\\ \text{ }\\ (2^{\tfrac{1}{3}}+3^{\tfrac{1}{2}})^{20}=\sum \limits_{k=0}^{20} \binom{20}{k}2^{\tfrac{20-k}{3}}3^{\tfrac{k}{2}}\\[/imath]
No, the question I have written on the paper still remains. How can he just switch places?? Also, Is my answer wrong? Would I get marked as right or wrong If I handed in the question done my way? If the question was what member is a rational number, then I would get four different members than my teacher, then what?
Dr.Peterson
Elite Member
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Of course you'd be marked as right. (In fact, I'd probably have done it your way.) You got the correct answer, by an equivalent method. I think that's what everyone has been saying.
He can switch the places because of the commutative property.
If you were asked to list the terms (that, rather than "member", is the appropriate word here), you would both list the same ones. You might number them differently, but that is only because you are listing them in a different order.
If you have been taught that an expansion must be listed in one particular order, so that you can number the terms in a specific way, then maybe your teacher would give a different answer; but he ought to know that not everyone around the world would do it the same. (And if he doesn't, then you shouldn't be asking your questions on an international site.)
Have you asked your teacher about this? You need to.
But you do NOT get a different answer. You get four terms that are rational. So does your teacher. You used different methods. I tried to explain why the methods give the same result. Remember that we are looking only for the terms that are rational.
Your way gives the following rational terms:
[math]k = 0 \implies \dbinom{20}{k} * (\sqrt{3})^{(20 - k)} * (\sqrt[3]{2})^k = \dbinom{20}{0} *3^{10} * 2^0.\\ k = 6 \implies \dbinom{20}{k} * (\sqrt{3})^{(20 - k)} * (\sqrt[3]{2})^k = \dbinom {20}{6} * 3^7 * 2^2.\\ k = 12 \implies \dbinom{20}{k} * (\sqrt{3})^{(20 - k)} * (\sqrt[3]{2})^k = \dbinom {20}{12} * 3^4 * 2^4.\\ k = 18 \implies \dbinom{20}{k} * (\sqrt{3})^{(20 - k)} * (\sqrt[3]{2})^k = \dbinom {20}{18} * 3^1 * 2^6.[/math]
Your teaacher's way gives the following rational terms:
[math]k = 20 \implies \dbinom{20}{k} * (\sqrt{3})^k * (\sqrt[3]{2})^{(20 - k)} = \dbinom{20}{20} * 3^{10} * 2^0 = \dbinom{20}{0}* 3^{10} * 2^0.\\ k = 14 \implies \dbinom{20}{k} * (\sqrt{3})^k * (\sqrt[3]{2})^{(20 - k)} = \dbinom{20}{14} * 3^7 * 2^2 = \dbinom{20}{6} * 3^7 * 2^2.\\ k = 8 \implies \dbinom{20}{k} * (\sqrt{3})^k * (\sqrt[3]{2})^{(20 - k)} = \dbinom{20}{8} * 3^4 * 2^4 =\dbinom{20}{12} * 3^4 * 2^4.\\ k = 2 \implies \dbinom{20}{k} * (\sqrt{3})^k * (\sqrt[3]{2})^{(20 - k)} = \dbinom{20}{2} * 3^1 * 2^6 = \dbinom{20}{18} * 3^1 * 2^6. [/math]
Same number of terms. Same value for the terms.
I haven't been present during lecture so that's why I am asking here.