Rational numbers
- Thread starter Loki123
- Start date
Harry_the_cat
Elite Member
- Joined
- Mar 16, 2016
- Messages
- 3,693
Can you please rewrite the question exactly as it was given to you? It doesn't make sense as is!
Steven G
Elite Member
- Joined
- Dec 30, 2014
- Messages
- 14,383
Suppose (90-12w)/5 is a whole number. How can that influence what type of a number x can be?
Basically you are saying that you can only raise a rational number to a whole number. That is not true. Sqrt(2) is irrational and 2 is a whole number yet I can compute (sqrt2)2. That result is 2
Basically you are saying that you can only raise a rational number to a whole number. That is not true. Sqrt(2) is irrational and 2 is a whole number yet I can compute (sqrt2)2. That result is 2
Here is the whole problem. The part written bellow is the explanation I was given which I do not understand. Why does (90-5k)/12 have to be a whole number?
D
Deleted member 4993
Guest
Can you think of a counter-example?
BigBeachBanana
Senior Member
- Joined
- Nov 19, 2021
- Messages
- 2,181
What's the domain of x? Positive real number? Positive integer?
If [imath]x \in \Z+[/imath], the answer is trivial.
If [imath]x \in \R+[/imath], then consider the irrational numbers like [imath]e[/imath] or [imath]\pi[/imath]. Is [imath]e^2[/imath] or [imath]\pi^2[/imath] a rational number?
pka
Elite Member
- Joined
- Jan 29, 2005
- Messages
- 11,971
Frankly I have seen that as a requirement for exponents.
It may be a particular usage by your text/instructor.
In fact for every integer [imath]k[/imath] the expression [imath]\dfrac{90-5k}{12}[/imath] is rational.
But it is a whole number for only [imath]k=6[/imath] this expansion. SEE THIS. Also THIS
BigBeachBanana
Senior Member
- Joined
- Nov 19, 2021
- Messages
- 2,181
In this example, x is positive integers 2 and 3. So as I said before, that requirement makes sense only when x is a positive integer.
What does integer mean? I haven't heard that one before?
BigBeachBanana
Senior Member
- Joined
- Nov 19, 2021
- Messages
- 2,181
A fancy word for whole numbers.
pka
Elite Member
- Joined
- Jan 29, 2005
- Messages
- 11,971
It appears that your real problem is caused by your having a poor English mathematical vocabulary.
D
Deleted member 4993
Guest
Use Google to learn all the intricacies that are associated with "integer".
BigBeachBanana
Senior Member
- Joined
- Nov 19, 2021
- Messages
- 2,181
Yes, No
BigBeachBanana
Senior Member
- Joined
- Nov 19, 2021
- Messages
- 2,181
That should answer your question in the OP: Why whole number in the exponent gives you a rational number?
For positive integers,
[imath]\text{integer}^\text{integer}=\text{integer}[/imath], which is a rational number. For example, [imath]2^2=4[/imath].
If the exponent is not an integer, then the result may or may not be not an integer.
For example, [imath]2^{1/2}=\sqrt{2}[/imath] <- irrational
But [imath]4^{1/2}=2[/imath] <- rational
Incredible.
Loki has been asked repeatedly to give complete and exact questions as specified in the guidelines. He failed to do that in this case until post 10 and even then it is not quite right. He means “terms” rather than “members” and “expanded” rather than “developed.” I think we could cope with slight deficiencies in the mathematical vocabulary of English if he would just follow the rules and provide complete questions exactly as given.
The only things required to solve this problem are the binomial theorem and some basic knowledge about irrational numbers.
[math]y = (\sqrt[3]{3}+ \sqrt{2})^5 \implies \\ y = (\sqrt[3]{3})^5 + 5(\sqrt[3]{3})^4(\sqrt{2})+ 10(\sqrt[3]{3})^3(\sqrt{2})^2 + 10(\sqrt[3]{3})^2(\sqrt{2})^3 + 5(\sqrt[3]{3})(\sqrt{2})^4 + (\sqrt{2})^5 \implies\\ y = 9\sqrt[3]{3} + 45\sqrt{2} + 60 + 20(\sqrt[3]{3})^2(\sqrt{2}) + 20\sqrt[3]{3} + 4 \sqrt{2}. [/math]
Now what is the answer?
Loki has been asked repeatedly to give complete and exact questions as specified in the guidelines. He failed to do that in this case until post 10 and even then it is not quite right. He means “terms” rather than “members” and “expanded” rather than “developed.” I think we could cope with slight deficiencies in the mathematical vocabulary of English if he would just follow the rules and provide complete questions exactly as given.
The only things required to solve this problem are the binomial theorem and some basic knowledge about irrational numbers.
[math]y = (\sqrt[3]{3}+ \sqrt{2})^5 \implies \\ y = (\sqrt[3]{3})^5 + 5(\sqrt[3]{3})^4(\sqrt{2})+ 10(\sqrt[3]{3})^3(\sqrt{2})^2 + 10(\sqrt[3]{3})^2(\sqrt{2})^3 + 5(\sqrt[3]{3})(\sqrt{2})^4 + (\sqrt{2})^5 \implies\\ y = 9\sqrt[3]{3} + 45\sqrt{2} + 60 + 20(\sqrt[3]{3})^2(\sqrt{2}) + 20\sqrt[3]{3} + 4 \sqrt{2}. [/math]
Now what is the answer?