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Help with math for my Computer Science class
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| creon444 |
I can't figure out these three problems from my take-home test:
3. Prove that
5^(1/2), log 7 (48) and tan (5 deg)
are irrational numbers.
6. Prove that the equation
x! + y! = 10z + 9
has no whole-number solutions.
7. Evaluate the floor function
y(x) = root (x^2 + root (4x^2 + root (16x^2 + 8x + 3)))
where x is a whole non-negative number.
Please help! I'm completely stuck! :( :conf:
Regards,
creon444
:nervous: |
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| Orbax |
www.math.com
it actually has a lot of stuff hehe, i checked it out before I posted. Not that Im actually trying to help. The last math I did was 5 years ago in highschool and we were doing SOH CAH TOA |
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| zarathustra |
This should help for 1 and 2
| quote: |
Taken from http://mathworld.wolfram.com/IrrationalNumber.html
Numbers of the form (n^1/m) are irrational unless n is the mth power of an integer. Numbers of the form logn(m), where log is the logarithm, are irrational if m and n are integers, one of which has a prime factor which the other lacks. (Niven 1956, Stevens 1999)
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| jdjd |
| 1 should be easy, 3 i dont feel like doing that much work, but for 2 try saying that you know the right side is odd (obvious). Factorials are always even (factor of 2 in it) unless 1! or 0! , in which case the left side would be 2. So that means the left is always even and the right is always odd... |
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| Noisician |
the ultimate way to solve the first three subproblems is by assuming the contrary and then proving its falseness. something like, let's assume that log7(48) is a rational number, then it follows that... blah blah blah... and then when u arrive at some sort of contradiction, it'll prove its irrationality. tis called "proof by contradiction"
for your second problem, compare the two sides of the equation, see what values x and y cannot take. once u figure out the limit for x nad y, u'll end up with only a few possible values that u can test (and then rule out).
the last problem i'm not quite sure about because i forgot what "floor function" stands for. |
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| jdjd |
| quote: | Originally posted by Noisician
the ultimate way to solve the first three subproblems is by assuming the contrary and then proving its falseness. something like, let's assume that log7(48) is a rational number, then it follows that... blah blah blah... and then when u arrive at some sort of contradiction, it'll prove its irrationality. tis called "proof by contradiction"
for your second problem, compare the two sides of the equation, see what values x and y cannot take. once u figure out the limit for x nad y, u'll end up with only a few possible values that u can test (and then rule out).
the last problem i'm not quite sure about because i forgot what "floor function" stands for. |
you dont need to sub values for 2, the left is even and the right is odd, done... i dont understand 3 either, evaluate usually means find a particular value.. yet it just defines a function..
the floor of a number is defined as the greatest integer <= that number... but that doesnt really apply here, i guess it means find the greatest integer <= that function, but that would be too easy, its just 1 (x=0)... |
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| Noisician |
| quote: | Originally posted by jdjd
the floor of a number is defined as the greatest integer <= that number... but that doesnt really apply here, i guess it means find the greatest integer <= that function, but that would be too easy, its just 1 (x=0)... |
ah yes, i do recognize it now. actually it does apply here. i think what we are asked to do is to find some function y1(x), such as for all values of x it defines the floor for all corresponding values of the y(x).
[edit]
got it! the answer is y1(x)=x+1. u can solve it in 3 steps by finding the limit for each radical and taking the lowest resulting boundary. |
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| creon444 |
| quote: | Originally posted by Noisician
the ultimate way to solve the first three subproblems is by assuming the contrary and then proving its falseness. something like, let's assume that log7(48) is a rational number, then it follows that... blah blah blah... and then when u arrive at some sort of contradiction, it'll prove its irrationality. tis called "proof by contradiction"
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Can you show me how to solve the tangent one? I kinda know how to do the first two of them, but the third one is different :conf: .
| quote: | | for 2 try saying that you know the right side is odd (obvious). Factorials are always even (factor of 2 in it) unless 1! or 0! , in which case the left side would be 2. So that means the left is always even and the right is always odd... |
Thanks, jdjd!
| quote: | | got it! the answer is y1(x)=x+1. u can solve it in 3 steps by finding the limit for each radical and taking the lowest resulting boundary |
But how do you know what the limits are?
Regards,
creon444 |
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| jdjd |
| i was wrong about 2 you can get odd on the left side, but then assume x! or y! is 1 and then show you cant get a factorial to be equal to 10Z + 8 |
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| Durafei |
| quote: | Originally posted by jdjd
i was wrong about 2 you can get odd on the left side, but then assume x! or y! is 1 and then show you cant get a factorial to be equal to 10Z + 8 |
If x and y > 1, then x! + y! are even.
So assume that y = 1.
if x > 10, x! = 10n.
if x >=5, x! = 10n. impossible.
So x<=4. Now just check 4 values. |
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| Z1D |
| For the tangent one you probably need to expand tan into its Taylor series (convert the 5 to radians) |
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| Noisician |
| quote: | Originally posted by creon444
Can you show me how to solve the tangent one? I kinda know how to do the first two of them, but the third one is different :conf: .
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let tan5º be a rational number, then it can be presented as a ratio of two coprime integers, i.e. m/n (|m|
then the number tan10º = tan(5º+5º) = (2tan5º)/(1-tan²5º) = (2(m/n))/(1-(m/n)²) = 2mn/(n²-m²) is also rational because it is merely a ratio of two integers (2mn) and (n²-m²). thereby, it follows that tan10º = p/q, where p and q are coprime integers (|p|
then the number tan20º = tan(10º+10º) = 2pq/(q²-p²) is also rational, i.e. tan20º = a/b, where a and b are coprime integers (|a|
finally, it follows that the number tan30º = tan(10º+20º) = (tan10º + tan20º)/(1-tan10ºtan20º) = (p/q+a/b)/(1-(p/q)(a/b)) = (pb+qa)/(qb-pa) is rational because it is merely a ratio of two integers (pb+qa) and (qb-pa).
but u can easily prove that tan30º equals (1/√3), using just the basics of planimetry and trigonometry. therefore, it is a fact that tan30º is an irrational number and thus cannot be expressed as the ratio above. because of this contradiction, the original supposition appears to be erroneous, and tan5º MUST be, in fact, irrational.
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for the third problem, the floor can be derived from a series of the following inequations (keep in mind that x is non-negative):
16x²+8x+1 < 16x²+8x+3 < 16x²+8x+4
or
(4x+1)² < 16x²+8x+3 < (4x+2)² (since 16x²+8x+4 < 16x²+16x+4)
after extracting the square root from all parts and adding a 4x² to all of them, we get
(4x+1)+4x² < 4x² +√(16x²+8x+3) < (4x+2)+4x²
or
(2x+1)² < 4x² +√(16x²+8x+3) < (2x+2)² (since 4x²+4x+2 < 4x²+8x+4)
after extracting the square root from all parts and adding an x² to all of them, we get
(2x+1)+x² < x² + √(4x² +√(16x²+8x+3)) < (2x+2)+x²
or
(x+1)² < x² + √(4x² +√(16x²+8x+3)) < (x+2)² (since x²+2x+2 < x²+4x+4)
after extracting the square root from all parts, we get
x+1 < √(x² + √(4x² +√(16x²+8x+3))) < x+2
since x is a whole non-negative number, the floor of the resulting radical is (x+1).
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