Decide whether α must be a rational number. HOW!?

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TobiWan

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Let [FONT=MathJax_Math-italic]α[/FONT] be the real number that tan([FONT=MathJax_Math-italic]α[/FONT] · [FONT=MathJax_Math-italic]π[/FONT]) = √2. Decide whether [FONT=MathJax_Math-italic]α[/FONT] must be a rational number.​







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Well, since you've shown no work of your own, I'm forced to assume you're stuck at the very beginning. A good place to begin would be to use what you know about the tangent function and about rational numbers. So, I'd graph tan(x) from x=0 to x=pi. Does it intersect the line y=sqrt(2)? If so, where? What about the graph of tan(x*pi)? Although eyeballing the graph probably won't give you an exact value for x, you can get an approximation. Does the value you got look like it might be a rational number?

Now that you've got at least an idea of what the answer might be, how would you verify that? Well, what do you know about rational numbers? A rational number can be written as a fraction a/b... but there's constraints placed on a and b. What are those constraints? Now, how would you solve the given equation for x? Is that expression a fraction? If so, does it meet the criteria for a rational number? Do you think there's any equivalent fraction that might be rational? Why or why not? If you think there exists such a fraction, how would you go about finding it? As an additional hint, you might consider the well-known proof of why sqrt(2) is irrational.

If you get stuck, that's okay. But when you reply, please include any and all work you've done on this problem, even if you know it's wrong. Thank you.
 
ksdhart2 said:
Does the value you got look like it might be a rational number?
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Can you please explain how you look at a point on the number line and tell if it might be rational? You do know that the rationals are dense in the Reals?[FONT=MathJax_Math-italic] [/FONT][FONT=MathJax_Math-italic][/FONT]

Instead of saying to graph from 0 to pi, its customary to say from -pi/2 to pi/2. In this case since sqrt 2 >0, say to graph from 0 to pi/2
 
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Jomo said:
Can you please explain how you look at a point on the number line and tell if it might be rational? You do know that the rationals are dense in the Reals?
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I do admit, I may have worded my statement poorly. By merely looking at a graph, one can of course never say for certain whether a given intersection point is rational or not, but, in my mind, certain values are "more likely" to be rational then others. For instance, if I eyeballed the graph and saw an it intersected the line y=sqrt(2) reasonably close to, say, x=2, then it would be a reasonable assumption that the corresponding x-value is a rational number. I merely asked it as a question to (hopefully) get the student thinking in the right direction, and get them to develop a hypothesis as to it's reasonable to assume the intersection point is rational, from which they might proceed to prove/disprove said hypothesis.
 
x+pi can't be written as a/b ,you said,we assume x=a/b,and what next?we have nothing

x+pi can't be written as a/b , but you said, we assume x=a/b, and what next? we have nothing, of course we know frome the task that x is a rational number
 
√2 there is between 54 and 55 degrees. α · π=(54;55), so α belongs (54/π ; 55/π). And we know, beetwen 2 irrationals numbers there is one irrational number.
 
TobiWan said:
Let \(\displaystyle \alpha\) be the real number that \(\displaystyle \tan(\alpha\, \cdot\, \pi)\, =\, \sqrt{\strut 2\,}.\) Decide whether \(\displaystyle \alpha\) must be a rational number.
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From what you know about the tangent function, if the tangent evaluates to the square root of two, then what must have been the argument of (that is, the input to) the tangent? If you set this equal to "alpha * pi", what do you get for alpha?

If you get stuck, please reply showing your work and reasoning in answering the above questions. Thank you! ;)
 
TobiWan said:
x+pi can't be written as a/b
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I'll assume you mean that a and b are integers and that b\(\displaystyle \neq\)0. Still what you wrote is not true! You can add two irrational numbers and get a rational number!

Try (-\(\displaystyle \pi\)) + (\(\displaystyle \pi\)) and you'll get 0 which is rational.

(4-\(\displaystyle \pi\)) + (\(\displaystyle \pi\))= 4, again a rational number.
 
stapel said:
From what you know about the tangent function, if the tangent evaluates to the square root of two, then what must have been the argument of (that is, the input to) the tangent? If you set this equal to "alpha * pi", what do you get for alpha?

If you get stuck, please reply showing your work and reasoning in answering the above questions. Thank you! ;)
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Stapel, isn't that the problem that this result you ask the op to find can not be determined to be rational or irrational??
 
Jomo said:
Stapel, isn't that the problem that this result you ask the op to find can not be determined to be rational or irrational??
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In the first quadrant, or from 0 to pi/2, if tan(A) = sqrt[2], then what must be the value of "A"? If A = a*pi, then what must be the value of "a"? ;)
 
stapel said:
In the first quadrant, or from 0 to pi/2, if tan(A) = sqrt[2], then what must be the value of "A"? If A = a*pi, then what must be the value of "a"? ;)
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OK, I'll state the answer to your question. I did not want to give it to the op.

if tan(A) = sqrt (2) than A = arctan (sqrt (2)), ie a*pi = arctan (sqrt (2)) and a= [arctan (sqrt (2))]/pi

tan (A). with A between 0 and pi/2, can take on any positive value (yeah we can bound it better for this example, but there is no need to do so).

The arctan (sqrt (2)) can be a rational number*pi, making a rational or the arctan (sqrt (2)) is not a rational number*pi making a irrational.

I still can't tell if a is rat or not.
 
Jomo said:
OK, I'll state the answer to your question. I did not want to give it to the op.

if tan(A) = sqrt (2) than A = arctan (sqrt (2)), ie a*pi = arctan (sqrt (2)) and a= [arctan (sqrt (2))]/pi

tan (A). with A between 0 and pi/2, can take on any positive value (yeah we can bound it better for this example, but there is no need to do so).

The arctan (sqrt (2)) can be a rational number*pi, making a rational or the arctan (sqrt (2)) is not a rational number*pi making a irrational.

I still can't tell if a is rat or not.
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Um... In the first quadrant, if tan(A) = sqrt(2), then A = pi/4. (This is a basic memorized reference angle relation.) Since A = a*pi, then A = (1/4)*pi, so a = ...? ;)
 
stapel said:
Um... In the first quadrant, if tan(A) = sqrt(2), then A = pi/4. (This is a basic memorized reference angle relation.) Since A = a*pi, then A = (1/4)*pi, so a = ...? ;)
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Wait, what? Unless I've forgotten everything I know about trig, that's not right either... for any x, tan(x) = sin(x)/cos(x), so:

\(\displaystyle \displaystyle tan\left(\frac{\pi }{4}\right)=\frac{sin\left(\frac{\pi }{4}\right)}{cos\left(\frac{\pi }{4}\right)}=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{ \sqrt{2}}}=1\)
 
stapel said:
Um... In the first quadrant, if tan(A) = sqrt(2), then A = pi/4. (This is a basic memorized reference angle relation.) Since A = a*pi, then A = (1/4)*pi, so a = ...? ;)
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Yes, I thought you were thinking that, but unfortunately it is not correct. I have been guilty of sloppy thinking in the past and sent myself to sit in the corner. Do you think that you need a break in the corner?
 
So in the end I hope that people here realize that this is not a simple problem and I feel that the op deserves a nice hint.
 
ksdhart2 said:
Wait, what? Unless I've forgotten everything I know about trig, that's not right either... for any x, tan(x) = sin(x)/cos(x), so:

\(\displaystyle \displaystyle tan\left(\frac{\pi }{4}\right)=\frac{sin\left(\frac{\pi }{4}\right)}{cos\left(\frac{\pi }{4}\right)}=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{ \sqrt{2}}}=1\)
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Oops! You're right! :oops: Dag-nabbit!
 
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