Can Rational and Irrational Numbers Multiply to Yield an Irrational Product?

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SUMMARY

The discussion confirms that there exist rational numbers x and irrational numbers y such that the product xy is irrational. Specifically, if x is a rational integer and y is an irrational number, then x^y can yield an irrational result. The example provided illustrates this with x = 2 and y = log2(e), demonstrating that y is irrational. The conversation also touches on the distinction between algebraic and transcendental numbers, emphasizing that while all transcendental numbers are irrational, not all irrational numbers are transcendental.

PREREQUISITES
  • Understanding of rational and irrational numbers
  • Familiarity with algebraic and transcendental numbers
  • Basic knowledge of exponentiation and logarithms
  • Conceptual grasp of geometric interpretations in mathematics
NEXT STEPS
  • Study the properties of algebraic and transcendental numbers
  • Explore the concept of irrational exponents in depth
  • Learn about the implications of the logarithmic function on irrational numbers
  • Investigate geometric interpretations of number theory problems
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Mathematicians, educators, and students interested in number theory, particularly those exploring the relationships between rational and irrational numbers and their properties.

Shackleford
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Prove or disprove that there is a rational number x and an irrational number y such that xy is irrational.

The book works out the case with x and y irrational and xy rational. They used the nonconstructive existence proof method with x = sqrt(2) and y = sqrt(2). If that's rational, then you're finished. If it's irrational, then you can simply raise it to the power of sqrt(2) to get 2. I'm not sure how to adapt this approach to this problem. If it is irrational, then you're finished. If it's rational, then I'm not sure how to show you manipulate it to be irrational.

Suppose that a and b are odd integers with a\neq{b}. Show there is a unique integer c such that |a - c| = |b - c|.

I know this is an easy problem, but I'm stumped for some reason. I've tried to find a general expression for c. The uniqueness method is appropriate here. I first need to show that an integer c with that property does exist. I then need to show that it's unique.
 
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Shackleford said:
The book works out the case with x and y irrational and xy rational. They used the nonconstructive existence proof method with x = sqrt(2) and y = sqrt(2). If that's rational, then you're finished. If it's irrational, then you can simply raise it to the power of sqrt(2) to get 2. I'm not sure how to adapt this approach to this problem. If it is irrational, then you're finished. If it's rational, then I'm not sure how to show you manipulate it to be irrational.
Do you know anything about transcendental numbers?

I know this is an easy problem, but I'm stumped for some reason. I've tried to find a general expression for c. The uniqueness method is appropriate here. I first need to show that an integer c with that property does exist. I then need to show that it's unique.
Try to think geometrically, i.e. draw a and b on a line. Where does c lie?
 
morphism said:
Do you know anything about transcendental numbers?Try to think geometrically, i.e. draw a and b on a line. Where does c lie?

I'm rusty on transcendental numbers.

I didn't think to consider the problem geometrically. Good idea. Of course, c is the value that gives the intersection of the two "functions."
 
Shackleford said:
I'm rusty on transcendental numbers.

I didn't think to consider the problem geometrically. Good idea. Of course, c is the value that gives the intersection of the two "functions."
Or more geometrically, c is the integer right smack in between the two odd integers a and b.
 
Mark44 said:
Or more geometrically, c is the integer right smack in between the two odd integers a and b.

Last night I came up with a few distance-related equations. One was [abs(a) + abs(b)]/2.
 
Shackleford said:
I'm rusty on transcendental numbers.
Okay, that's fine. If you were comfortable with transcendental numbers, then there's a conceptual way of seeing how x^y could be irrational. (Here is the sketch: if both x and y were rational, then x^y would be algebraic. So if x is rational and x^y is transcendental, then y would necessarily be irrational. With these observations you can produce tons of examples of rational x and irrational y such that x^y is irrational - indeed transcendental.)

Anyway, here's another approach. For simplicity let's suppose that x is an integer, and let's look at x^y. Choose your favorite irrational number z. If x^y=z, what can you say about y?
 
morphism said:
Okay, that's fine. If you were comfortable with transcendental numbers, then there's a conceptual way of seeing how x^y could be irrational. (Here is the sketch: if both x and y were rational, then x^y would be algebraic. So if x is rational and x^y is transcendental, then y would necessarily be irrational. With these observations you can produce tons of examples of rational x and irrational y such that x^y is irrational - indeed transcendental.)

Anyway, here's another approach. For simplicity let's suppose that x is an integer, and let's look at x^y. Choose your favorite irrational number z. If x^y=z, what can you say about y?

I just made myself familiar with the terms algebraic and transcendental numbers. All transcendental numbers are irrational, but not all irrational numbers are transcendental. The square root of 2 is an example of an irrational algebraic number. I'll give an example. It's probably a bit nasty. Haha.

x^y = e
2^y = e

y = log2(e) which is clearly irrational.

For the second part, if z is irrational and x is rational, then y could be rational or irrational.
 
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