SOLUTION: Jacinta says the product of a rational number and an irrational number is always irrational. is that true, does it work for 0?
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Question 1164515: Jacinta says the product of a rational number and an irrational number is always irrational. is that true, does it work for 0?
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
No, her statement is not true when the rational number is 0.
This is because 0*x = x*0 = 0.
Multiply 0 with any number (x) and you'll get 0 as a result.
So multiplying 0 (which is rational) with any irrational number you want, and you'll get 0
If Jacinta said that the rational number was not zero, then her claim would be correct.
The phrasing "not zero" is the same as "nonzero".
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Extra Info:
Let's prove the claim "the product of a nonzero rational number and an irrational number is always irrational". We'll use a proof by contradiction.
Let x be rational and y be irrational.
Since x is rational, we can use integers p,q to say x = p/q where q is nonzero.
We'll make p nonzero as well to avoid x = p/q being zero.
We cannot write y as a fraction of integers since it is irrational. This is the definition of what it means to be irrational. It means "not rational".
Keep this in mind for later.
Next, let's assume the original claim is flipped to say that multiplying any nonzero rational number with an irrational number is rational.
In short, let's assume x*y = r, where r is a rational number.
If x*y was rational, then
x*y = m/n
for some integers m,n and n is nonzero.
Because x = p/q, we can say the following
x*y = m/n
(p/q)*y = m/n
y = (m/n)*(q/p)
y = (m*q)/(n*p)
y = (some integer)/(some other integer)
Side note: multiplying any two integers results in some other integer.
This shows y is rational, but this contradicts the earlier definition where we made y irrational.
This contradiction makes the claim "the product of a nonzero rational number and an irrational number is rational" to be false.
Therefore, the opposite claim "the product of a nonzero rational number and an irrational number is irrational" must be true.
This concludes the proof.
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