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About LinkBacksHow do you prove this question?
Q: Prove that if x is irrational, then 1/x is irrational?
I am thinking Proof by contradiction but I am not sure. Even where to begin?
Contradiction works.Originally Posted by Grillakis
Hint: If we were to assume that 1/x WAS rational, then it could be written as a ratio of integers p/q.
If x is irrational, and 1/x is rational, then we are saying that the rational number q can be written as a product of an irrational number x, and a rationa number. Is this valid?
it is not valid b/c our assumption of 1/x being rational is false. Is this correct?Contradiction works.
Hint: If we were to assume that 1/x WAS rational, then it could be written as a ratio of integers p/q.
If x is irrational, and 1/x is rational, then we are saying that the rational number q can be written as a product of an irrational number x, and a rationa number. Is this valid?
Not quite.
Proof by contradiction works when we assume the opposite of what we want to proove is true. We then derive an erroneous result... a contradiction.
Our problem is... if x is irrational, prove that 1/x is irrational. The first thing we do is assume the opposite. Assume that x is irrational, but conversely, 1/x is rational. From this we will try to derive an impossibility.
A property of a rational number is that it can be written as a ratio of integers. ANY rational number can be written in the form, where p and q are both integers.
So if we assume that 1/x is rational then:
Now multiply both sides by xq
What this statement says is that the integer q can be written as a product of the integer p, and the irrational number q. THIS is invalid, because multiplying an integer by an irrational number gives an irrational number. Hence q can't be an integer, and our initial assumption MUST be wrong.
Hence 1/x is not rational. And any number that is not rational is irrational.
I see now...thanks MushNot quite.
Proof by contradiction works when we assume the opposite of what we want to proove is true. We then derive an erroneous result... a contradiction.
Our problem is... if x is irrational, prove that 1/x is irrational. The first thing we do is assume the opposite. Assume that x is irrational, but conversely, 1/x is rational. From this we will try to derive an impossibility.
A property of a rational number is that it can be written as a ratio of integers. ANY rational number can be written in the form
So if we assume that 1/x is rational then:
Now multiply both sides by xq
What this statement says is that the integer q can be written as a product of the integer p, and the irrational number q. THIS is invalid, because multiplying an integer by an irrational number gives an irrational number. Hence q can't be an integer, and our initial assumption MUST be wrong.
Hence 1/x is not rational. And any number that is not rational is irrational.
The next step should be to observe that your assumption thatContradiction works.
Hint: If we were to assume that 1/x WAS rational, then it could be written as a ratio of integers p/q.
If x is irrational, and 1/x is rational, then we are saying that the rational number q can be written as a product of an irrational number x, and a rationa number. Is this valid?is rational implies that there exist integers
and
such that:
.
But it does not imply that. If we were to assume that x was rational, then you would indeed be correct. But our method maintains that x is irrational, but assumes that 1/x is irrational.The next step should be to observe that your assumption that
.
So the next step is to observe that 1/x = p/q, such that p and q are integers. And this leads the contradiction q = px, which cannot be.
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