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If r is rational number , x is irrational , how can I prove r+x and rx are irrational?
Should I consider this problem to be a Field and Set problem? Help me out~
Numbers are the essence of the Universe
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I have no idea how you prove something doesn't repeat with a pattern or not.
Good Luck...
igloo myrtilles fourmis
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Well, I would ask you to experiment with rational and irrational numbers for each one.
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Contradiction.
Proving something is irrational is the same as proving something is not rational (assuming the number is in fact real). The "natural" way to prove that some number does not hold a property is to simply assume that it does, and show how this reaches a contradiction.
The first proof would look something like this:
Use the same exact format for your second proof, just change up the operations. Notice how explicit I was to state that bd does not equal 0. This is very important.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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the second one , let r be rational , x be irrational , r =a/b for some a,b(Z) ,
Assume rx rational , rx=c/d for some c , d
x=bc/ad , contradiction. Thanks Ricky.For helping me out~
Numbers are the essence of the Universe
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Actually , I misunderstood the definition of Irrational number , I though like 3.33333333333...... is also irrational , however I was wrong .lol
Numbers are the essence of the Universe
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the second one , let r be rational , x be irrational , r =a/b for some a,b(Z) ,
Assume rx rational , rx=c/d for some c , dx=bc/ad , contradiction. Thanks Ricky.For helping me out~
The above is incorrect. I tried to hint at this after my proof. If a = 0, ad = 0, and thus, bc/ad is not a rational number. However, if a = 0, it is simply a special case. All you do is add:
Let a = 0. Then a/b = 0, and so a/b * x = 0, which is rational. Now assume that a is not 0. And then go on with your proof.
Edit:
On second thought, bc/0 is not even an irrational number, still reaching your contradiction. This seems a bit iffy to me however. I'd personally go with the 2 different cases (a = 0 and a does not equal 0), however, I believe you can still claim your proof to be valid.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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3.33333333333...... is the real equivalence of 1/3. Reals could have infinite digits to accomodate both rationals and irrationals at the same time. However, I find contradiction within infinite digits. If you ignore this contradiction you can accept reals and regard 0.333... indeed the equivalence of 1/3.
X'(y-Xβ)=0
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A rational is an integer divided by a non-zero integer.
X'(y-Xβ)=0
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hmmm, non-zero integer
Numbers are the essence of the Universe
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Basically a rational is just an integer divided by an integer, like 3/2 or 198/2018
But we must say "divided by a non-zero integer" to avoid n/0.
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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