Is the Square Root Function Bijective in all Branches of Mathematics?

AI Thread Summary
The square root function can be bijective if its domain and codomain are specified as the non-negative real numbers. To prove this, one must demonstrate that the function is injective, meaning if √x = √y, then x must equal y. Additionally, it must be shown to be surjective, meaning for every non-negative number y, there exists a non-negative number x such that √x = y. The discussion clarifies that the square root function is not bijective if the codomain is set to all real numbers. Understanding these properties is essential for establishing the bijectiveness of the square root function in mathematics.
aaaa202
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In what branches of mathematics is this proven.. I have never seen a proof, so I wonder if anyone can give me the basics of what is done to proove it or got a link to a proof..

Edit: By square root I mean the positive square root.
 
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to prove square root function to be bijective you need to specify the domain and codomain of the function.
if codomain is set of real numbers, then its not bijective.
 
aaaa202 said:
In what branches of mathematics is this proven.. I have never seen a proof, so I wonder if anyone can give me the basics of what is done to proove it or got a link to a proof..
Is this a homework assignment? You should know by know that we won't do your assignments for you, but we'll guide you while you do the work.
aaaa202 said:
Edit: By square root I mean the positive square root.
 
This is definitely not a homework assignment. I'm on holidays and this speculation came to me. I study physics and thus my courses do not focus on how to get to the mathematical results rigorously but rather their applications. Therefore I have to do this work on my own and I have no idea whatsoever what tools will be needed for the above proof. The domain of the positive square root are all positive integers including zero. But to go from here to show that the square root takes one number into another number uniquely I don't know how to show. Thus I ask you.
 
Then why post it in "Homework and Classwork"? In any case "the square root takes one number into another number uniquely" is NOT "bijective". That is "injective". You can't say "bijective" without, as pcm said, specifying the domain and codomain.

If you intend the domain and codomain as "the non-negative real numbers" then, yes, the square root function is bijective. To show that you show it is "injective" ("one to one"): if \sqrt{x}= \sqrt{y} then x= y. That's easy to show. And "surjective" ("onto") is equally easy- given any non-negative number, y, show that there exist a non-negative number x such that \sqrt{x}= y.
 
aaaa202 said:
The domain of the positive square root are all positive integers including zero.
HallsOfIvy alluded to this, but to make it more explicit - the domain of this function is all nonegative real numbers, not just the nonnegative integers.
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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