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Prove for any positive number $t$, there is a solution for $x^2=t$.

So we want to show that $x^2=t$ for $t\geq0$.

We can break this into two cases:

Case 1: Assume $t=0$, then we have $x^2=0$ $\Rightarrow$ $x=0$, and we are done.

Case 2: Assume $t>0$, then we have $x^2=t$ $\Rightarrow$ $x=\sqrt{t}$

And here is where I'm stuck, not sure what I should do next...?

Michelle
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    As it stands, your question doesn't contain enough information. Are you trying to prove this from the basic properties of $\mathbb{R}$, or can you use tools like the intermediate value theorem? It seems that you don't yet have the existence of $\sqrt t$. –  Aug 01 '14 at 22:40
  • The first sentence is the exact question I was given, I'm thinking I have all of $\mathbb{R}$ to work with. – Michelle Aug 01 '14 at 22:42
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    But the answer depends on what context you've encountered this problem in. What is the source of this, and what earlier results can you use? Are you trying to prove this from the completeness of $\mathbb{R}$? –  Aug 01 '14 at 22:42
  • We have been discussing Pythagorus...perhaps somehow using his theorem would help, but I'm not sure how to do that either. – Michelle Aug 01 '14 at 22:45
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    You should add the source of the problem, and exactly what you can use; until this is added, the question is unanswerable. –  Aug 01 '14 at 22:46
  • Its a HW question, there is no source, it was made up by the professor, who has been unhelpful thus far this semester and I have provided all the information that he has given to me. – Michelle Aug 01 '14 at 22:47
  • "It's a HW question, there is no source, it was made up by the professor ...". You might at least give the name of the course in question, perhaps also the textbook you're using and how far you've progressed into it. – Blue Aug 01 '14 at 23:20

3 Answers3

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We have for $t\ge0$

$$x^2=t\iff x^2=(\sqrt t)^2\iff x^2-(\sqrt t)^2=0\\\iff (x-\sqrt t)(x+\sqrt t)=0\iff x=\pm\sqrt t$$

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    I didn't downvote, but I don't think this is what the OP is looking for. For example, how do you know that $\sqrt{t}$ is a real number by itself, actually what is $\sqrt{t}$ to begin with? – Lord Soth Aug 01 '14 at 22:45
  • @LordSoth From what he is saying, I believe that $x$ and $t$ are both real numbers... So if $t > 0$, then $\sqrt t$ is also defined in the real numbers, meaning there is a solution for $x^2=t$. However OP was unclear in his question, so I'm not sure – Anonymous Computer Aug 01 '14 at 22:49
  • Have a nice time Sami. :-) – Mikasa Aug 02 '14 at 06:26
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$f(x) = x^2$ is continuous and unbounded above. So, there is some $a$ such that $f(a) > t$. Since $t \geq 0 = 0^2$, by the IVT, there is an $0 \leq x \leq a$ such that $x^2 = t$.

MRicci
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Consider the Theorem "for every positive real number A there is a real number B such that B^2=A"

If you take that to be given i.e. proven, then your proof is (trivially) concluded. I assumed you do because you used "sqrt(t)".

If you are looking for a proof of that Theorem it goes like this, invoking the completeness property of the reals: http://planetmath.org/existenceofsquarerootsofnonnegativerealnumbers

ps1: If you have studied the construction of the reals from the axiomatic definition of the set of natural numbers then this proof hardly comes as a surprise. The reals were constructed exactly with the motivation to fill in those gaps in what we have in our head as the "line of numbers".

ps2: Excuse my not using mathematical symbols im new to the website and i came across this question before i had a chance to fin dout how that works.

Nikos Δr
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