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Prove that $1, \sqrt{2}$ are linearly independent over the rational numbers.
Calculus 3
Chapter 5
Vector Spaces and Modules
Section 1
Vector Spaces and Bases
Vectors
Johns Hopkins University
University of Michigan - Ann Arbor
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Ah, good day, ladies and gentlemen. Today we're looking at a problem number 29 from section 4.7. And it is asking us to consider too linearly independent solutions to the to the differential equation like this and are and we want to show that s o It's a linear differential differential equation of this form. On some interval you need to be. And we want to prove that, um, both Why wanted White? You cannot both be zero at some value of tea. Okay, so this is a question that is asking us to prove something. And the proof method we use again, eyes a proof by contradiction, which means that we assume Okay, so we assume both that. Why want it? Why, too your are, in fact, linearly independent. So we're assuming this is true. And we're also assuming that at some value of tea, there they are, both zero. Okay. And now what happens then? Well, if they're both zero, let's let's just calculate the wrong skin and evaluate at that poin t. Well, um, this is zero, and this is zero and zero times. Any number, of course, is equal to zero. So we have zero minus zero and, of course, us equal to zero. Now the key here is that the Ron skiing one of the properties of the Ron skiing is if it is zero. If w why wonder why, too? Is zero at any value of tea, Just any value, just one value. It means that the two functions here are linearly dependent. Okay, so if you don't know that again, that's something you need to know about the round skin. That it's it's one of the important. It's is one of the most important properties of the round ski in here. And this pertains on Lee to, um, solutions off differential equations. Amazingly, um, that that works. So what we've done now is we've shown that there exists a fixed value of tea such flat. The ron skin is zero. And of course what? Like I said, that means that they are linearly depended or in particularly not linearly independent. So you can't you can't assume something is literally independent. And that, um, is that you can't assume that they're both linearly independent and this together because we get a contradiction. We we've shown that they're they're not linearly independent So you can't. I mean, you can't be literally independent and not not linearly independent at the same time. So something has to be wrong. And, of course, what it is is it's this for a year because this is our second assumption of people. Well, um, that this cannot be true. So therefore, they cannot both be zero at the same point. Um, yeah. So So this is really just a nice I mean, Proust by contradiction are really nice. Can be really nice little proofs, but it just turns out, uh, that therefore, since they cannot, this cannot hold. That's the idea. So, uh, that is it for here? Um, sometimes proust, like contradiction and stuff like that take ahh. Bit of getting your head around. So I might be worthwhile to think about it in your own time. And I just want to remind you again that the basic idea is we made two assumptions on the functions in our proof. We said, Well, we're gonna assume, and this is this is the what I might call, like the first major assumption. So the one that we have to I mean, we have to have something. Has to be absolutely true, which is that they are. We're assuming that they are linearly independent solutions to this differential equation. Okay. And then in the course of the proof I'm saying, Well, I'm gonna assume that I give this statement here, all right? And then what it leads me to is a contradiction. And since this leads me to a contradiction, it cannot be true with this assumption. And therefore, these two statements cannot beach both through at the same time. So that that's basically the idea here. And, um, that's it for today. Thank you very much.
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