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I have done the following exercise.

Show that if $x_{1}$ and $x_{2}$ are linearly independent and $z>0$ then $zx_{1}$ and $zx_{2}$ are linearly independent.

Now I am not sure how to regard this in general. One way would be;

Let $x_{1}$, $x_{2}$ be linearly independent and $z>0$. Suppose $c_{1}zx_{1}+c_{2}zx_{2}=0$ and deduce that $c_{i}=0$

The other;

Suppose $x_{1}$, $x_{2}$ be linearly independent and $z>0$. Suppose $c_{1}zx_{1}+c_{2}zx_{2}=0$ then $c_{i}=0$

or

Suppose $x_{1}$, $x_{2}$ be linearly independent and $z>0$ let $c_{1}zx_{1}+c_{2}zx_{2}=0$ then $c_{i}=0$

Which of these are the most suitable for the above problem? I have in some sense to prove and implication which contains an implication and thats what confuses me.

Update

After copying the style of another proof it seems one can do as follows;

Let $c_{1}zx_{1}+c_{2}zx_{2}=0$, then use the hypoteses w.o stating them(they are of course present in the statement of the theorem) and conclude.

It seems there is no given way how to go about this unfortunately! Anyone experienced care to share their toughts?

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    What aspect of these three statements (which to me look as alike as peas in a pod) do you want us to focus on? They are all contrapositive arguments. Some use the word let where others use the word suppose, but aren't these synonyms for assume? – kimchi lover Aug 05 '17 at 13:56
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    Would you have become excessively tired if you had written “linearly independent” instead of “L.I”? – José Carlos Santos Aug 05 '17 at 13:57
  • @kimchilover I dont think they are the same. "Let" defines something while assume is refering to a hypothesis or starts a "supposition". But here I am proving an implication which involves and implication and I am not sure how to express it or think about it. –  Aug 05 '17 at 14:03
  • @kimchilover https://math.stackexchange.com/questions/483544/difference-in-terminology-between-let-and-assume –  Aug 05 '17 at 14:04
  • "Let", "suppose " and "assume" have different meanings. The wording for a proof can change (of course), because the point of view presented for a proof by contradiction or contrapositive can be different. But the proof behind (contradiction or contrapositive) are equivalent regardless of these differences in their presentation. – Masacroso Aug 05 '17 at 14:09
  • @Masacroso when I think about this I think of it as a "direct proof". I simply use the two hypotesis to prove the linear indepedance of $zx_{i}$. As i interpret it it should be shown that linear indepedance of $x_{i}$ and positivity of $z$ is sufficent for linear indepedance of $zx_{i}$. –  Aug 05 '17 at 14:10
  • They are all fine. Maybe in your second presentation I will say "suppose also" for the second "suppose". My personal preference in these kind of wording is using "let" that presents something like a fact, like an axiom, in contrast to some hypothesis, in some context. The third wording doesnt like me so much, Idk if the expression is correct (Im not english speaker so Im not sure :S). – Masacroso Aug 05 '17 at 14:14
  • @Masacroso right that is what I think I am gonna make my preference to once I got a good grip of this. But you are saying that you would do it as in the second presentation then? It confuses me since I say "suppose" three things and dedude something. It is not clear why I could not "commute" the supposes. Or what the diffrence is of using one suppose. I.e I dont think its clear that a second "supposition" starts –  Aug 05 '17 at 14:17
  • @Masacroso right that is what I think I am gonna make my preference to once I got a good grip of this. But you are saying that you would do it as in the second presentation then? It confuses me since I "suppose" three things and dedude something. It is not clear why I could not "commute" the supposes. Or what the diffrence is of using one suppose. I.e I dont think its clear that a second "supposition" starts –  Aug 05 '17 at 14:23
  • I would use something similar to your first presentation, seems clearer to me eyes. Anyway Im not english speaker and my "english" is not the best, so I cant help you very much. Seeing how others mathematicians write proofs is the best way to learn about the matter. – Masacroso Aug 05 '17 at 14:29

2 Answers2

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Isn't this a proof of contradiction? Suppose $zx_1\ and\ zx_2$ are not linearly independent and follow your deduction. Then we find that $x_1$,$x_2$ are also not linearly independent, which is contradict to the condition that $x_1$ and $x_2$ are linearly independent.

uPhone
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  • What's wrong with a proof by contradiction? – fleablood Aug 05 '17 at 15:15
  • Nothing wrong with it. He said he's confused by the logic flow of the proof. I thought my answer may help him organise the flow, without playing with those logic words.@fleablood – uPhone Aug 05 '17 at 15:20
  • right we get an "and" statement this way. But there should exist a direct proof right? Or is it possible that there is a contratidtion proof but not a direct proof? Maybe thats a weird question i.e if there always exists a direct proof of any true statement. –  Aug 05 '17 at 16:01
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    There are three paths of logic: 1) assume x1,x2 are l.i. => zx1, zx2 are linearly indepedent. 2)assume zx1, zx2 are not l.i => x1, x2 are not l.i. 3) assume x1, x2 are li, zx1, zx2 are not => contradiction. All are fine. We have a bit of a subtlty that linear independence is negative definition of "not dependent" so the the "direct" proof is x1,x2 not =>zx1,zx2 are not, is actually a proof be contradiction on dependence. So the best logic would be....cont.... – fleablood Aug 05 '17 at 16:17
  • Suppose zx1 and zx2 or linearly dependant => x1 and x2 are linearly dependent (trivially easy: zx1 = wz2 so x1 = (wz/z)x2). Hence x1 and x2 are linearly independent only if zx1 and zx2 are independent. As for "suppose", "let" and "assume", they may have different meanings but their logical implications are not significantly different for this post. I think Assume x1,x2 l.i. and let czx1 + czx2=0 is best as x1,x2 li. is hypothesis and czx=0 is a specific value to manipulate. But honestly, none are wrong and no-on will notice. – fleablood Aug 05 '17 at 16:27
  • Gödel's first incompleteness theorem says that there are true statements that are not provable or no (direct) proofs exist. Does this answer your question? Check more on this https://math.stackexchange.com/questions/468043/examples-of-statements-which-are-true-but-not-provable @user21312 – uPhone Aug 05 '17 at 17:26
  • @uPhone but there exist proof by contradiction? –  Aug 05 '17 at 17:28
  • @user21312 No. Not always. The thing of "no proofs at all for true statements" exists. – uPhone Aug 05 '17 at 17:30
  • @uPhone Im sorry I mean "there does not exist a direct proof but there exists an by contradiction?" –  Aug 05 '17 at 17:36
  • @user21312 Probably. Here is a not-so-formal example: prove god is not omnipotent. You can not directly prove this, right? Use contradiction: suppose it is, then it can create something it can't do, which is contradict to that it is omni. – uPhone Aug 05 '17 at 17:42
  • Empirical scientists like to say "you can prove a negative". Proving something is not the case directly is often (but not always) not possible. However proving that something is the case wold result is something false, is usually adequate. – fleablood Aug 05 '17 at 17:49
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I honestly think it makes no difference.

Although "let" and "Suppose" have different meanings, their logical conclusions are the same.

"let" means assign an concrete instance. However, this instance must be general and have no specific known characteristics. Therefore anything we can conclude from the specific instance with no known characteristics, would have to be a conclusion that applies to any other hypothetical instance. So "let" $\implies$ "suppose".

And likewise "suppose" means to take a hypothetical instance and then reach a general conclusion that applies to all instances. So it applies to all specific instances. So "suppose" $\implies$ "let".

For style I'd say "Suppose $x_1$ and $x_2$ are linearly dependent and let $c_1x_1 + c_2x_2 = 0$". We are trying to prove something in all hypothetical cases, but we are using a equation to get a specific result which we apply to all cases.

But honestly, no-one will notice and even if they did, they could not deny that the logic of any choice would work.

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On the other hand, it could be argued "let" and "suppose" are not significantly different.

If we are walking through a forest full of vectors and we trip over a pair, $x_1$ and $x_2$, and it just so happens they are linearly independent. That is "suppose".

If instead we get on our hands and knees and pick a pair, $w_3$ and $y_7$ and they are not linearly independent so we throw them away, and we keep searching until we find $x_1$ and $x_2$ that are linearly independent and we say "you'll do". That's "let".

In both cases, though, we end up with the same thing; a pair of linearly independent vectors.

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By the way, this isn't pertinent to you question, if your definition of linear independence is: $\{x_i\}$ are linearly independent if--- $\sum c_i x_i = 0 \iff c_i = 0$, then you do not need the second "let/suppose $c_1zx_1 + c_2zx_2 =0$".

We'd simply say: Let/suppose $x_1$ and $x_2$ are linearly independent. Then $c_1zx_1 + c_2zx_2 = 0 \iff c_iz = 0 \iff c_i= 0$. So $zx_1$ and $zx_2$ are linearly independent.

fleablood
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  • A proof in my book seem to just skip the first sentance in your last paragraph. They simply write(if we translate into the theorem we have here) "Let zcx+zcx=0" then they use the hypothesis to conclude that $zx$ are l.i., without mentioning the hypothesis.I guess there is no given standard or way of doing these things. –  Aug 06 '17 at 13:03
  • The seems fair. Linear independence of $x_1$ and $x_2$ means $kx_1 + jx_2 =0$ implies $j$ and $k$ are zero. So $jzx_1 + kzx_2$ imply $kz$ and $jz$ equal $0$ which implies $j,k$ are equal to zero. So $zx_1$ and $z_x2$ are linearly independent. – fleablood Aug 06 '17 at 15:24
  • I guess. The whole point with the post was to find a soild way to think about how one would strucutre a proof so it becomes clear whats going on. But I guess there is no answer to that –  Aug 06 '17 at 16:46