- #1

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In proving theorem I.2, how is theorem I.1 used to assert that 'there is at most one such x'? The first image below gives the background text. The text I have trouble with is highlighted in the second image below.

Thank you.

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- Thread starter omoplata
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- #1

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In proving theorem I.2, how is theorem I.1 used to assert that 'there is at most one such x'? The first image below gives the background text. The text I have trouble with is highlighted in the second image below.

Thank you.

- #2

Samy_A

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That means that:

a + x = b

a + y = b

What can you deduce from that about a + x and a + y?

- #3

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Assume that x and y satisfy theorem 1.2.

But I can't assume that. Theorem I.2 is the one I'm trying to prove.

- #4

Samy_A

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You are questioning the part of theorem 1.2 that 'there is at most one such x'.But I can't assume that. Theorem I.2 is the one I'm trying to prove.

You prove that by

That in itself indeed doesn't prove that "there exists" an x satisfying theorem 1.2, but that if one exists, it is unique.

Do you see how that uniqueness follows from theorem 1.1?

- #5

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So I have come up to the point saying a + x = b.

But I do not know yet if x is unique. So I assume there is a another y such that a + y = b.

Then, a + x = a + y

From theorem I.1, x = y.

Therefore x is unique.

Thank you!

- #6

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Had the same problem as omoplata! Thanks for answering :)

That means that:

a + x = b

a + y = b

What can you deduce from that about a + x and a + y?

- #7

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If you have taken linear algebra. A good example would be the uniqueness of the additive identity of a vector space, the 0 vector.

Your post #5, is a great idea to think about when we want to show uniqueness.

- #8

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One more thing. If you are able to show existence and uniqueness, we get an iff statement.

- #9

member 587159

If you have taken linear algebra. A good example would be the uniqueness of the additive identity of a vector space, the 0 vector.

Your post #5, is a great idea to think about when we want to show uniqueness.

Rather assuming that if two things satisfy a given property, then they are equal. Sometimes contradiction can be used, but most of the times it isn't necessary (in your example of the additive identity it isn't necessary: if 0,0' are identities, it follows that 0 + 0' = 0 = 0' + 0 and 0+0' = 0' = 0' + 0, in particular thus 0=0')

- #10

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That is why i said usually. However, you were correct with the "rather assuming ..." part. I chose the example of the additive identity, because it shows how to start a uniqueness proof. The contradiction part was added, because there are a few theorems and exercises coming up in Apostol that the OP will need to know this idea.Rather assuming that if two things satisfy a given property, then they are equal. Sometimes contradiction can be used, but most of the times it isn't necessary (in your example of the additive identity it isn't necessary: if 0,0' are identities, it follows that 0 + 0' = 0 = 0' + 0 and 0+0' = 0' = 0' + 0, in particular thus 0=0')

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