Recent content by issacnewton

Ok, fresh_42, taking hints from you, following is my proof. Assume to the contrary that ##f## is not one to one. Then ##\exists a,b \in \mathbb{N}## such that ##f(a) = f(b) ## and ## a \ne b ##. I am going to use a Lemma here, which I have already proven. Lemma: Let ##x \in \mathbb{N}##...

Now, we are given that ##\mathbb{N}'## is a set and ##1' \in \mathbb{N}'## and ##s' :\mathbb{N}' \rightarrow \mathbb{N}' ## is a function. So, using the recursion theorem, there is a unique function ##f : \mathbb{N} \rightarrow \mathbb{N}'## such that ##f(1) = 1'## and ## f\circ s = s' \circ...

Awesome !!. I had studied the book "How to Prove It: A Structured Approach" by Daniel Velleman. He is a set theorist, so the language of the book is very precise. So, I am very comfortable with quantifiers. Proofs become easier to follow with them.

Ok. I think I can put your arguments together and following is my proof. Let ##H## be defined as follows $$H = \Bigl\{a \in \mathbb{N} \;| \forall n \in \mathbb{N}\; (n \in G \to n \geqslant a) \Bigr \} $$ Now, for arbitrary ##n \in \mathbb{N}##, suppose that ##n \in G##. Since ##n \in...

Thanks.

I am trying to understand the proof given in Ethan Bloch's book "The real numbers and real analysis". I am posting snapshot of the proof in the book. I am also posting theorem 1.2.9 given in the book. Here author is trying proof by contradiction. First, I don't understand why specific...

Thanks nuuskur. If I write many equations like (1) and (2), it will take lot of space and I may not get any responses.

with this background, we proceed to the proof. Let us define a set $$ G = \{ x \in \mathbb{N} | \; y, z \in \mathbb{N}\; \text{ if } (x \cdot z) = (y \cdot z) \text{ then } x = y \} $$ We want to prove that ##G = \mathbb{N} ##. For this purpose, we will use part 3) of Peano postulates...

fresh_42, thanks for your input. I am learning this stuff on my own. so, it would be difficult to write as compact proofs as you did. Also, for somebody who is learning this topic on his or her own, detailed explanation would be helpful. All future readers of my post will be able to understand...

with this background, we proceed to the proof. Let us define a set $$ G = \{ z \in \mathbb{N} | \; x, y \in \mathbb{N}\; (x \cdot y) \cdot z = x \cdot (y \cdot z) \} $$ We want to prove that ##G = \mathbb{N} ##. For this purpose, we will use part 3) of Peano postulates given above...

with this background, we proceed to the proof. Let us define a set $$ G = \{ z \in \mathbb{N} | \mbox{ if } y \in \mathbb{N}, y\cdot z = z \cdot y \} $$ We want to prove that ##G = \mathbb{N} ##. For this purpose, we will use part 3) of Peano postulates given above. Obviously, ## G...

Thanks for the input fresh_42. Even I noticed that. I was wondering how could I improve the readability of my proof. I will try to follow your advice from now on.

I want to prove that ##(a+b)\cdot c=a\cdot c+b\cdot c## using Peano postulates where ##a,b,c \in \mathbb{N}##. The book I am using ("The real numbers and real analysis" by Ethan Bloch ) defines Peano postulates little differently. Following is a set of Peano postulates I am using. (Axiom 1.2.1...

I have to prove ##a \cdot 1 = a = 1 \cdot a## for ##a \in \mathbb{N}##. The book I am using ("The real numbers and real analysis" by Ethan Bloch) defines Peano postulates little differently. Following is a set of Peano postulates I am using. (Axiom 1.2.1 in Bloch's book) There exists a set...

I have to prove that ##1 + a = s(a) = a + 1## using Peano postulates if ##a \in \mathbb{N}##. The book I am using ("The real numbers and real analysis" by Ethan Bloch) defines Peano postulates little differently. Following is a set of Peano postulates I am using. (Axiom 1.2.1 in Bloch's book)...