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For any natural numbers $x,y,p$

$x<y$ if and only if $x+p<y+p$. It is asked to prove by induction.

My attempt: Base case: Let $p=1$. Then $1+x < y+1$.

Inductive step: suppose $x<y$

Then, $S(x)<S(y)$

Adding $p$ to both side of inequality,

$S(x)+p < S(y)+p$

Thanks in advance

Joffan
  • 39,627
Natasha J
  • 825

3 Answers3

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As a proof? No, it's is not correct.

The form of induction is this:

  • First you prove the base case. Here you have stated it but not proved it.

  • Then you assume that the statement is true for some $k$ and using that assumption (the "inductive hypothesis") prove that it is also true for $k+1$.You haven't done this either.

It seems likely that the intention was for you to run induction on values of $p$. It's not clear what foundational truths you are relying on here but your use of $S()$ suggests that you can rely on $S(x) = x+1$ and $S(x)>x$.

Then the base case for example might need two cases: $y=S(x)$ and $y>S(x)$.

The inductive step could use associativity, $x+(p+1) = (x+p)+1$.

Hope this is enough to get you on track.

Joffan
  • 39,627
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No, it is not correct, because you use the theorem inside the proof (doing $S(x)<S(y)\implies S(x)+p<S(y)+p$).

The inductive proof requires $x<y\implies S(x)<S(y)$, as well as $S(a+b)=a+S(b)$. If this is taken for granted, the inductive step is

$$x+p<y+p\implies S(x+p)<S(y+p)\implies x+S(p)<y+S(p).$$

Then as the claim is true for $p=0$, it is true for all $p$.

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ok! so here is another try. I am new to induction so please bear with me! Thanks.

Base case: for p=1

x+1<y+1

this implies x<y.

inductive step: Assume it is true for p.

then x+p<y+p this implies S(x+p)<S(y+p)

this implies (x+p)+1<(y+p)+1

this implies x+(p+1)< y+(p+1)

Is this correct? Thanks for your time!

Natasha J
  • 825
  • Somehow you have equals signs where you should have inequalities? And you are still stating a lot things rather than proving them. I think you are perhaps drawing on knowledge/truths which are subsequent to this proof, ideas like "subtract from both sides" which isn't yet available at this level of proof. – Joffan Mar 12 '21 at 15:50
  • It might be worth practicing induction proof in a less restricted environment than abstract analysis. For example, prove the formula for triangular numbers $T(n) =\sum_{k=1}^n k= n(n+1)/2$ by induction. – Joffan Mar 12 '21 at 15:54
  • I added such a proof to an old triangular numbers formula question here as a demonstration. – Joffan Mar 12 '21 at 16:15
  • It was a typing mistake, that equal sign. – Natasha J Mar 12 '21 at 16:45
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    This is getting closer. You should recall though that you are aiming to prove both directions (if and only if): both $x<y \implies x+p<y+p$ and $x+p<y+p \implies x<y$. – Joffan Mar 12 '21 at 22:00
  • So I have to prove both ways by induction? – Natasha J Mar 13 '21 at 01:50
  • Somehow, both implications should be proved - perhaps at the same time, I haven't considered the details. – Joffan Mar 13 '21 at 16:12