# Trying to figure out how to use "proof related notations" correctly

Let $$a$$ and $$b$$ be positive numbers. Prove that $$a+b=1$$ if and only if $$a=2t/(1+t)$$ and $$b=(1-t)/(1+t)$$ for some number $$t$$, $$0.

My Proof:

Let $$P$$ denote: "$$a=2t/(1+t)$$ and $$b=(1-t)/(1+t)$$ for some number $$t$$, $$0"

Let $$Q$$ denote: "$$a+b=1$$"

Let $$G$$ denote: "$$a$$ and $$b$$ are positive numbers"

Let $$S$$ := "$$G$$ and $$Q$$"

We want to show that $$P => S$$ and $$\neg P => \neg S$$, then it follows that that $$S$$ is true iff $$P$$ is true. First we prove $$P => Q$$ and $$P => G$$:

$$a+b = 2t/(1+t) + (1-t)/(1+t) = (1+t)/(1+t)=1.$$ We see that $$a+b=1$$ for all reals (except -$$1$$) and we know that $$a$$ and $$b$$ $$\in \mathbb R$$ whenever $$t$$ $$\in \mathbb R$$, and $$0 are certainly real numbers. Hence $$P => Q$$.

If $$t$$ is less than one but greater than zero, we find that the denominator $$(1+t)$$ will be positive no matter what. Same goes for $$2t$$. We also find that the numerator of $$b$$, $$(1-t)$$, will remain positive. Thus we have shown that the numerator and denominator of both $$a$$ and $$b$$ will always be positive in the intervall $$0, and so $$a$$ and $$b$$ will always be positive aswell in $$0, i.e. $$P => G$$ and so we have successfully proved that $$P => S$$.

Next we want to show that $$\neg P => \neg S$$. Expressing $$\neg P$$ in words we have:

"$$a=2t/(1+t)$$ and $$b=(1-t)/(1+t)$$ for some number $$t$$, $$t \leq 0$$ or $$1 \leq t$$"

By simple using the values for $$t$$ you will easily find that $$a$$ and $$b$$ won't be positive. That is, $$\neg P => \neg G$$. And since "$$\neg G$$ and $$Q$$" is not the same as $$S$$ defined to be "$$G$$ and $$Q$$", we have $$\neg P => \neg S$$. ∎

Questions: I am quite sure that this proof has some mistakes to learn from that will be a valuable lesson. First of all I would like to have the flaws of the argument pointed out if found. I would like to mention that deinfing $$S$$ to be two cases at the same time is nothing I have seen nor done before, can one do that? Also, I have read that $$A => B$$ is "logically equivalent" with $$\neg A => \neg B$$, wouldn't that imply that we only have to show $$P => S$$ and then we are done? Thanks.

• Your negative of $P$ is wrong. The correct would be there is no $t \in (0, 1)$ such that $a = 2t/(1 + t)$ and $b= (1-t)/(1+t)$. Commented Aug 14, 2019 at 16:10
• Also, answering your question, it is true that $A \implies B$ is equivalent to $\neg B \implies \neg A$. In this case It is easier to prove the implication $S \implies P$. Just assume $S$ and find out which is the value of $t$ that makes the claim true. Commented Aug 14, 2019 at 16:12
• Assuming S to find out which is the value of t that makes the claim true - wouldn't I then have showed that "if and only if a+b=1 and a & b are positive will we have a=2t/(1+t) and b=(1−t)/(1+t) with t between 0 and 1". Isnt that the opposit/the reverse of what we are asked to prove here? Commented Aug 14, 2019 at 17:22
• Sketch of proof for $S \implies P$: let $a, b > 0$ with $a + b = 1$. Notice that $1 = a + b > a$ and also $b < 1$. Now, define $t = a/(2 - a)$. We can check that $t \in (0, 1)$ since $a \in (0, 1)$. Also, we have $a, b$ are expressed like we want in $P$. Commented Aug 14, 2019 at 17:40
• Just to get it very clear in my brain. If P => Q, then it is correct to say "if and only if Q then P" yes?. Commented Aug 14, 2019 at 18:11

## 1 Answer

I decided to write an answer resuming the comments.

On the correctness of your proof. You defined statements $$P, Q, G, S$$. You want to show that $$P \iff S$$, where $$S$$ is "$$Q$$ and $$G$$". The proof of $$P \implies S$$ is correct and the strategy of expressing $$S$$ as a compound statement and proving each part separately is ok.

However, the proof of $$S \implies P$$ (which is indeed equivalent to $$\neg P \implies \neg S$$) is incorrect. The problem is that negating the statement $$P$$ is not

"$$a=2t/(1+t)$$ and $$b=(1−t)/(1+t)$$ for some number $$t$$, $$t\le 0$$ or $$1\le t$$"

but the statement

"there is no $$t\in (0,1)$$ such that $$a=2t/(1+t)$$ and $$b=(1−t)/(1+t)$$".

Proof of $$S \implies P$$: Let $$a,b>0$$ with $$a+b=1$$. Notice that $$1=a+b>a$$ since $$b>0$$. Analogously we have $$0. Now, define $$t:=a/(2−a)$$, that is always well defined since $$a \neq 2$$. The definition of $$t$$ above is not arbitrary, it simply comes from expressing $$a$$ in terms of $$t$$ in our target expression. Notice that $$0 < a < 1\ \text{ and }\ 1 < 2 - a < 2$$ Since both are positive, so is $$t = a/(2-a)$$. Also, we have $$t = \frac{a}{2-a} < \frac{1}{1} = 1$$ and we conclude that $$0 < t < 1$$. Now that we have a candidate $$t$$ we only have to check that the expressions for $$a$$ and $$b$$ are correct. The expression for $$a$$ surely is correct, since we derived $$t$$ so that it would be so. Also, notice that $$b = 1 - a = 1 - \frac{2t}{1+t} = \frac{(1+t) - 2t}{1 + t} = \frac{1 - t}{1 + t},$$ and we showed that the expression for $$b$$ is also correct, implying $$P$$.