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 $t$, $0<t<1$.

My Proof:

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

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 $0<t<1$ 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 $0<t<1$, and so $a$ and $b$ will always be positive aswell in $0<t<1$, 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.

  • $\begingroup$ Your negative of $P$ is wrong. The correct would be there is no $t \in (0, 1)$ such that $a = 2t/(1 + t)$ and $b= (1-t)/(1+t)$. $\endgroup$ Commented Aug 14, 2019 at 16:10
  • $\begingroup$ 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. $\endgroup$ Commented Aug 14, 2019 at 16:12
  • $\begingroup$ 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? $\endgroup$
    – Hassebae
    Commented Aug 14, 2019 at 17:22
  • 1
    $\begingroup$ 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 $P$. $\endgroup$ Commented Aug 14, 2019 at 17:40
  • $\begingroup$ 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?. $\endgroup$
    – Hassebae
    Commented Aug 14, 2019 at 18:11

1 Answer 1


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 $0<b<1$. 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 $P$.


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