How to prove $x + 3 = (x + 1)^2 \Leftarrow x=-2\;\text{or}\;1 \,? \quad \sqrt{x+3}=x+1 \Leftarrow x=1 \, ? \quad x^2 = x \Leftarrow x=0 \; $?

Question

I can't deduce the $\color{blue}\impliedby$ of the 3 blue biconditionals below. How do I explicitly wrest or wring out the red functions in terms of x, from the 1 or 2 integers on the RHS?

Doubtless, I know that the RHS's roots satisfy the LHS! For example, for the first equation, I know $-2 + 3 \equiv \color{red}{(-2 + 1)^2}, 1 + 3 \equiv \color{red}{(3 + 1)^2}$. But it doesn't suffice to evaluate functions with 1 or 2 integers, because this merely verifies the $\color{blue}\impliedby$ for merely 2 integers of $x$! We need to prove the $\color{blue}\impliedby \forall \, x$. Hence our conclusion must express the red functions in terms of $x$!

1. $x+3=\color{red}{(x+1)^2} \color{blue}{\iff} x=-2\;\text{or}\;1\;$

   $x = -2 \iff x \color{limegreen}{+3} = -2 \color{limegreen}{+3} \iff x + 3 = 1.$

   OR $\quad x = 1 \iff x \color{limegreen}{+3} = 1 \color{limegreen}{+3} \iff x + 3 = 4.$ Then how to deduce $\color{red}{(x+1)^2}$?

2. $\color{red}{\sqrt{x+3}}=x+1 \color{blue}{\iff} x=1 $

   $x = 1 \iff x \color{limegreen}{+1} = 1 \color{limegreen}{+1} \iff x + 1 = 2.$ Then how to deduce $\color{red}{\sqrt{x+3}}$?

3. $\color{red}{x^2 = x} \color{blue}{\iff} x=0\;\text{or}\;1$

   $x = 1 \iff x \color{limegreen}{\cdot x} = 1 \color{limegreen}{\cdot x} \iff x^2 = x.$ I have no question on this case, which works!

   But $x = 0 \iff x \color{limegreen}{\cdot x} = 0 \color{limegreen}{\cdot x} \iff x^2 = 0.$ Then how to deduce $\color{red}{x^2 = x}$?

Answer 1

1. $$x=p \:\:\text{or}\:\:q$$ is abbreviation for $$x=p \quad\text{or}\quad x=q.$$
2. The statement $$P(x)\impliedby \big(R(x)\quad\text{or}\quad S(x)\big)$$ implicitly means that whenever either $R(x)$ or $S(x)$ is true for some value of $x,$ $P(x)$ is also true for the same value. Logically, this means $$\text{for each }x,\big(R(x){\implies} P(x)\big)\quad\textbf{and}\quad\text{for each }x,\big(S(x){\implies} P(x)\big).$$

So, to prove $$\color{red}{x+3}=\color{violet}{(x+1)^2}\impliedby x=-2\:\:\text{or}\:\:1\tag1$$ just means to argue this:

• for $x=-2,$ $$\color{red}{x+3}=1=\color{violet}{(x+1)^2};$$
• while for $x=1,$ $$\color{red}{x+3}=4=\color{violet}{(x+1)^2}.\tag2$$

$x+3=\color{red}{(x+1)^2} \color{blue}{\iff} x=-2\;\text{or}\;1\;$

$x = -2 \iff x \color{limegreen}{+3} = -2 \color{limegreen}{+3} \iff x + 3 = 1.$

OR $\quad x = 1 \iff x \color{limegreen}{+3} = 1 \color{limegreen}{+3} \iff x + 3 = 4.$ Then how to deduce $\color{red}{(x+1)^2}$?

1. Replace the logically incorrect or with and or but or while or somesuch.
2. Are you just working in the reverse direction? Then display ⟸ instead of ⟺. Not only is this less confusing for the reader, since you aren't paying attention to the forward direction you may inadvertantly assert false facts by displaying a stronger symbol than intended.
3. Writing x+3 = 1+3 = 4 instead of x+3 = 1+3 , x+3 = 4 makes it easier for both the reader and yourself.

---

Addendum corresponding to the OP's new comment and edit

Doubtless, I know that the RHS's roots satisfy the LHS!

What you're saying here is exactly what it means to prove LHS ⟸ RHS: saying that the root $a$ satisfies the equation $x^3+7=11$ means that plugging in $x{=}a$ makes the equation true.

merely verifies the $\color{blue}\impliedby$ for merely 2 integers of $x$! We need to prove the $\color{blue}\impliedby \forall \, x.$

Let's make explicit the key point of the above answer: to prove $$\color{red}{x+3}=\color{violet}{(x+1)^2}\impliedby x=-2\:\:\text{or}\:\:1\tag1$$ means to argue that $$\text{for each }x,\big(x=-2{\implies} \color{red}{x+3}=\color{violet}{(x+1)^2}\big)\\\quad\textbf{and}\quad\text{for each }x,\big(x=1{\implies} \color{red}{x+3}=\color{violet}{(x+1)^2}\big).$$ In particular, say for $x{=7},$ we have $$\big(\text{False}{\implies} \text{False} \big)\\\quad\textbf{and}\quad\big(\text{False}{\implies} \text{False} \big),$$ that is, two true implications. So, statement $(1)$ is true for $-2$ and $1$ and trivially true for every other real number; that is, statement $\boldsymbol{(1)}$ is true for every real value of $\boldsymbol x.$ The word 'trivially' means that that portion of the proof is tacit; a succinct proof is more welcome than a tedious proof.

Hence our conclusion must express the red functions in terms of $x$!

Alright, then let's make explicit this line from above: $$\color{red}{x+3}=4=\color{violet}{(x+1)^2}.\tag2$$ This means $$\color{red}{x+3}=4\\\text{and}\quad\color{violet}{(x+1)^2}=4;\\\text{therefore}\quad \color{red}{x+3}=\color{violet}{(x+1)^2}.$$ We are allowed to read an equality from either side, due to the symmetric property of equality.

I don't understand your first sentence "Replace the logically incorrect or with and or but or while or somesuch."

I'm saying that there's a difference between asserting that statements $A$ and $B$ are true, and asserting that either statement $A$ or $B$ is true. In the former, both statements are true; in the latter it could be that $A$ is false while $B$ is true. (The former implies the latter, but not vice versa.) In our above proof, we do require that both statements be true.

To reiterate what I said before: (A or B) implies C is logically equivalent to (A implies C) and (B implies C), not to (A implies C) or (B implies C).

Answer 2

To deduce $$f(x)=g(x)\impliedby x=a$$ in case 2, simply check that $$f(a)=g(a).$$

And to deduce $$f(x)=g(x)\impliedby (x=a\lor x=b)$$ in case 1 and 3, simply check that $$f(a)=g(a)\text{ and }f(b)=g(b).$$

1. $f(x)=x+3,$ $g(x)=(x+1)^2,$ $f(-2)=1=g(-2),$ $f(1)=4=g(1).$

2. $f(x)=\sqrt{x+3},$ $g(x)=x+1,$ $f(1)=2=g(1).$

3. $f(x)=x^2,$ $g(x)=x,$ $f(0)=0=g(0),$ $f(1)=1=g(1).$

Answer 3

The deduction from $x=0$ to $x^2=x$, for example, is quite intuitive. After all, $0^2=0$. If you are wondering what the formal reasoning behind this deduction is, it is the axiom schemas of equality, specifically the axiom schema of substitution for formulas:

For all variables $x$ and $y$, and any formula $\varphi(x)$, if $\varphi'$ is obtained by replacing any number of free occurances of $x$ in $\varphi$ with $y$, such that these remain free occurances of $y$, then $$ x=y \implies (\varphi \implies \varphi'). $$

This does require some familiarity in logic to understand fully, but we can look at an example. Suppose we want to show that $x=0\implies x^2=x$. First, we have $x=0\implies 0=x$ (this is known as the symmetry of equality, which can be deduced from the axiom schemas of equality). From the axiom of substitution, we know that $$ 0=x \implies (0^2=0\implies x^2=x). $$ We know that $0^2=0$ is true, so $0^2=0\implies x^2=x$ is logically equivalent to $x^2=x$. Chaining together all those implications, we have $x=0\implies x^2=x$.

Now, you can apply the axiom schemas of equality to prove all the implications in question, but you can also show that $a=b\implies f(a)=f(b)$ and apply this instead to avoid all the hassles.

Answer 4

The explicit answer is to your questions :

The algebraic steps you used to construct the original equations are correct, however these algebraic steps do not allow us to derive the original equations .

---

For instance :

$$\begin{align}\color{red}{\sqrt{x+3}}&=x+1 \color{blue}{\iff} x=1 \end{align}$$

and

$$\begin{align}x = 1 &\iff x \color{limegreen}{+1} = 1 \color{limegreen}{+1} \\ &\iff x + 1 = 2.\end{align}$$

Then how to deduce $\color{red}{\sqrt{x+3}} \thinspace\thinspace\thinspace ?$

---

Recall that, you have :

$$ \begin{align}&\sqrt {x+3}=x+1\\\\ \iff &\begin{cases}x+3=(x+1)^2\\ x+1≥0\end{cases}\\\\ \iff &\begin{cases}x^2+x-2=0\\ x+1≥0\end{cases}\\\\ \iff &\begin{cases} x\in\{-2,1\}\\ x≥-1\end{cases}\\\\ \iff &x=1 \quad\tiny{\blacksquare}\end{align} $$

You can now reverse the steps :

$$ \begin{align}x=1&\iff\begin{cases} x\in\{-2,1\}\\ x≥-1\end{cases}\\\\ &\iff \begin{cases}x^2+x-2=0\\ x+1≥0\end{cases}\\\\ &\iff \begin{cases}x+3=(x+1)^2\\ x+1≥0\end{cases}\\\\ &\iff \sqrt {x+3}=x+1 \\\\ &\iff x+1=\sqrt {x+3} \quad\tiny{\blacksquare}\end{align} $$

Note that the following equivalence relations are also valid :

$$ \begin{align}x=1&\iff x+1=1+1\\ &\iff x+1=2\end{align} $$

As you can see, the algebraic steps are all consistent and there is no contradiction between them . However, depending on the result you want to achieve, i.e. for our purposes, it is important what algebraic steps you will take .

Answer 5

(Adapted from my comments.)

Let's analyse the situation on one of your examples: $x+3=(x+1)^2\impliedby x=-2\text{ or }x=1$.

It seems that the real question here is: obviously we can substitute $x=-2$ and $x=1$ into the LHS and see that the equation is satisfied. But, why is this enough? Why don't we need to substitute any other $x$ or provide some generic proof that would work for every $x$? It feels a bit like cheating, doesn't it?

In fact, it is not cheating and is completely legitimate. To prove the converse implication $\impliedby$, you only need to prove the LHS is true when the RHS is true. (Which in our case means only checking $x=-2$ and $x=1$.) It is irrelevant whether the LHS is true when the RHS is false.

Recall the truth table for implication: $$\begin{array}{c|c|c}p&q&p\implies q\\\hline\bot&\bot&\top\\\bot&\top&\top\\\top&\bot&\bot\\\top&\top&\top\end{array}$$

The entire implication is true when the premise $p$ is false. The only way for the implication to be false is when the premise $p$ is true and the conclusion $q$ is false.

This closely matches natural logic. If you want to check if the sentence "All fish live in the water" is true, you only need to check the fish. (And, if you find one that does not live in the water, then it is false; otherwise it is true.) You don't need to check any other non-fish animals, say mammals. Whether they live in the water is irrelevant (and, as a matter of fact some do, and some don't). The same as it is irrelevant, in your case, whether the LHS is true for the $x$ that is neither $1$ nor $−2$.