trying to solve a contradiction equation

Question

${2x + 1 = 2x + 3}$

I know its a contradiction (correction) because its false for all values of X. What I'm trying to do is figure out why.

${3x + 6 = 3 (x + 2)}$

this is an identity. still want to know why. as in what result would tell me that its an identity.

( sorry my math teacher skipped over those because they're basics I guess ? )

my attempts :

${2x + 1 = 2x + 3}$

what I did for this one is (correction) subtract 2x on both sides so I'm left with

${ 1 = 3 }$

do I subtract 1 from 3 ? then its = 2
or subtract 3 from 1 then it's = -2
or is it impossible to subtract and its just 1 = 3 ?

${3x + 6 = 3 (x + 2)}$

as for this one I divided by 3x on both sides and I'm left with

${6 = 6}$ so if the solution is a 0 its an identity ( at least that's what I got ).

"what I did for this one is divide by $2x$ on both sides". Actually, you subtracted $2x$ on both sides. — paw88789, Sep 14 '23 at 13:30
Why don't you just multiply out the right hand side for the second expression ? And for the first, subtract $2x$ from both sides ? — true blue anil, Sep 14 '23 at 13:32
$1=3$ is a false statement and because it does not have $x$ in it, it is false for any $x$. One other thing: you can divide an equation by something that is never zero; it looks like you subtracted the same from both sides. Adding, subtracting or multiplying is OK. — Vasili, Sep 14 '23 at 13:35
"3x + 6 = 3 (x + 2)
this is an identity. still want to know why. as in what result would tell me that its an identity." This is the distributive property of multiplication and addition... — 5xum, Sep 14 '23 at 13:39
Clearly there is no real number $x$ that satisfies $2x+1=2x+3$, as the existence of such a real number would give rise to a proof of the false statement "1 = 3". On the other hand, every real number $x$ satisfies $3x+6=3(x+2)$, because of the distributive law. Thus by definition, $2x+1=2x+3$ is a contradiction; and $3x+6=3(x+2)$ is an identity. — paw88789, Sep 14 '23 at 13:43
For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. — Henrik supports the community, Sep 14 '23 at 14:27
@Henriksupportsthecommunity I stand corrected, It makes everything bigger and easier to read I assume. — Moath, Sep 14 '23 at 14:31
Your question does need MathJax. It looks terrible (to me) because it doesn't use it. — Henrik supports the community, Sep 14 '23 at 14:35
yea figured out it made everything bigger I guess ? I used MathJax in my first question because it needed it, Will use it next time for sure, thanks. — Moath, Sep 14 '23 at 14:36

score 1 · Accepted Answer · answered Sep 14 '23 at 13:41

In both examples you are asked to find out for which values of $x$ the left and right sides of the equation represent the same number.

In the first case you correctly conclude that if for some number $x$, $$ 2x + 1 = 2x + 3 $$ then subtracting $2x$ from both sides tells you $$ 1 = 3 . $$ Since that is false, the answer is that the original equation is never true.

You need not call the equation a contradiction. All you need say is that there is no value of $x$ that makes it true.

In the second case the distributive law tells you that whatever $x$ may be, $$ 3(x+2) = 3x +6 $$ so that equation is always true.

An equation like the second one, that's always true whatever $x$, is called an identity.

@Moath Yes. The equal sign in mathematics means just that the two expressions are different names for the same thing. See https://math.stackexchange.com/questions/2738360/what-exactly-is-an-equation/2738382#2738382 — Ethan Bolker, Sep 14 '23 at 14:07

score 1 · Answer 2 · answered Sep 14 '23 at 14:09

You successfully reached a falsehood, essentially $1=0$, which we just take for granted as false. But then why should that be false?

In a sense, you did the mathematical equivalent of repeatedly asking "why A? (because B) ...why B? (because C) ...why C?" and so forth. At some point, logically tracing back to the source, you will eventually reach a rock bottom, where the only available answer is a "just because". See Münchhausen's trilemma. We want to rest our logical system on some foundation in order to avoid it relying circularly on itself or resting on an infinite chain. But, as we unfortunately lack an oracle who can provide us with any fundamental truths, we have to assume that foundation ourselves. We call such assumptions 'axioms'.

Here, those relevant "just because" axioms may be the essential properties of the integers or real numbers.

So why is $1=0$ false? Because we assume the real numbers to have an order or to at least have some distinct elements. A number line needs, at the very least, different numbers on it! Assume that $1=0$ were true. Then all other numbers would satisfy $x=x\cdot 1=x\cdot 0 = 0$; every number would then be identically zero and equal to every other number. In order to accept $1=0$, we must give up another essential assumed property, such as having distinct numbers. So we take for granted that $1=0$ is false.

This shows also that there are systems we could devise in which we may equate the elements "1" and "0" (eg, a Field with one element), but the point is, that is not what we typically want for our number system.

trying to solve a contradiction equation

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