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I've seen identities, which means they have infinite solutions.$\longleftarrow$This is incorrect; see the comments below. Examples are$$4x+6=2(2x+3)$$$$9q-6\lt9q+3$$$$12u\le3(4u)$$$${x\over 9}={x\over 3\cdot3}$$$$15a\ge15a-2$$The first example evaluates to $4x+6=4x+6$ using the Distributive Property. How is it possible that (maybe some) of these equations or inequalities are identities? Maybe they have the same variable terms on both sides of each example and the constants are the same in an equation and one constant is bigger/smaller than the other constant on the other side depending on the inequality symbol. Maybe that's what makes them identities! I want to hear from your comfortable answers!

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  • $\begingroup$ I'd rather call such relations "tautologies", to express that they are always true (regardless the values of the unknowns). I reserve "identities" for equal expressions. $\endgroup$ – Yves Daoust Oct 28 '14 at 23:59
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    $\begingroup$ It is not correct to say "infinite solutions" when you mean "infinitely many solutions". "Infinite solutions" means "solutions, each one of which is infinite". If there were such a thing as an infinite solution, and an equation has six of those, and no finite solutions, then there would be infinite solutions, but there would not be infinitely many solutions. Moreover, it is incorrect that if an equation has infinitely many solutions, then it is an identity. For example, $\sin x=0$ has infinitely many solutions, but it is not an identity since most numbers are not solutions of it. ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 29 '14 at 1:45
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To make your own identity, start with something that you know is definitely true, such as: $$ 1 + 1 = 2 $$ Now make the identity more complicated by doing legal (and reversible) operations to both sides of the equation. For example, we can add $x$ to both sides: $$ 1 + 1 + x = 2 + x $$ then multiply both sides by $3$: $$ 3(1 + 1 + x) = 6 + 3x $$ and so on. Notice that at any point, we may reverse our steps to get back to our initial identity.

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    $\begingroup$ The steps you show are reversible, but that doesn’t really matter. Starting with $3(1+1+x)=6+3x$ to get to $1+1=2$ doesn’t prove that $3(1+1+x)=6+3x$ is an identity. (You could similarly derive $1+1=2$ from the false “identity” $3x+4=2x-7$: multiply both sides by zero, then add 2 to both sides.) Anything implies $(1+1=2)$, so that’s not a very helpful thing to know. Things that follow from $1+1=2$ are sure to be true. Things from which you can derive that $1+1=2$, who knows? $\endgroup$ – Steve Kass Oct 29 '14 at 2:12
  • $\begingroup$ Reversible steps absolutely matter. In your example, you multiply both sides by zero. That is definitely not reversible, as we can't divide by zero. $\endgroup$ – Adriano Oct 29 '14 at 3:31
  • $\begingroup$ Steps need to be reversible only if you start with an identity you hope to prove and end with something like $1+1=2$. In your example, you started with $1+1=2$ and ended with an identity. That alone is a proof of the identity, and the steps need not be reversible. There’s nothing wrong with your derivation of the true identity $3(1+1+x)=6+3x$. Your remark that you can reverse the steps, however, shouldn’t be taken as meaning only reversible steps are allowed. (You could take the $\tan$ of both sides and still have an identity, even though $\tan(A)=\tan(B)$ does not imply that $A=B$. $\endgroup$ – Steve Kass Oct 29 '14 at 4:18
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For the second identity, subtracting $9q$ from both sides results in the true statement $-6<3$, so this is an identity.

For the third, note first that the statement is true for $u=0$. Otherwise, dividing both sides by $u$ gives the true statement $12 \le 3\cdot 4$.

The arguments are similar for the other two.

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  • $\begingroup$ Be careful. Just because you can reach $-6<3$ from an equation doesn’t mean it’s an identity. You want to start with $-6<3$ (or any true statement) and get your identity. For example, you can get $-6<3$ from any starting point, true or false, because it’s a fact. Start with $1=2$. If $1=2$, then whenever $x<y$, you know that $1+x<2+y$. Do this for $x=-7$ and $y=1$, which is okay, because $-7<1$. Then you have $-6<3$. Are you convinced that $1=2$? $\endgroup$ – Steve Kass Oct 29 '14 at 2:16

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