Let’s say I have $x = x + 1$, which is a false statment for real $x$; why can I solve for real $x$ when I square both sides of the equation, giving $x^2=(x+1)^2$?
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Why can't you square both sides of an equation?  I've been asked this many times and can never quite give a good, clear, concise answer (for beginning algebra students) in plain language. I just searched the web and still couldn't find a simple-to-understand answer for why squaring both sides gives you extraneous solutions.
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Is it possible to prove that $1+1 = 2$? Or rather, how would one prove this algebraically or mathematically?
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As far as I see, $\sqrt{x^2}$ is not $x$, but $|x|$, meaning the "absolute". I totally get this, because $x^2$ is positive, if $x$ is negative, so $\sqrt{y}$, whether $y = 10^2$ or $y = -10^2$: $y$ is positive.  But then I remember that $\sqrt(x)$ is the same as $x^{1/2}$ and thus, $\sqrt{x^2}$ is the same as $x^{1}$.   So, as far as I get it, $\sqrt{x^2} = x^{2/2}$.  But then, I can cancel $\frac{2}{2}$ to $1$. So: $\sqrt{x^2} = x^1 = x$.  Where does the absolute value in this derivation of the calculation come from?  I get it why its there: Because $(-x)^2 = x^2$, so the radicand is always positive. But from $\sqrt{x^n} = x^{n/2}$ I dont understand how it gets there.
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I'm pretty confident that this is a valid proof, but I still need people who are better than me to confirm it, here's the proposition: Proposition: $2^x=0$ has no real solutions. Proof: By contradiction Assume $2^x=0$ has a real solution  (i.e. $2^x=0$ when $x=\alpha$), then $$2^\alpha=0 \implies 2^\alpha \cdot 2=0 \implies 2^{\alpha+1}=2^\alpha \implies \alpha=\alpha+1$$ Which is a contradiction because $\alpha=\alpha+1$ has no real solutions. $\therefore 2^x=0$ has no real solutions.
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Before, the concept of imaginary numbers, the number $i = \sqrt{-1}$ was shown to have no solution among the numbers that we had. So we declared $i$ to be a new type of number. How come we don't do the same for other "impossible" equations, such as $x = x + 1$, or $x = 1/0$?  Edit: OK, a lot of people have said that a number $x$ such that $x = x + 1$ would break the rule that $0 \neq 1$. However, let's look at the extension from whole numbers to include negative numbers (yes, I said that I wasn't going to include this) by defining $-1$ to be the number such that $-1 + 1 = 0$. Note that this breaks the "rule" that "if $x \leq y$, then $ax \leq ay$", which was true for all $a, x, y$ before the introduction of negative numbers. So I'm not convinced that "That would break some obvious truth about all numbers" is necessarily an argument against this sort of thing.
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So I was going through  my 11th class package on Quadratic equations and I saw a question to prove that a polynomial of $4$th degree with all real roots cannot have $\pm 1$ as all its coefficients.   I tried proving it using calculus, by showing that at least one consecutive maxima  and minima will lie either above or below the x axis, but couldn't solve it using that.  I also tried using Descartes Rule of Signs but couldn't solve it with that too. Any help?
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Possible Duplicate:          I'm trying to understand the exact point of failure in the following reasoning:  \begin{equation*} 1 = \sqrt{1} = \sqrt{(-1)(-1)} = \sqrt{\sqrt{-1}^2\sqrt{-1}^2} = \sqrt{(\sqrt{-1}\sqrt{-1})^2} = \sqrt{-1}\sqrt{-1} = \sqrt{-1}^2 = -1. \end{equation*}  I've been previously told that the problem is due to square root not being a function in C; which I found totally unhelpful. Could someone please explain the problem here in simpler terms.  Edit:  Thank you all for your comments in trying to help me understand this. I finally do. Following is the explanation on my problems in understanding this, in case it'll be of any help to anyone else.  My problem was really due to using an incorrect definition of i: $i = \sqrt{-1}$. While the correct definition would be: $i^2 = -1$.  My incorrect definition led me reasoning, such as (which superficially seemed to give expected results. I see now that this is incorrect, too):  \begin{equation*} \sqrt{-9} = \sqrt{9 * (-1)} = \sqrt{\sqrt{9}^2 \sqrt{-1}^2} = \sqrt{(\sqrt{9} \sqrt{-1})^2} = \sqrt{9} \sqrt{-1} = 3i. \end{equation*}  Instead, had I used the correct definition of i:  \begin{equation*} {(xi)}^2 = x^2i^2 = -x^2 = -9, \\ x^2 = 9, \\ x = +- 3. \end{equation*}  Now, analyzing the equations in the original problem, I can see at least the following two errors:  1) In the third =, I'm relying on $-1 = {\sqrt{-1}}^2$, while I should be relying on: $-1 = (+-\sqrt{-1})^2$ which would of course give two different branches. Hmm.. on the second reading, this isn't really a problem, as even with the two separate branches, both of them will lead to the result in the next step.  2) In the fifth =, I'm relying on $\sqrt{i^4} = i^2$, which would be correct, if i was a non-negative number in R. But as i is the imaginary unit and in C: $\sqrt{i^4} = \sqrt{i} = +-(1/\sqrt{2})(1 + i)$.
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Find all real polynomials with $P(0)=0$ and $P(x^2+1)=(P(x))^2+1$  I' like a solution for this problem  and proof verification:  My proof:      Let $a_0=0$ and $a_{n+1}=a_n^2+1$, clearly $P(a_0)=a_0$ and if $P(a_n)=a_n$ then $P(a_{n+1})=P(a_n^2+1)=(P(a_n))^2+1=a_n^2+1=a_{n+1}$. So $P(x)=x$ for infinitely many values of $x$ (because $a_n$ is strictly increasing), hence $P$ is the identity polynomial, and it clearly works.
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I saw a less formal version of this fake proof that claimed to prove $2=1$ but because it assumed $a=b$ from the start I knew why it was wrong. It does seem however that the proof can be used to prove an certain implication I find very odd however. Clearly $\forall a. \forall b. a = b \to 0 = 1$ is false because I could pick $a$ and $b$ to be $1$ and then I would have derived 0 = 1 with no other conditions. So there must be something wrong with this proof that I am not seeing.  proof of $\forall a. \forall b. a = b \to 0 = 1$:  $a = b \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (assume)\\ a+a=a+b  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (x=y \to f(x) = f(y), f(x)=a+x)\\ 2a=a+b \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,  (a+a=2a)\\ 2a-2b=a+b-2b \,\,\, (x=y \to f(x) = f(y), f(x)=x-2b)\\ 2(a-b)=a+b-2b \,\, (distributive)\\ 2(a-b)=a-b \,\,\,\,\,\,\,\,\,\,\,\,\,\, (-b=b-2b)\\ 2=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (x=y \to f(x) = f(y), f(x) =x/(a-b))\\ 1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (x=y \to f(x) = f(y), f(x) =x-1) $  now that I have "proven" $\forall a. \forall b. a = b \to 0 = 1$ I can now pick a and b to be 1.  $ 1 = 1 \to 1=0 \,\,\,\,\,\, (a=1, b=1, above)\\ 1 = 1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (\text{reflexivity})\\ 1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (\text{modus ponens}) \\ $  So now it seems that I have managed to derive $1=0$. Clearly I messed up given that no one has been able to show $ZFC$ inconsistent and ostensibly this fits well within the most studied aspects of $ZFC$. Either that or a random guy that makes meme's online found the most profound result in mathematics ever.  So where have I missed the mistake?
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This is a classic problem, but I haven't seen it on the site before.     Suppose $x$ and $y$ are real numbers that both lie in the interval   $(0,1)$. Prove  $$x^y+y^x&gt;1.$$   The following spoiler box contains a hint that indicates the solution I know. Does anyone know a more natural solution?       Use Bernoulli's inequality to show $$x^y\ge \frac{x}{x+y}.$$   Edit: I recently learned the source is the 1996 French math olympiad.
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So everyone knows that when $ax^2+bx+c=0$,$$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$$ But why does it equal this? I learned this in maths not 2 weeks ago and it makes no sense to me
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Example: Let's presume one was attempting to isolate m below:  A common mistake would be: $k^2 = m^2 + n^2 \to k = m +n$  Even though: $k^2 = m^2 + n^2 \to k \neq m +n$  If you apply a square root to both sides of the equation, you will have an inequality.  Why is this true?
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Prove by contradiction (not using a calculator) that $\sqrt6 + \sqrt2 &lt; \sqrt{15}$.   How do you approach such a problem?  I need to admit that I'm completely new to proof writing and I have completely no experience in answering that kind of questions. I tried to square it as I read online but I'm stuck with having $\sqrt3$ and not knowing how to get rid of it to have the answer as clear as the sun. squaring them yielded: $(\sqrt2 + \sqrt6)^2 =  8 + 4\sqrt3 &lt; 15$ which indeed holds but the problem is that I am not allowed to use calculator to check the little difference.  My final answer was $4\sqrt3 &lt; 7$.
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$$ Ae^{Bx} + Cx = D $$  Solve for $x$, where $A, B, C, D$ are real constants. This popped up in a run of the mill high school first-order DE question. No matter what I try with logarithms it doesn't seem like I can isolate an $x$ onto one side of the equation. Is this equation unsolvable analytically? If not, how does one generally go about solving something like this?
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If $x$ is irrational then $3x+2$ is irrational.  My Working:   Proof by contraposition:   $3x + 2$ is rational and $x$ is rational  $x = a/b$  $y = c/d$  $3x + 2 = c/d$  $3x = (c/d) - 2$  $x = c-2d/3d$  $a/b = c-2d / 3d$  I am having a hard time understanding as to why do we end up solving for $x$ in this equation to determine weather $3x+2$ is rational or not if $x$ is rational.
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I am trying to understand the proof above, Proposition 7, given by the author, but I got stuck at some points.   My explanation:  Now, suppose for contradiction there is an $\epsilon &gt; 0$, for all $x &gt; 0$ such that  $NOT (x^2 &lt; 2 &lt; (x+ \epsilon)^2)$,  Since $NOT (x^2 &lt; 2 &lt; (x+ \epsilon)^2) \ \ iff \ \ (x^2 \geq 2 \ \ OR  \ \ (x+ \epsilon)^2 \leq 2)$,  We must show that for some $\epsilon&gt;0$ and for all $x &gt; 0$, the statement $x^2 \geq 2 \ \ OR  \ \ (x+ \epsilon)^2 \leq 2$ leads to a contradiction.  I think in the proof above, author tries to get a contradiction using right hand side of $OR$ clause, namely $(x+ \epsilon)^2 \leq 2$. But no contradicition is achieved by him for the left hand side side of $OR$ clause namely, $x^2 \geq 2$.  First Question is, am I correct at this point, i.e. the need for a contradiction for the left hand side side of $OR$ clause?  Second question is, I don't understand how he derived the following sentence in his proof above,     Since $0^2 &lt; 2$, we thus have $\epsilon^2 &lt; 2$, which then implies $(2 \epsilon)^2 &lt; 2$ and indeed $(n \epsilon)^2 &lt; 2$   Why does he feel a need to multiply $\epsilon$ constantly by $n$
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Calculate the real solutions $x\in\mathbb{R}$ to      $$ \sqrt{5-x} = 5-x^2 $$   My Attempt:  We know that $5-x\geq 0$ and thus $x\leq 5$ and  $$ \begin{align} 5-x^2&amp;\geq 0\\ x^2-\left(\sqrt{5}\right)^2&amp;\leq 0 \end{align} $$  which implies that $-\sqrt{5}\leq x \leq \sqrt{5}$. Now let $y=\sqrt{5-x}$. Then  $$ \tag1 y^2=5-x $$  and the equation converts into  $$ \begin{align} y &amp;= 5-x^2\\ x^2 &amp;= 5-y\\ y^2-x^2 &amp;= 5-x-(5-y)\\ y^2-x^2 &amp;= y-x\\ (y-x)(y+x)-(y-x) &amp;= 0\\ (y-x)(y+x-1) &amp;= 0 \end{align} $$  So either $y=x$ or $x+y=1$.  Case 1 ($y=x$):  We can plug this into $(1)$ to get  $$ \begin{align} y^2 &amp;= 5-x\\ x^2 &amp;= 5-x\\ x^2+x-5 &amp;= 0\\ x &amp;= \frac{-1\pm \sqrt{1+20}}{2} \end{align} $$  Since $-\sqrt{5}\leq x\leq \sqrt{5}$, the only solution is  $$ x = \frac{-1+\sqrt{21}}{2} $$  Case 2 ($y=1-x$):  We can plug this into $(1)$ to get  $$ \begin{align} y^2 &amp;= 5-x\\ (1-x)^2 &amp;= 5-x\\ 1+x^2-2x &amp;= 5-x\\ x^2-x-4 &amp;= 0\\ x &amp;= \frac{1\pm\sqrt{17}}{2} \end{align} $$  Since $-\sqrt{5}\leq x\leq \sqrt{5}$, the only solution is  $$ x = \frac{1-\sqrt{17}}{2} $$  So final solution is  $$ x \in \left\{\displaystyle  \frac{1-\sqrt{17}}{2}, \frac{-1+\sqrt{21}}{2} \right\} $$  Is it possible to solve this problem using geometry? For example, could we use the properties of a half-circle and a parabola?