As a student with no history of proof-based math courses, I am having some difficulty understanding the proof by contradiction by assuming $\sqrt{2} = a/b$ and that a and b are in least common denominators.   Could you explain how this disproves that it is irrational? Doesn't it just merely disprove that our initial assumption that a and b are even, is wrong?  I'm speaking of the popular proof that is given here:
121472
A popular proof of the irrationality of $\sqrt2$ is to first assume that the number is rational. This means that $\sqrt2=a/b$ where $a$ and $b$ are integers. Another assumption is that $a$ and $b$ are coprime. It turns out that this leads to a contradiction. And the conclusion is that $\sqrt2$ is not rational (because the assumption of rationality led to a contradiction). But what about the second assumption? Logically, the conclusion could also be that $a$ and $b$ are not coprime..  How can this be resolved?  Edit:  Thank you all for your comments. It cleared up my mind.  Three aspects of this kind of proof are: mathematical, logical, and didactical.  Mathematically, it is the case that any rational number can be written as a fraction with coprime numerator and denominator. But this point is often described in a way that is either literally a logical assumption, or in a way that is natural to interpret as a hidden logical assumption. In this case, the proof will have a flawed logic, because any of the two assumptions could be the cause of the contradiction. Didactically, this can cause confusion in the mind of the student, who might not be so accepting of the proofs conclusion. I used to be that student.  So, how to resolve it? Many of you provided explanations on the mathematical aspect, which are all correct. But my main point is not of not understanding the proof, but of the use of flawed logic which leads to not understanding the proof. I am sure that a lot of teachers think or say something along the lines of WLOG, but for many students, this makes the proof weaker, because it is not so convincing. When we present this proof to our students, we need to have this in mind.  Thank you.
91022
I had a little back and forth with my logic professor earlier today about proving a number is irrational. I proposed that 1 + an irrational number is always irrational, thus if I could prove that 1 + irrational number is irrational, then it stood to reason that was also proving that the number in question was irrational.  Eg. $\sqrt2 + 1$ can be expressed as a continuous fraction, and through looking at the fraction, it can be assumed $\sqrt2 + 1$ is irrational. I suggested that because of this, $\sqrt2$ is also irrational.  My professor said this is not always true, but I can't think of an example that suggests this.  If $x+1$ is irrational, is $x$ always irrational?   Actually, a better question is: if $x$ is irrational, is $x+n$ irrational, provided $n$ is a rational number?
23099
I am trying to prove this by contradiction. So if $3$ doesn't divide $a$ or $3$ doesn't divide $b$, then the remainder is either $1$ or $2$. I am struggling on what to do next. How do I get the remainder of $a^2$ and $b^2$ for these cases?  Any help is greatly appreciated. Thank you!
7195
The proof that shows the square root of 2 is irrational starts by assuming, for a contradiction, that it is rational. It starts with the assumption that it can be written as p/q where p and q have no factors in common.  Why do we start by assuming p and q have no factors in common? If we went on to find that they have factors in common, we could cancel them later on?
74872
I have proved in earlier exercises of this book that $\sqrt 2$ and $\sqrt 3$ are irrational. Then, the sum of two irrational numbers is an irrational number. Thus, $\sqrt 2 + \sqrt 3$ is irrational. My first question is, is this reasoning correct?  Secondly, the book wants me to use the fact that if $n$ is an integer that is not a perfect square, then $\sqrt n$ is irrational. This means that $\sqrt 6$ is irrational. How are we to use this fact? Can we reason as follows:  $\sqrt 6$ is irrational  $\Rightarrow \sqrt{2 \cdot 3}$ is irrational.  $\Rightarrow \sqrt 2 \cdot \sqrt 3$ is irrational  $\Rightarrow \sqrt 2$ or $\sqrt 3$ or both are irrational.  $\Rightarrow \sqrt 2 + \sqrt 3$ is irrational.  Is this way of reasoning correct?
25312
How can I prove that $\sqrt{2}$ is irrational with the ideal $$\{n \in \mathbb{Z} \mid n\times \sqrt{2} \in \mathbb Z\}?$$
77954
Let $a$ and $b$ be positive integers. Show that $\sqrt{2}$ always lies between $\dfrac{a}{b}$ and $\dfrac{a+2b}{a+b}$.  Please give the easy solution as possible.
29865
The contradiction method given in certain books to prove that sqare root of a prime is irrational also shows that sqare root of $4$ is irrational, so how is it acceptable? e.g. Suppose $\sqrt{4}$ is rational, $$\begin{align}        \sqrt{4} &amp;=p/q        \qquad\text{where pand q are coprimes} \\         4 &amp;=p^2/q^2\\       4q^2&amp;=p^2       \tag{1} \\          4&amp;\mid p^2\\          4&amp;\mid p\\      \text {let }p&amp;=4m  \qquad\text{for some natural no. m}  \\        p^2&amp;=16m^2\\       4q^2&amp;=16m^2   \qquad\text{(from (1) )}\\        q^2&amp;=4m^2\\          4&amp; \mid q^2\\          4&amp;\mid q \end{align} $$ but this contradicts our assumption that $p$ and $q$ are coprime since they have a common factor $p$. Hence  $\sqrt{4}$ is not rational. But we know that it is a rational. Why?
119230
I'm taking 'Introduction to Mathematical Thinking' on Coursera platform and following proof steps are given :      Proof of $\sqrt{2}$ is irrational.      Assume $\sqrt{2}$ is rational. $\sqrt{2}=p/q$      p and q have no common factors.   Why do p and q have no common factors? Is this a consequence of a property of the rational numbers? As p and q can be rational numbers we can set p = 6, q = 9 so p, q have common factors?
55469
I have seen the proof that  $\sqrt2$ is irrational However I'm a bit confused. The proof goes something like this. first assume $m$ and $n$ are integers such that $$ \frac{m}{n}  = \sqrt{2} $$ $$ \frac{m^{2}}{n^{2}}  = 2 $$ $$m^{2}  = 2 n^{2}$$  $\implies m^{2} $ is divisible by $2 \implies m$ is divisible by 2. Let $2q = m$. Then $$m^{2}  = 4 q^{2} = 2n^{2}$$ $$ n^{2}  = 2 q^{2}$$ This means that $n^{2}$ is divisible by $2 \implies n$ is divisible by $2$...... Hence $\sqrt{2}$ is irrational. $$$$If $ m^{2} $ is divisible by $2$ how do we know that $m$ is also divisible by $2$? If you say it's a prime number then I can argue that I didn't need to do all these steps. I could have just said that its factors don't multiply to $2$ hence $\sqrt{2}$ is irrational
21934
The most common way to prove the irrationality of a number is to do it by  contradiction. However, in this proof the contradiction arises from having two numbers, say p\q that are assumed to not have any common factors but are later proved to share the number under the radical as a factor. But for instance if the question so arises:  Is √4 rational or not? Answer with a proof.  At first glance, it is obviously rational, but assuming I do not know that √4 can be simplified to 2, what would stop me from arriving at irrationality as my conclusion?   This question seems quite trivial and probably stupid. Is this meant to be by definition or am I missing something fundamental to the aforementioned fallacy?
910
I stumbled across this relationship while I was messing around. What's the proof, and how do I understand it intuitively? It doesn't really make sense to me that the sum of odd numbers up to $2x + 1$ should equal $x^2$.
25556
Possible Duplicate:          I have this homework problem that I can't seem to be able to figure out:Prove: If $n\in\mathbb{N}$ is not the square of some other $m\in\mathbb{N}$, then $\sqrt{n}$ must be irrational. I know that a number being irrational means that it cannot be written in the form $\displaystyle\frac{a}{b}: a, b\in\mathbb{N}$ $b\neq0$ (in this case, ordinarily it'd be $a\in\mathbb{Z}$, $b\in\mathbb{Z}\setminus\{0\}$) but how would I go about proving this? Would a proof by contradiction work here?  Thanks!!
90592
how would you prove that $2\sqrt5 +\sqrt{11}$ is irrational? I started with a proof by contradiction that assumes that $2\sqrt5 +\sqrt{11}$ is rational and therefore there exist integers $a$ and $b$ such that $\frac{a}{b}=2\sqrt5 +\sqrt{11}$ and squaring both sides yields $\frac{a^2}{b^2}=31 +4\sqrt5\sqrt{11}$ and from this point im stuck as i dont know how to continue to arrive at a contradiction.
100900
Prove $a^2$ is even $\implies$ $a$ is even.  It was proven via contradiction by my friend.  Here is the proof in question, which uses proof by contradiction  Assume towards a contradiction that $a$ is odd.  Let $a = 2k+1$, then $a^2 = 4k^2 + 4k + 1 = 2(2k^2+2k)+1$ Therefore $a^2$ is odd.  Contradiction.  To me, this proof in question seems incorrect because it shows that if $a$ is odd then $a^2$ is odd. We have shown that the converse of what we want to contradict is a contradiction.  I feel like this proof by contradiction is not mathematically correct or vigorous enough, and for it to be correct we would have to start from $a^2$ being even, working our way down to $a$ being odd, and then finding a contradiction.    Am I correct? Or am I just confused about something?
104097
We're all familair with this beautiful proof whether or not an irrational number to an irrational power can be rational. It goes something like this:  Take $(\sqrt{2})^{\sqrt{2}}$  If it's rational, then you proved it, if it's irrational, take $((\sqrt{2})^{\sqrt{2}} ){^\sqrt{2}} = 2$ and you've proved it.  I'm wondering if you can raise $\pi$ or $e$ to a certain non-trivial real power to make it rational? And if not, where is the proof that it can't be done?   p.s. - I almost left out the real part, but then I realized that $e^{i\pi} = -1$.
95006
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?