Linked Questions

0
votes
2answers
915 views

Prove that there do not exist nonzero integers $a$ and $b$ such that $a^2=3b^2$. [duplicate]

Well, by intuition, of course there is doesn't exist any nonzero integers, but how would you prove that? I was thinking of doing the GCD of $a$ and $b$ is $1$, but that leads me to nowhere.
0
votes
1answer
658 views

Can a^2 = 2b^2 have a solution where a, b are in Z but not zero? [duplicate]

Possible Duplicate: How can you prove that the square root of two is irrational? Can $a^2 = 2b^2$ have a solution where $a, b$ are in $\mathbb{Z}$ but not zero? $\mathbb{Z}$ = positive and ...
0
votes
1answer
380 views

How to prove $\sqrt3$ is irrational? [duplicate]

Possible Duplicate: How can you prove that the square root of two is irrational? How to prove $\sqrt3$ is irrational using Fermat's infinite descent method? Like says in Carl Benjamim Boyer's ...
1
vote
1answer
97 views

Positive rationals satisfying: $a^2+a^2=c^2$? [duplicate]

If there are none why not? Thanks in advance.
-1
votes
2answers
188 views

Direct Irrationality Proof for $\sqrt{3}$ and $\sqrt{6}$ [duplicate]

I am having trouble with proving this directly. I am currently learning about greatest common divisors and know that this has a role in the proof. However, I can only prove the two through ...
-4
votes
1answer
210 views

Show that there is no rational number whose square is $2$ or $8$ [duplicate]

Show that there is no rational number whose square is $2$ or $8$
0
votes
1answer
75 views

Proof using method of contradiction. Use the method of contradiction to prove that √2 is irrational. [duplicate]

Use the method of contradiction to prove that √2 is irrational. I don't understand how to prove that √2 is irrational using this method. And I feel difficult to form the contradiction.
-1
votes
3answers
72 views

Proof of why $\sqrt{2}$ is an irrational number [duplicate]

I am studying the proof by contradiction below. But I am confused on why the proof is valid. It first assumes that $p, q$ have no common factor, and then arrives at a conclusion where $p, q$ are both ...
0
votes
0answers
40 views

How to show that $(a/b)^2$ in this situation cannot be an integer [duplicate]

Let $a$,$b$ be relatively prime integers which are both greater than or equal to $􏰀2$ . Show that $(a/b)^2$ cannot be an integer. I have tried to use contradiction to prove this result, but I am not ...
1
vote
0answers
22 views

Is there a general template for Proving the Irrationality of some Radical by Contradiction [duplicate]

I’ve consistently been made to solve a rather absurd number of problems that involve proving the irrationality of some radical (Nth Root of X) by Contradiction, and since these problems often use a ...
131
votes
34answers
20k views

Examples of mathematical results discovered “late”

What are examples of mathematical results that were discovered surprisingly late in history? Maybe the result is a straightforward corollary of an established theorem, or maybe it's just so simple ...
101
votes
18answers
8k views

What is the most unusual proof you know that $\sqrt{2}$ is irrational?

What is the most unusual proof you know that $\sqrt{2}$ is irrational? Here is my favorite: Theorem: $\sqrt{2}$ is irrational. Proof: $3^2-2\cdot 2^2 = 1$. (That's it) That is a ...
156
votes
12answers
38k views

How to prove: if $a,b \in \mathbb N$, then $a^{1/b}$ is an integer or an irrational number?

It is well known that $\sqrt{2}$ is irrational, and by modifying the proof (replacing 'even' with 'divisible by $3$'), one can prove that $\sqrt{3}$ is irrational, as well. On the other hand, clearly ...
28
votes
8answers
42k views

If $a^2$ divides $b^2$, then $a$ divides $b$ [duplicate]

Let $a$ and $b$ be positive integers. Prove that: If $a^2$ divides $b^2$, then $a$ divides $b$. Context: the lecturer wrote this up in my notes without proving it, but I can't seem to figure out why ...
11
votes
4answers
4k views

Proof by contradiction: $r - \frac{1}{r} =5\Longrightarrow r$ is irrational?

Prove that any positive real number $r$ satisfying: $r - \frac{1}{r} = 5$ must be irrational. Using the contradiction that the equation must be rational, we set $r= a/b$, where a,b are positive ...

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