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We all know what a proof is, right? We proved that humans can't fly. We proved that slavery is immoral. We use reasoning to show that a statement is either true or false (in rare cases, neither, but I'll get to that later). But proving in mathematics can get pretty complex. Some things, like the irrationality of \(\sqrt{2}\), can be proven in a few simple steps. Some, like Andrew Wiles' proof of Fermat's Last Theorem, get extreme. Furthermore, there are different kinds of proofs, not just in mathematics, but anywhere. Here, I will enumerate three types of proofs that are most popular, and why they are used. So, let us begin, shall we?

 

1. Algebraic proof

 

An algebraic proof is a simple, straight-forward, brute-force approach. You use algebra to prove that a fact is true. Here is an example.

 

Proof that negative*negative=positive

 

\(-a(b-b)=0\) because \(b-b=0\).

\(-ab+(-a)(-b)=0\)

\((-a)(-b)=ab\). Q.E.D.

 

 

​​​​This type of proof is very useful because it uses elementary algebra, so it can be understood by most of the community (who are mathematically inclined). But usually this type of proof fails to get you anywhere in problems like, oh, I don't know, the Fold-and-Cut theorem (xD). But in all seriousness, elementary algebra sometimes just isn't enough. Anyway, on to the next one.

2. Reductio Ad Absurdum or Proof by Contradiction

 

This should be easy to understand. Proof by contradiction assumes the statement is false, then proves that it cannot be false, therefore the statement is true. An example of this is the famous proof that \(\sqrt{2}\) is irrational, but I won't put that here. Instead, I'll give you another example.

 

Proof that ALL roots greater than 2 of 2 are irrational

 

Assuming \(\sqrt[n]{2}\) is rational for \(n>2\), \(\sqrt[n]{2}={a\over b}\) such that \(a\) and \(b\) are integers and \({a\over b}\) is fully simplified.

\(2=({a\over b})^n\)

\(2={a^n\over b^n}\)

\(2b^n=a^n\)

\(b^n+b^n=a^n\)

By Fermat's Last Theorem, this statement is impossible, therefore \(\sqrt[n]{2}\) is irrational.

Q.E.D.

 

Did you like that? Proof by contradiction is a very powerful tool. It's easy to understand, and usually you get to an answer right away. But, this proof fails too. An example is "This statement is false." Prove it.

 

3. Proof by induction

 

Some of you probably have not heard of this before, but it is exetremely useful. To prove by induction, first show that the statement is true for 0 or 1, then show that if the statement is true for \(n\), it is also true for \(n+1\). Doing this, you create a domino effect that proves the statement for all natural numbers. Here is an example:

 

Proof that \(3^n-1\) is always a multiple of 2

 

Show that it is true for 1:

\(3^1-1=2\) 

Show that if it is true for \(n\), it is true for \(n+1\):

Assume \(3^n-1\) is a multiple of 2.

\(3^{n+1}-1\)

\(=3*3^n-1\)

\(=(2*3^n+3^n)-1\)

\(=2*3^n+3^n-1\)

\(2*3^n\) is a multiple of 2 (multiplying by 2), and \(3^n-1\) is also a multiple of 2 (our assumption), therefore \(3^{n+1}-1\) is a multiple of 2, therefore the statement is true for all natural numbers.

 

Nice! Proof by induction can save us a lot of time, right? It pretty much does the work for us so we don't have to prove it infinitely many times. But of course, it has its own downsides. If you are not careful, you can end up with ridiculous claims. I'd give you an example, but this question is getting long.

 

 

 

These three are just a few ways you can prove things in mathematics, and they are all very useful indeed, if we use them correctly. Now, this wouldn't be a web2.0calc post without a question, eh? So...

 

Using any type of proof, prove the Pythagorean identity:  \(\sin^2\theta+\cos^2\theta=1\).

 

wink

​​

Mathhemathh  Oct 19, 2017
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7+0 Answers

 #1
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Thanks, Mathhemathh  !!!!!!

 

 

cool cool cool

CPhill  Oct 19, 2017
 #2
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For your "Proof by Contradiction", you should have used the famous proof of Euclid of "Infinity of Primes!" That is one of the most elegant proofs in the history of Math.

Guest Oct 19, 2017
 #3
avatar+90982 
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Why don't you demonstrate :)

Melody  Oct 19, 2017
 #4
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Good Point! I thought everybody interested in Math knew it! Here is a brief proof of it.

 

https://math.stackexchange.com/questions/1017218/euclids-proof-for-infinitely-many-prime-numbers

Guest Oct 19, 2017
 #6
avatar+90982 
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You 'assumed' all people interested in mathematics knew. 

As someone interested in mathematics you should know that assumptions usually lead to errors.

That is why in mathematics all things must be proven. :)

And proofs must be demonstrated :)

Melody  Oct 19, 2017
edited by Melody  Oct 19, 2017
edited by Melody  Oct 19, 2017
 #7
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True what you say! Here is my take on it:

Suppose we ONLY had these Primes: 2 x 3 x 5 x 7 x 11 x 13 x 17 x 19=9,699,690. Now, we add 1 to it and we have: 9,699,690 + 1=9,699,691. Now, this new number is either a prime or has Prime Factors that are NOT included in our list, because if you divide it by any of them, it will always leave a remainder of 1. It, therefore, must either be a Prime Number or has Prime Factors > 19 listed above.  And if we check it, we see that 9,699,691 = 347 * 27953, which are its Prime factors and both, in this case, are larger than our listed largest prime, or 19!. And that proves there are infinitely MANY PRIMES!!.

Guest Oct 19, 2017
edited by Guest  Oct 19, 2017
 #8
avatar+90982 
+1

Yes, I did accept your website proof but it is good that you explained it in your own words :)

Melody  Oct 19, 2017

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