^Travis has a very interesting post over here. It's a fun geometric problem that calls for a two-part proving - but let me just tackle the first part. It states that if given three sides of a right triangle with the expressions n

Before we even start, let's have quick brush up about the parts of the right triangle. The hypotenuse is the side directly opposite the right-angle of that triangle. Since the right-angle happens to be the largest angle in the triangle, it also means that the hypotenuse is the longest side of the triangle. With this in mind, we need to find out which among the three expressions is the largest.

If we look at it intuitively, n

But of course we can't just say, "It's intuitive!" on paper, right? So let's check the first statement.

Suppose:

n

Therefore:

n

Hence:

0 > -2

But since 0 > -2 is always true, therefore n

Same logic applies to the second statement.

Suppose:

n

Therefore:

n

By factoring the left side of the inequality, we get:

(n-1)

This statement is true because the square of any number, in this case (n-1), is always positive (read: it's greater than zero). Hence, n

So now, we finished proving that n

^{2}+ 1, n^{2}– 1 and 2n (where n > 1), what expression represents the hypotenuse of the right triangle?Before we even start, let's have quick brush up about the parts of the right triangle. The hypotenuse is the side directly opposite the right-angle of that triangle. Since the right-angle happens to be the largest angle in the triangle, it also means that the hypotenuse is the longest side of the triangle. With this in mind, we need to find out which among the three expressions is the largest.

If we look at it intuitively, n

^{2}+ 1 should be the longest side because 1 added to any square of any number (n) is always greater than 1 subtracted from the same square of any number (n). Also, the addition of 1 to the square of any number (n) would always be greater than the same number multiplied by 2.But of course we can't just say, "It's intuitive!" on paper, right? So let's check the first statement.

Suppose:

n

^{2}+ 1 > n^{2}– 1Therefore:

n

^{2 }– n^{2}> -1 – 1Hence:

0 > -2

But since 0 > -2 is always true, therefore n

^{2}+ 1 > n^{2}– 1 is always true.Same logic applies to the second statement.

Suppose:

n

^{2}+ 1 > 2nTherefore:

n

^{2}+ 1 - 2n > 0By factoring the left side of the inequality, we get:

(n-1)

^{2}> 0This statement is true because the square of any number, in this case (n-1), is always positive (read: it's greater than zero). Hence, n

^{2}+ 1 > 2n is always true, too.So now, we finished proving that n

^{2}+ 1 is bigger than n^{2}– 1 or 2n. We can then conclude that n^{2}+ 1, by definition, should the hypotenuse of the right triangle.Oo na. Ako na. Ako na si Shaira Luna! Tseeeeeeeeh >:P hihihihihihi

[ma]all these solving of inequalities and factoring of binomial equations is SO turning me on. LOLS.[/ma]

ReplyDeletenosebleed

ReplyDeletewit ata teh... n raised to the second power + 1 cannot be the hypotenuse.. keber kung mas malaki siya sa dalawa... a2 + b2 = c2 diba formula? try mo nga kung yung n2 + 1 = c2... ayoko iprove dahil hindi ako alam kung pano itype yung mga mathematical symbols na yan sa comment (excuses excuses) heheh

ReplyDeletePotahkelya, NKKLK!!!

ReplyDeletebff, tara na i-treat kita sa resto ni waiter!

ReplyDeleteang Ganda ko siguro. wala akong nagets eh. diba patas ang Diyos?

ReplyDeleteThis is so...

ReplyDeletehighschool... ;))tissue please! hahaha

ReplyDelete^Travis: look for my infinite limits post.

ReplyDeletebaka labasan ka! :))

Coño: eto, tissue o ;))

Ewan:

(n^2-1)^2 + (2n)^2 = (n^2+1)^2

n^4 - 2n^2 + 1 + 4n^2 = n^4 + 2n^2 + 1

n^4 + 2n^2 + 1 = n^4 + 2n^2 + 1

crap. pinatulan ko talaga :))

Caridad: PAK! ahahahaha

Spiral: ikaw na tseeeeeeeeh :-L

Xian: hihihihi

ok. *faints* lol

ReplyDeleteLee: ay ambot! :p

ReplyDeleteCity: PAK!@ganda :))

ReplyDeleteThe converse is applicable to the Pythagorean Triples. Being a converse, it states a partial truth and, unlike the law, is not always right.

ReplyDelete-yan din sinabi ko sa blog ni ^travis. ok gow. malelate na talaga ako sa first class.

Green: actually, yun pythagorean triples ang una ko chineck for the converse of the statement. i'm not sure though if it holds true if and only it's the pythaogrean triples.

ReplyDeletenakakatamad kasi maghanap ng ibang instances hahahaha

[co="green"]i did! and the output was copious as the limit approached orgasm. lols. kayo na! kayo na ni pareng the green breaker! usapang PT/triad algorithms na ito.hahaha.[/co]

ReplyDeletemy gad, may answer na po ang question, wag nang hanapan pa ng ibang instances at baka madisprove at baka andami daming bunggang solution at papel pa ang masayang! mga...math geeks!

ReplyDelete^Travis: *sampal* wag ka ngang hysterical.

ReplyDeletelolz

Spiral: tseeeeeeeeh :-L

oh! aside from math, I like violence, too.

ReplyDeletegood evening ladies and gentlemen! uhm, can you repeat the question pls? - chin

ReplyDeleteTravis: spank me. aw! ;))

ReplyDeleteChin: :))

p.s. haymishuuuuuuuuuuuuu

You lied! Nasa #7 pa ako. haha Top 10, i am so there. lol

ReplyDeleteKaso, nagpalit nanaman ako ng pangalan so back to 0 ako? shezzzz..

Huy Ternie, asan ka na? It's been about a month since you published this. ;p