Proof: Triangle Inequality Theorem | Real Analysis

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The absolute value of a sum is less than or equal to the sum of the absolute values for any two real numbers. That is: |a+b| is less than or equal to |a|+|b|. This is called the triangle inequality. It's very useful in real analysis and we'll prove it in today's lesson! The name of the theorem is explained a little below.

If we considered a triangle made of vectors x, y, and z, with x and z starting at the origin, then the vector z is the vector x+y, and thus the theorem says this third side (z) of the triangle must be no greater than the sum of the other two sides: |z| is less than or equal to |x| + |y| and since the vector z is the vector x+y we have |x+y| is less than or equal to |x| + |y|. Speaking strictly in terms of geometry, we'd often use a strictly less than inequality for this theorem because the equality case describes a "degenerate" triangle, where the "triangle" is just a line. However, the "less than or equal to" inequality is perfectly valid and reasonable for real numbers.

I hope you find this video helpful, and be sure to ask any questions down in the comments!

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The outro music is by a favorite musician of mine named Vallow, who, upon my request, kindly gave me permission to use his music in my outros. I usually put my own music in the outros, but I love Vallow's music, and wanted to share it with those of you watching. Please check out all of his wonderful work.

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+WRATH OF MATH+

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Support the production of this course by joining Wrath of Math to access exclusive and early videos, original music, plus the real analysis lecture notes at the premium tier!

WrathofMath
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why is it called triangle inequality thou

ryantrynda
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This is how a person who truly understands logic, can beautifully and precisely communicate it . Thank you.

thedeathofbirth
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We really love you here in south Africa 🇿🇦😭💜💜💜

Alan-svym
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Beautiful and concise!! thank you for helping me understand this proof better!

red
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Thanks for explaining this clearly. If you are able to, please do some work from Abbott's "Understanding Analysis." It would be awesome.

valeriereid
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This guy really did a 2 hour lecture in 5 minutes

shhaque
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Can you do an entire video on absolute value identities as building blocks for other theorems? I would also illustrate these both on a 1D number line and on a 2D plane as vectors.

punditgi
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if 2:44 wasn't directly obvious to you as it was for me you can use this in the proof:

a1 <= b1 AND a2 <= b2
a1 + a2 <= b1 + a2 AND a2 +b1 <= b1 +b2
a1 + a2 <= b1 + b2

benjtheo
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An excellent AND simple way to grasp a crucial theorem for the understanding of analysis. Txs.

androidandroid
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Wrath, once i get this mfing masters degree ima invite you to a carne asada thx G

existentialrap
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Very nice and neat. Thank you. Got yourself a new sub!!!

TahaCodes
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Very good ' but plz prove it again when â and b are real numbers.

anasshuaibu
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Can you please do the same proof with a, b and c

Divine_wsk
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It's very good sir. I am from Assam, Indian.

BhaskarD
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Sir, could I know why this is called as "triangular inequality" ?
Is this related to triangle when viewed geometrically ?

jeswintitusjv
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great! the only thing that bothered me a little bit is that we didnt take into account when a number is negative and the other is positive. but that case is trivial so i guess doesnt matter that much ahahaha

augustus
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I Asked question but you didn't answer me

muhammadasghar
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My man, you're awesome! Much love from Toronto, Canada.

johnsally
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-(a+b) does not equal (-a) + (-b) by the way

daniellin