Triangle Inequality for Real Numbers Proof

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Triangle Inequality for Real Numbers Proof
  1. The Math Sorcerer
  2. Real Number
  3. Triangle Inequality
  4. Math
  5. Mathematics (Field Of Study)
  6. mathematical proof
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Комментарии
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This series is incredibly helpful and well-explained.

Subscribed.

thetedmang
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Thank you! You explained better than my textbook did!! I am new to your channel and saw this vid and subscribed! Thanks, Sorcerer!!!

kazmafaheem
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Quick and straight forward, nice. Subscribed!

observever
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Super helpful thank you!!! Quick, clear, and easy to follow!

YeetaIta
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Very nicely done. Thank you so much!

sugongshow
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thank you so much! I was so discouraged when my teacher just told me to memorize this property. Thanks for helping me out!!!

elonmusk
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Wow, this is really ingenious! Thank you for making this video!

harry
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short, straightforward, nice done indeed

pfxvrhb
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You just expanded it and then recondensed it? What was proven

tylerscetta
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Interesting video> Can you prove it with the axioms O1-O5? That's what I have the most trouble with. Also can you prove the triangle inequality geometrically? I'm in first year math, so I have a ton to learn!!!

Thanks for the video.

csabour
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I believe you still need to look at three cases: a and b have different signs; a and b are positive; a and b are negative.

robertmadeo
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Grazie mille per questo video, short e anche facile da capire

carolinarojano
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Great video!
What happens with the +/- when you take the square root at the end?

antonioprgomet
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Thank you .
I was seriously about to drop my Advanced Calculus class .

liranekm
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Thhhank you very much finally I am understand will 😩💜

JustMe-mqyp
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Hi Math Sorcerer:
Could u explain the following?

Prove
Lim x tends to infinity
4x2-3x+2/8x2-6x+1
=1/2
Tried to understand it with the help of someone else, it’s difficult to clearly understand.

Thank you

qjrvmob
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at 02:36
if you take the square root of an inequality shouldn't you take into consideration also the possibility of
abs(a) + abs(b) <= abs(a+b)?
because there are always two options when you take a square root of an inequality.
suppose you have x²>=9
x could be greater than 3 but it could also be smaller than minus 3.

omnipotentfish
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U r right but u should add one more step at the end
X^2>y^2 THEN WE CANT say directly that x>y it is mathematically incorrect
Correct statement is |x|>y it means x<-y or x >y
So, here |a|+|b|<-|a+b| or |a|+|b|>|a+b|
But |a|+|b| is always positive so first one is wrong so, |a|+|b|>|a+b|

untitledgenius
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when proving 2ab <= 2|a||b| in 1:42 - 1:59, I'm not seeing what inequality property you exactly applied to be able to "replace |a| with a and |b| with b". Personally how I did it was I applied your hint of |x| >= x for any real number x, then (|ab| = |a|*|b|) >= ab, so 2|a||b| >= 2ab. Then adding a^2 + b^2 = |a|^2 + |b|^2 to both sides of the inequality gives the desired intermediate step result!

bobmarley
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can you tell me how to derive this inequality from the orignal inequality in the video |a-b|>= |a|-|b|

mowafkmha