If a² = b² + c² then how to cal calculate a = b + C?

Indranil

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Feb 22, 2018
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If a² = b² + c² then how to cal calculate a = b + C?

If we all know a = √b² + c² = a + b but I want to clarify the methods of getting a = b + c.
I have done below:
a² = b² + c², a = √b² + c² = (b²)¹⁄² + (c²)¹⁄² = b²⁄¹⋅¹⁄² + c²⁄¹⋅¹⁄² = b + c Am I correct? Pleace check my steps
 

lev888

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Jan 16, 2018
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Please post complete text of the problem.
If a² = b² + c² then it does not follow that a = b + c. E.g. straight triangle sides satisfy a² = b² + c², but definitely not a = b + c.
 

Denis

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Feb 17, 2004
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I have done below:
a² = b² + c², a = √b² + c² = (b²)¹⁄² + (c²)¹⁄² = b²⁄¹⋅¹⁄² + c²⁄¹⋅¹⁄² = b + c
Am I correct?
a² = b² + c², a = √b² + c²

HOW d'heck did you get that?

Simple example: a=5, b=4, c=3

a^2 = b^2 + c^2
25 = 16 + 9
25 = 25

a = √b² + c²
5 = b + 9
5 = 4 + 9
5 = 13

Suggestion: start checking your work yourself using examples....

Anyhooooo: if a^2 = b^2 + c^2, then a = sqrt(b^2 + c^2).
Have you not been exposed to right triangles yet?
 
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Dr.Peterson

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If we all know a = √b² + c² = b + c[?] but I want to clarify the methods of getting a = b + c.
I have done below:
a² = b² + c², a = √(b² + c²) = (b²)¹⁄² + (c²)¹⁄² = b²⁄¹⋅¹⁄² + c²⁄¹⋅¹⁄² = b + c Am I correct? Pleace check my steps
This is the false idea I was correcting in the other thread. It is not true that \(\displaystyle \sqrt{a + b} = \sqrt{a} + \sqrt{b}\). Many beginners imagine it to be true, but it is false! They will never get anywhere in algebra until they learn that what feels right is not always right. You must only do what you have been taught is valid.

It is true that \(\displaystyle \sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}\); the root of a product is the product of the roots. But this is not true for addition.

Your claimed proof is based on a more general assumption that power of a sum is the sum of the powers. It is not true that \(\displaystyle (a + b)^n = a^n + b^n\), which you are assuming when you say "√(b² + c²) = (b²)¹⁄² + (c²)¹⁄²", since you are implying that "(b² + c²)1/2 = (b²)1/2 + (c²)1/2".
 

Jomo

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Dec 30, 2014
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If we all know a = √b² + c² = a + b but I want to clarify the methods of getting a = b + c.
I have done below:
a² = b² + c², a = √b² + c² = (b²)¹⁄² + (c²)¹⁄² = b²⁄¹⋅¹⁄² + c²⁄¹⋅¹⁄² = b + c Am I correct? Pleace check my steps
What puzzles me when students say this is that why would we not simply state the Pythagorean as a + b = c?
 

Indranil

Junior Member
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Feb 22, 2018
Messages
215
This is the false idea I was correcting in the other thread. It is not true that \(\displaystyle \sqrt{a + b} = \sqrt{a} + \sqrt{b}\). Many beginners imagine it to be true, but it is false! They will never get anywhere in algebra until they learn that what feels right is not always right. You must only do what you have been taught is valid.

It is true that \(\displaystyle \sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}\); the root of a product is the product of the roots. But this is not true for addition.

Your claimed proof is based on a more general assumption that power of a sum is the sum of the powers. It is not true that \(\displaystyle (a + b)^n = a^n + b^n\), which you are assuming when you say "√(b² + c²) = (b²)¹⁄² + (c²)¹⁄²", since you are implying that "(b² + c²)1/2 = (b²)1/2 + (c²)1/2".
Do you mean that
1. a = √(b² + c²) or a = √(b² + c²) = (b² + c²)¹⁄² but not a = √b² + √c² or a = (b²)¹⁄² +( c²)¹⁄²
2. If a² = b² + c², then a ≠ b + c (not possible)
 

Dr.Peterson

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Nov 12, 2017
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3,097
Do you mean that
1. a = √(b² + c²) or a = √(b² + c²) = (b² + c²)¹⁄² but not a = √b² + √c² or a = (b²)¹⁄² +( c²)¹⁄²
2. If a² = b² + c², then a ≠ b + c (not possible)
Correct.

If a = √(b² + c²), then you can't conclude that a = b + c.

In fact, the only way these can both be true is if b or c is zero.

As others have pointed out, you can easily determine this by trying examples. You don't need to guess at it.
 

JeffM

Elite Member
Joined
Sep 14, 2012
Messages
3,244
Do you mean that
1. a = √(b² + c²) or a = √(b² + c²) = (b² + c²)¹⁄² but not a = √b² + √c² or a = (b²)¹⁄² +( c²)¹⁄²
2. If a² = b² + c², then a ≠ b + c (not possible)
\(\displaystyle (a + b)^2 = (a + b)(a + b) = a(a + b) + b(a + b) \implies\)

\(\displaystyle (a + b)^2 = a^2 + ab + ba + b^2 = a^2 + ab + ab + b^2 = a^2 + 2ab + b^2.\)

Any questions so far?

\(\displaystyle a \ne 0 \text { and } b \ne 0 \implies ab \ne 0 \implies 2ab \ne 0 \implies\)

\(\displaystyle a^2 + 2ab + b^2 \ne a^2 + b^2 \implies (a + b)^2 \ne a^2 + b^2.\)

Still following?

\(\displaystyle \therefore a \ne 0 \text { and } b \ne 0 \implies \sqrt{(a + b)^2} \ne \sqrt{a^2 + b^2}\)

\(\displaystyle \text {THUS, } a + b \ne \sqrt{a^2 + b^2} \text { if } a \ne 0 \text { and } b \ne 0.\)

EDIT: To expand a bit on what Dr. Peterson wrote.
You can prove, for any pair of real numbers, the truth of:

\(\displaystyle a = 0 \text { or } b = 0 \implies \sqrt{a^2 + b^2} = |a| + |b|.\)

It is false however that, for all real numbers a and b,

\(\displaystyle a = 0 \text { or } b = 0 \implies \sqrt{a^2 + b^2} = a + b.\)
 
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