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A binary operation * is defined on the set R of real numbers by:a*b=a+b+ab where a, b € R.Calculate 5*(-2)*5. Find the identity...Asked by Anonymous
A binary operation * is defined on the set R of real numbers by: a+b+ab where a, b€R .Calculate 5*(-2)*5.Find the identity element of R under the operation*. Determine the inverse under * of a general element a €R
Answers
Answered by
oobleck
I shall assume that * associates left to right, like addition. If so, then
5*(-2)*5 = (5*(-2))*5 = (5 + -2 + 5(-2))*5 = (-13)*5
= -13 + 5 + (-13)(5) = -73
What is the identity element under *? It must be zero, since a*0 = a
If b = a<sup><sup>-1</sup></sup> under * then a*b = b*a = 0
So, what do you think a<sup><sup>-1</sup></sup> is?
5*(-2)*5 = (5*(-2))*5 = (5 + -2 + 5(-2))*5 = (-13)*5
= -13 + 5 + (-13)(5) = -73
What is the identity element under *? It must be zero, since a*0 = a
If b = a<sup><sup>-1</sup></sup> under * then a*b = b*a = 0
So, what do you think a<sup><sup>-1</sup></sup> is?
Answered by
oobleck
oops. 5*-2 = -7
fix the 2nd step.
fix the 2nd step.
Answered by
Akingbade
Respond
Answered by
John De Beloved
This is another way to solve it:
2.b = a + b + ab where a, b ER
5.(-2).5
solution
5+(-2)=5+(-2)+5X(-2)
= 5-2-10
= 3-10 =-7
(-7).5=- 7+5+(-7) 5
= 7+5 35
=-2-35
=-37
(ii) a. era
ate+aera
etae = a-a
e(1+a)=0
e = 0
Let a' be the inverse of a, then
ata' + aa' = 0
a' taa' =-a
a'(1+a)=-a
2.b = a + b + ab where a, b ER
5.(-2).5
solution
5+(-2)=5+(-2)+5X(-2)
= 5-2-10
= 3-10 =-7
(-7).5=- 7+5+(-7) 5
= 7+5 35
=-2-35
=-37
(ii) a. era
ate+aera
etae = a-a
e(1+a)=0
e = 0
Let a' be the inverse of a, then
ata' + aa' = 0
a' taa' =-a
a'(1+a)=-a
Answered by
Taiwo
A binary operation * is defined on the set R of real number by a * b = a + b + ab ( where a, b belong to R ). Calculate 5 * ( - 2 ) *, and find the identity element e of R under the operation. Then determine the inverse under * of a general element a belong to R and state which element has no inverse.
Answered by
Moyin
Answer to the question
Answered by
Anonymous
first 5×-2+(5+-2) then...
Answered by
Marvellous
Akingbade is Correct
but is the remaining part
a^-1(1+a)=~a
since we are looking for a^-1 then we make a^-1 the subject of Formula
a^-1=-a/(1+a)
but is the remaining part
a^-1(1+a)=~a
since we are looking for a^-1 then we make a^-1 the subject of Formula
a^-1=-a/(1+a)
Answered by
Ayomide
Ogunremi
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