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The binary operation is defined on the set of real numbers, R, by x y = x + y + 10.

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2 parts
a
-10
b
-23
-15

Step-by-step solution

5 steps

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The binary operation Δ\Delta is defined on the set of real numbers, RR, by xΔy=x+y+10x \Delta y = x + y + 10.

Part a· 2 steps

Find the identity element

  1. Define the identity element.

    An element ee is the identity element if for any x∈Rx \in R, xΔe=xx \Delta e = x and eΔx=xe \Delta x = x.

  2. Use the definition of the operation to find ee.

    Using xΔe=xx \Delta e = x:

    x+e+10=xx + e + 10 = x

    Subtract xx from both sides:

    e+10=0e + 10 = 0

    Solve for ee:

    e=−10e = -10

    The identity element is −10\boxed{-10}.

Part b-23

Find the inverses of 3 and -5 under Δ\Delta

  1. Define the inverse element.

    The inverse of an element xx, denoted x−1x^{-1}, is such that xΔx−1=ex \Delta x^{-1} = e and x−1Δx=ex^{-1} \Delta x = e. We found the identity element e=−10e = -10.

  2. Find the inverse of 3.

    Let 3−13^{-1} be the inverse of 3. Using 3Δ3−1=e3 \Delta 3^{-1} = e:

    3+3−1+10=−103 + 3^{-1} + 10 = -10 3−1+13=−103^{-1} + 13 = -10

    Subtract 13 from both sides:

    3−1=−10−133^{-1} = -10 - 13 3−1=−233^{-1} = -23

    The inverse of 3 is −23\boxed{-23}.

  3. Find the inverse of -5.

    Let (−5)−1(-5)^{-1} be the inverse of -5. Using (−5)Δ(−5)−1=e(-5) \Delta (-5)^{-1} = e:

    −5+(−5)−1+10=−10-5 + (-5)^{-1} + 10 = -10 (−5)−1+5=−10(-5)^{-1} + 5 = -10

    Subtract 5 from both sides:

    (−5)−1=−10−5(-5)^{-1} = -10 - 5 (−5)−1=−15(-5)^{-1} = -15

    The inverse of -5 is −15\boxed{-15}.

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