Solved: the test cross ab/ab x ab/ab is performed

     
I am currently studying Halmos" "Linear Algebra Problem Book" & am stuchồng on problem 21(4).

Let $V$ be the phối $acsantangelo1907.combbR_+$, & let $F$ be the phối $acsantangelo1907.combbR$.

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Let"s define the sum of two positive numbers: $$a+b = ab,$$and the product of a positive number $a$ and a real number $b$: $$a*b = b^a.$$Prove sầu that $V$ is a vector-space.

Proving that addition is commutative sầu, associative, as well as the existence of an additive sầu identity (which is equal khổng lồ $1$) and an inverse (which is equal lớn $1/x$) wasn"t problematic.

However, multiplication & distributivity caused a few problems. Is the multiplicative identity also $1$, since $x^1 = x$?

Is $(ab) x = x^b^a$, in which case isn"t $a (bx) = x^b^a = x^ba$?


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edited Jul 15 "15 at 11:40
M Grande
asked Jul 15 "15 at 0:42
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M GrandeM Grande
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Yes, the scalar $1 in Bbb R$ is the multiplicative identity for the vector space, because $x^1=x$, as you wrote.

Actually, $(ab)x eq x^a^b$ because since $a,b$ are scalars, their product is the usual multiplication on $Bbb R$, not the scalar multiplication as defined for a scalar & a vector.

Hence, $(ab)x=x^ab=x^ba=(x^b)^a=a(x^b)=a(bx)$, so scalar multiplication is associative sầu.

For distributivity, what you need is $(a+b)x=ax+bx$ & $a(x+y)=ax+yx$.

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The first one: $(a+b)x=x^a+b=x^ax^b=ax+bx$

The second one: $a(x+y)=a(xy)=(xy)^a=x^ay^a=ax+ay$.


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answered Jul 15 "15 at 0:59
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coldnumbercoldnumber
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Distributivity is just $(ab)^c=a^cb^c$. Associativity (of the action) is $(a^b)^c=a^bc$.

$(ab)x=x^ab$, by the definition of the action. $a(b(x))$ is $(x^b)^a=x^ba$, which you should be careful khổng lồ distinguish from $x^b^a$, since exponentiation is not associative.


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answered Jul 15 "15 at 0:58
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Kevin ArlinKevin Arlin
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Different approach: note that$$a+_V b = exp(log(a) + log(b))\a*_V b = exp(bcdot log(a))$$


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answered Jul 15 "15 at 1:52
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Ben GrossmannBen Grossmann
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