Proof of property of given homomorphism
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Given that $f$ to be group homomorphism from $mathbb R^*tomathbb R^*$ and I want proof of property "If $x>0$ then $f(x)>0$". Please give me hint or something so I can proceed further.
abstract-algebra group-homomorphism
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up vote
1
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Given that $f$ to be group homomorphism from $mathbb R^*tomathbb R^*$ and I want proof of property "If $x>0$ then $f(x)>0$". Please give me hint or something so I can proceed further.
abstract-algebra group-homomorphism
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given that $f$ to be group homomorphism from $mathbb R^*tomathbb R^*$ and I want proof of property "If $x>0$ then $f(x)>0$". Please give me hint or something so I can proceed further.
abstract-algebra group-homomorphism
Given that $f$ to be group homomorphism from $mathbb R^*tomathbb R^*$ and I want proof of property "If $x>0$ then $f(x)>0$". Please give me hint or something so I can proceed further.
abstract-algebra group-homomorphism
edited Aug 23 at 5:53
an4s
2,0632417
2,0632417
asked Aug 23 at 5:03
Vaibhav Kalia
204
204
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2 Answers
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If $x>0$, then
$$x=sqrtxsqrtx$$ so $$f(x)=f(sqrtx.sqrtx)=f(sqrtx)f(sqrtx)=Big(f(sqrtx)Big)^2 >0$$
Additional exercise: Prove in general that every automorphism of $BbbR^*$ maps positive numbers to positive numbers and negative numbers to negative numbers
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up vote
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Hint: For $a in Bbb R^*$, we have $f(a^2) = f(a)^2$
Ok this is homomorphism property but I didn't get how to use it.
â Vaibhav Kalia
Aug 23 at 5:14
Another hint: if $a in Bbb R^*$, which numbers can $a^2$ be?
â Omnomnomnom
Aug 23 at 5:26
If a in R* then $a^2$ will be greater than zero and since f(a^2)=(f(a))^2 this says f(a^2)>0. Correct?
â Vaibhav Kalia
Aug 23 at 5:31
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
If $x>0$, then
$$x=sqrtxsqrtx$$ so $$f(x)=f(sqrtx.sqrtx)=f(sqrtx)f(sqrtx)=Big(f(sqrtx)Big)^2 >0$$
Additional exercise: Prove in general that every automorphism of $BbbR^*$ maps positive numbers to positive numbers and negative numbers to negative numbers
add a comment |Â
up vote
2
down vote
If $x>0$, then
$$x=sqrtxsqrtx$$ so $$f(x)=f(sqrtx.sqrtx)=f(sqrtx)f(sqrtx)=Big(f(sqrtx)Big)^2 >0$$
Additional exercise: Prove in general that every automorphism of $BbbR^*$ maps positive numbers to positive numbers and negative numbers to negative numbers
add a comment |Â
up vote
2
down vote
up vote
2
down vote
If $x>0$, then
$$x=sqrtxsqrtx$$ so $$f(x)=f(sqrtx.sqrtx)=f(sqrtx)f(sqrtx)=Big(f(sqrtx)Big)^2 >0$$
Additional exercise: Prove in general that every automorphism of $BbbR^*$ maps positive numbers to positive numbers and negative numbers to negative numbers
If $x>0$, then
$$x=sqrtxsqrtx$$ so $$f(x)=f(sqrtx.sqrtx)=f(sqrtx)f(sqrtx)=Big(f(sqrtx)Big)^2 >0$$
Additional exercise: Prove in general that every automorphism of $BbbR^*$ maps positive numbers to positive numbers and negative numbers to negative numbers
answered Aug 23 at 8:49
Chinnapparaj R
1,916318
1,916318
add a comment |Â
add a comment |Â
up vote
0
down vote
Hint: For $a in Bbb R^*$, we have $f(a^2) = f(a)^2$
Ok this is homomorphism property but I didn't get how to use it.
â Vaibhav Kalia
Aug 23 at 5:14
Another hint: if $a in Bbb R^*$, which numbers can $a^2$ be?
â Omnomnomnom
Aug 23 at 5:26
If a in R* then $a^2$ will be greater than zero and since f(a^2)=(f(a))^2 this says f(a^2)>0. Correct?
â Vaibhav Kalia
Aug 23 at 5:31
add a comment |Â
up vote
0
down vote
Hint: For $a in Bbb R^*$, we have $f(a^2) = f(a)^2$
Ok this is homomorphism property but I didn't get how to use it.
â Vaibhav Kalia
Aug 23 at 5:14
Another hint: if $a in Bbb R^*$, which numbers can $a^2$ be?
â Omnomnomnom
Aug 23 at 5:26
If a in R* then $a^2$ will be greater than zero and since f(a^2)=(f(a))^2 this says f(a^2)>0. Correct?
â Vaibhav Kalia
Aug 23 at 5:31
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint: For $a in Bbb R^*$, we have $f(a^2) = f(a)^2$
Hint: For $a in Bbb R^*$, we have $f(a^2) = f(a)^2$
answered Aug 23 at 5:08
Omnomnomnom
122k784170
122k784170
Ok this is homomorphism property but I didn't get how to use it.
â Vaibhav Kalia
Aug 23 at 5:14
Another hint: if $a in Bbb R^*$, which numbers can $a^2$ be?
â Omnomnomnom
Aug 23 at 5:26
If a in R* then $a^2$ will be greater than zero and since f(a^2)=(f(a))^2 this says f(a^2)>0. Correct?
â Vaibhav Kalia
Aug 23 at 5:31
add a comment |Â
Ok this is homomorphism property but I didn't get how to use it.
â Vaibhav Kalia
Aug 23 at 5:14
Another hint: if $a in Bbb R^*$, which numbers can $a^2$ be?
â Omnomnomnom
Aug 23 at 5:26
If a in R* then $a^2$ will be greater than zero and since f(a^2)=(f(a))^2 this says f(a^2)>0. Correct?
â Vaibhav Kalia
Aug 23 at 5:31
Ok this is homomorphism property but I didn't get how to use it.
â Vaibhav Kalia
Aug 23 at 5:14
Ok this is homomorphism property but I didn't get how to use it.
â Vaibhav Kalia
Aug 23 at 5:14
Another hint: if $a in Bbb R^*$, which numbers can $a^2$ be?
â Omnomnomnom
Aug 23 at 5:26
Another hint: if $a in Bbb R^*$, which numbers can $a^2$ be?
â Omnomnomnom
Aug 23 at 5:26
If a in R* then $a^2$ will be greater than zero and since f(a^2)=(f(a))^2 this says f(a^2)>0. Correct?
â Vaibhav Kalia
Aug 23 at 5:31
If a in R* then $a^2$ will be greater than zero and since f(a^2)=(f(a))^2 this says f(a^2)>0. Correct?
â Vaibhav Kalia
Aug 23 at 5:31
add a comment |Â
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