[maths27]

I need to show that the question number 3 is a subgroup.

I have to go all the axioms of the subgroup.

Can you please help?

[maths27]

For the closure under addition

if x,y are in H where tanx, tay are in Q then I have to show that

x+y=xy ?

[Hathlan]

(Original post by maths27)
For the closure under addition

if x,y are in H where tanx, tay are in Q then I have to show that

x+y=xy ?

Dude this makes no sense. To show closure under addition, you want to show that x+y is in H.

The fastest way to show something is a subgroup is often the two step test:

check H is non-empty. (if H is empty it ain't a subgroup)
check that if x and y are in H, then x-y is in H

you should be able to show that these two conditions are equivalent to H being a subgroup (i.e. closure under addition and inverses). It's not much different but a tad neater/faster.

[maths27]

x+y in H but tanx and tany but be also in Q, How are we going to show that/

[DFranklin]

(Original post by maths27)
x+y in H but tanx and tany but be also in Q, How are we going to show that/

No offense, but it would help if you could write this in intelligible English.

[maths27]

For closure:

x,y ∈H, then tanx ∈Q and tany ∈Q.

Now we have to check that x+y ∈H, which means we have check whether tan(x+y)∈Q

from the formula:

tan(x+y)= (tanx+tany)/(1-tanx*tany)

so (rational + rational)/ (1-rational*rational)= 2rational / (1-rational^2) which gives rational ?

[DFranklin]

(Original post by maths27)
For closure:

x,y ∈H, then tanx ∈Q and tany ∈Q.

Now we have to check that x+y ∈H, which means we have check whether tan(x+y)∈Q

from the formula:

tan(x+y)= (tanx+tany)/(1-tanx*tany)

so (rational + rational)/ (1-rational*rational)= 2rational / (1-rational^2) which gives rational ?

Usually, but not always.

Hint: division by 0...

[maths27]

if we say that 1-rational^2 must not be equal to 0.

then we find that rational is not equal to 1.

just this?

[DFranklin]

Two things: The denominator is (1 - AB) where A = tan x, B = tan y, so you should only think of (1-rational^2) as shorthand; it might be that A = 1/2 and B = 2, say.

In any event, if you choose x, y s.t. AB = 1, then tan(x+y) isnt in Q. It's not hard to see that this can happen.

[maths27]

yes. when tanx*tany then=1 the tan(x+y) is not in Q. So the closure does not hold?

[DFranklin]

(Original post by maths27)
yes. when tanx*tany then=1 the tan(x+y) is not in Q. So the closure does not hold?

Correct. In particular, is in H, but what happens when we add ?

[maths27]

tan(pi/2) goes to infinity. So it is not a subgroup of real numbers with addition?

[maths27]

Our lecturer told as that it is a subgroup of this group.......

[nuodai]

(Original post by maths27)
Our lecturer told as that it is a subgroup of this group.......

Your lecturer's wrong. Certainly since . If were a subgroup of then we'd have by closure under the group operation; but this is and we certainly don't have !

[DFranklin]

(Original post by nuodai)
Your lecturer's wrong. Certainly since . If were a subgroup of then we'd have by closure under the group operation; but this is and we certainly don't have !

I have to say, the "hint" leads you towards a "yes it does form a subgroup" answer, so I'm not totally surprised to see the lecturer saying this.

But as you say, he's wrong.