Help with simple proof

[samboxell]

Hi,

Appreciate some help with a proof that I think should be quite simple but can't seem to get it out.

Let t0=0 ,t1>t0 and t2>t1. a>0 and 0<d<1

If (1+t2 . a).d^t2 > (1+t1.a).d^t1, show that (1+t1.a).d^t1 > (1+t0.a).d^t0 =1


Thanks

 

[Steven G]

Hi,

Appreciate some help with a proof that I think should be quite simple but can't seem to get it out.

Let t0=0 ,t1>t0 and t2>t1. a>0 and 0<d<1

If (1+t2 . a).d^t2 >(1+t1.a).d^t1, show that (1+t1.a).d^t1 > (1+t0.a).d^t0 =1
Thanks

Hi, The policy of the forum is to help students with their problems not just do it for them. What have you tried? Can you show us your work?
Can you at least show us that (1+t0.a).d^t0 = 1? It really is easy.

 

[samboxell]

Hi, The policy of the forum is to help students with their problems not just do it for them. What have you tried? Can you show us your work?
Can you at least show us that (1+t0.a).d^t0 = 1? It really is easy.

Yes I do know that that's 1 since t0=0 and so its (1+t0.a).d^t0 = (1+0).1 . That I have no problem. But I just can't show the proof using the condition given.
I have been rearranging the condition given, but still I don't see how it implies the proof.

 

[Steven G]

Hi,

Appreciate some help with a proof that I think should be quite simple but can't seem to get it out.

Let t0=0 ,t1>t0 and t2>t1. a>0 and 0<d<1

If (1+t2 . a).d^t2 >(1+t1.a).d^t1, show that (1+t1.a).d^t1 > (1+t0.a).d^t0 =1
Thanks

If you raise a number more than 1 to ANY positive real number you will get more than 1.

 

[samboxell]

If you raise a number more than 1 to ANY positive real number you will get more than 1.

Maybe the expression wasn't clear, but its (1+t1.a). [d^t1 ] . Clearly (1+t1.a) >1 but d<1 so [d^t1 ] <1 and it is not clear that (1+t1.a). [d^t1 ]
will be greater than one.

Supposedly the first expression should help, but I can't see how.

 

[Steven G]

Maybe the expression wasn't clear, but its (1+t1.a). [d^t1 ] . Clearly (1+t1.a) >1 but d<1 so [d^t1 ] <1 and it is not clear that (1+t1.a). [d^t1 ]
will be greater than one.

Supposedly the first expression should help, but I can't see how.

I am sorry but I stick by my statement that If you raise a number more than 1 to ANY positive real number you will get more than 1.

You are confused about what you get by raising a number that is more than 1 to a positive power less than 1.

Suppose for example d=1/2. Let us consider 1.2^.5. Either 1.2^.5<1, 1.2^.5=1 or 1.2^.5 >1.

If 1.2^.5 =k then k*K=1.2. Clearly k >1 as if you multiply a number less than 1 by itself you will NOT get more than 1 and if you multiply 1 by itself you do not get more than 1.

OK?

 

[samboxell]

I am sorry but I stick by my statement that If you raise a number more than 1 to ANY positive real number you will get more than 1.

You are confused about what you get by raising a number that is more than 1 to a positive power less than 1.

Suppose for example d=1/2. Let us consider 1.2^.5. Either 1.2^.5<1, 1.2^.5=1 or 1.2^.5 >1.

If 1.2^.5 =k then k*K=1.2. Clearly k >1 as if you multiply a number less than 1 by itself you will NOT get more than 1 and if you multiply 1 by itself you do not get more than 1.

OK?

I am not confused by that. And I totally agree with your statement "If you raise a number more than 1 to ANY positive real number you will get more than 1" . But that is not what my question is asking. Pls relook at it.


In my question the first expression (1+t1a) is clearly > 1 since t1>0 and a>0 but d<1. So suppose d=0.5 as you suggest, (0.5)^t1 will definitely be <1
And the multiplication of a term less than 1 ([d^t1 ]) and a term more than 1 ((1+t1.a)) is not necessary more than 1. E.g 2*[(0.5)^2] =0.5

 

[samboxell]

I am sorry but I stick by my statement that If you raise a number more than 1 to ANY positive real number you will get more than 1.

You are confused about what you get by raising a number that is more than 1 to a positive power less than 1.

Suppose for example d=1/2. Let us consider 1.2^.5. Either 1.2^.5<1, 1.2^.5=1 or 1.2^.5 >1.

If 1.2^.5 =k then k*K=1.2. Clearly k >1 as if you multiply a number less than 1 by itself you will NOT get more than 1 and if you multiply 1 by itself you do not get more than 1.

OK?

No I'm not confused by that. And I completely agree with your statement that "If you raise a number more than 1 to ANY positive real number you will get more than 1."

But that's not what my question is asking. Pls relook at it. In the question I am raising d to a positive real number and d is not more than 1.

So the expression is (1+t1.a). [d^t1 ]
First term on left (1+t1.a) is clearly >1 since t1>0 and a>0
Second term on right [d^t1 ] is clearly <1. I am raising a number less than 1 to a positive real number e.g. (0.5^n) is always less than 1
So the multiplication of both terms may not necessarily be more than one, and I have to prove that it is, from inferring from the first expression.

 

[stapel]

Let t0=0 ,t1>t0 and t2>t1. a>0 and 0<d<1

Would it be correct to assume that numbers after letters are meant to be subscripts (so "t0" actually means "t₀" rather than "t times zero" or "t⁰"), and that you are using the decimal point (".") to represent multiplication ("*")? I cannot guess the meaning of the extra spaces around some of your decimal points...?

(FYI: Using standard formatting might help with communication.)

If (1+t2 . a).d^t2 >(1+t1.a).d^t1, show that (1+t1.a).d^t1 > (1+t0.a).d^t0 =1

Under the stated assumptions, I will guess that your exercise is as follows:

. . .Let t₀ = 0, t₁ > t₀, t₂ > a*t₁ > 0, and 0 < d < 1.

. . .If (1 + a*t₂)(d^t₂) > (1 + a*t₁)(d^t₁), then show that (1 + a*t₁)(d^t₁) > (1 + a*t₀)(d^t₀) = 1

You are given that t₀ = 0, so a*t₀ = a*0 = 0, and 1 + a*t₀ = 1 + 0 = 1. Since any positive value to the zero power is just 1, then d^t₀ = d⁰ = 1. So this part of the "then" is already known to be true. All that remains is to show that (1 + a*t₁)(d^t₁) > 1. I think this is the part for which you are asking assistance.

We know that t₁ > t₀ = 0, so t₁ > 0. We know that t₂ > a*t₁ > 0, so a > 0 (else we'd have had a*t₁ < 0 for a < 0 or a*t₁ = 0 for a = 0) and also that a > 1 (else we'd have had t₂ < a*t₁).

At the moment, I'm drawing a blank on where to go next, but then, I haven't had my morning caffeine. Let me return to this later.

 

[samboxell]

Would it be correct to assume that numbers after letters are meant to be subscripts (so "t0" actually means "t₀" rather than "t times zero" or "t⁰"), and that you are using the decimal point (".") to represent multiplication ("*")? I cannot guess the meaning of the extra spaces around some of your decimal points...?

(FYI: Using standard formatting might help with communication.)


Under the stated assumptions, I will guess that your exercise is as follows:

. . .Let t₀ = 0, t₁ > t₀, t₂ > a*t₁ > 0, and 0 < d < 1.

. . .If (1 + a*t₂)(d^t₂) > (1 + a*t₁)(d^t₁), then show that (1 + a*t₁)(d^t₁) > (1 + a*t₀)(d^t₀) = 1

You are given that t₀ = 0, so a*t₀ = a*0 = 0, and 1 + a*t₀ = 1 + 0 = 1. Since any positive value to the zero power is just 1, then d^t₀ = d⁰ = 1. So this part of the "then" is already known to be true. All that remains is to show that (1 + a*t₁)(d^t₁) > 1. I think this is the part for which you are asking assistance.

We know that t₁ > t₀ = 0, so t₁ > 0. We know that t₂ > a*t₁ > 0, so a > 0 (else we'd have had a*t₁ < 0 for a < 0 or a*t₁ = 0 for a = 0) and also that a > 1 (else we'd have had t₂ < a*t₁).

At the moment, I'm drawing a blank on where to go next, but then, I haven't had my morning caffeine. Let me return to this later.

Yes sorry for the formatting, but you understood it completely right. I'll just point out that a>0 is given.