Exponential size
If ( 2 1 ) x > ( 2 1 ) y , which of the following must be true about x and y ?
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6 solutions
Thanks very helpful
=> 2 x 1 > 2 y 1 => 2 y > 2 x => y > x
Why does 2 y > 2 x imply that y > x ?
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It's logic
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It seems trivial, but can you give a rigorous proof?
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@Jesse Nieminen – Take log on both sides give,
ylog2>xlog2
log2 get cancelled
Therefore, y>x
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@Genis Dude – Why is taking logarithms justified? Why does it preserve the inequality as it is?
For example, multiplying both sides of the inequality by − 1 flips the inequality other way around.
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@Jesse Nieminen – If u multiply by negative number for example -1 , we will get log(-1) which is not defined.So we cannot cancel it
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@Genis Dude – Yes, but that does not explain why taking logarithms is justified.
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@Jesse Nieminen – We took log so that we will get the inequality btw x and y
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@Genis Dude – But why can we take logarithm of both side of the inequality and keep the inequality as it is?
Why does a > b imply that lo g 2 a > lo g 2 b ?
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@Jesse Nieminen – I did not said logx(base 2).I said xlog2<ylog2 Log2 is a constant value(0.301) So it get cancelled
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@Genis Dude – Don't refer me to a wiki. Yes, x lo g 2 < y lo g 2 ⟹ x < y is true, when the base of the logarithm is greater than 1 , but why is lo g 2 x < lo g 2 y ⟹ x lo g 2 < y lo g 2 true, when the base is greater than 1 ?
Give me a proof that taking logarithm (of base greater than one) of both sides of an inequality is allowed. Hint: Logaritmic functions (with base greater than 1 ) are strictly increasing functions.
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@Jesse Nieminen – log2^x=log(2 2 2. .. .)till x times log2+ log2+log2..... (till x times) xlog2
@Jesse Nieminen – Also taking log doesn't change the inequality
@Jesse Nieminen – https://brilliant.org/wiki/logarithmic-inequalities/
Also, I have a counterexample for you, ( − 1 ) 2 > ( − 1 ) 3 is true, but 2 > 3 is false.
Define F ( a ) = x a for 0 < x < 1 , then F ′ ( a ) = x a ln x < 0 which is obvious since 0 < x < 1 . So F is decreasing and thus if ( 2 1 ) x > ( 2 1 ) y which implies x < y .
if y > x then y = x + z and z > 0.
0 < ( 1 / 2 ) z < 1 for z > 0 .
( 1 / 2 ) y = ( 1 / 2 ) x + z = ( 1 / 2 ) x ( 1 / 2 ) z
It follows that ( 1 / 2 ) x > ( 1 / 2 ) x ( 1 / 2 ) z = ( 1 / 2 ) y
0.5 * 0.5 = 0.25. 0.25 < 0.5. Thus, adding powers to 0.5 makes it smaller.
Guys it is simple 2^(-x)>2^(-y) -x>-y x<y(from inequalities) Hope it helps
Why does 2 − x > 2 − y imply that − x > − y ?
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It's logic
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It seems trivial, but can you give a rigorous proof?
between 0-1, if we give power in any number it will decrease.
so, easily ,0.5 * 0.5 = 0.25.
0.25 < 0.5.
Thus, adding powers to 0.5 makes it smaller.