💡Okay, so I have this proposition: "In a plane, if a line is perpendicular to the projection of a skew line in this plane, then it is also perpendicular to the skew line itself." And I need to find the inverse of this proposition and determine if it's true or false.First, I need to understand what the inverse of a proposition is. From what I remember, the inverse of a conditional statement "If P, then Q" is "If not P, then not Q." So, in logical terms, if the original statement is P → Q, the inverse would be ¬P → ¬Q.Alright, so applying that to the given proposition. Let's break it down:Original proposition: If a line is perpendicular to the projection of a skew line in this plane (P), then it is also perpendicular to the skew line itself (Q).So, P is "a line is perpendicular to the projection of a skew line in this plane," and Q is "it is also perpendicular to the skew line itself."Therefore, the inverse would be: If a line is not perpendicular to the projection of a skew line in this plane (¬P), then it is not perpendicular to the skew line itself (¬Q).So, the inverse statement is: "In a plane, if a line is not perpendicular to the projection of a skew line in this plane, then it is not perpendicular to the skew line itself."Now, I need to determine if this inverse statement is true or false.To do that, I should probably recall some properties of skew lines and their projections. Skew lines are lines in three-dimensional space that are neither parallel nor intersecting. When we project a skew line onto a plane, the projection is another line in that plane.If a line in the plane is perpendicular to the projection of the skew line, does that imply it's perpendicular to the skew line itself? The original proposition says yes. But we're looking at the inverse: if it's not perpendicular to the projection, then it's not perpendicular to the skew line.I think I need to visualize this. Imagine a skew line and its projection onto a plane. If a line in the plane is perpendicular to this projection, it means it forms a right angle with the shadow or the projection of the skew line. But does that necessarily mean it's perpendicular to the actual skew line in 3D space?I think in 3D space, being perpendicular to the projection doesn't automatically make it perpendicular to the original line because the skew line has a component out of the plane. So, the projection only captures part of the skew line's direction.Therefore, if a line is not perpendicular to the projection, it might still have some component that could make it perpendicular to the skew line in 3D. Wait, no, actually, if it's not perpendicular to the projection, which is a component of the skew line, then it can't be perpendicular to the entire skew line because the projection is part of the skew line's direction.Hmm, this is a bit confusing. Maybe I need to think about it more carefully.Let me consider the relationship between the line in the plane and the skew line. If they are perpendicular, then their dot product should be zero. The projection of the skew line onto the plane would have a certain direction, and the line in the plane would have to be perpendicular to that direction as well.But if the line in the plane is not perpendicular to the projection, does that mean their dot product isn't zero? And if their dot product isn't zero, then the original lines can't be perpendicular either because the projection is a component of the skew line.Wait, so if the line in the plane is not perpendicular to the projection, then it can't be perpendicular to the skew line itself because the projection is part of the skew line's direction. Therefore, the inverse statement would be true.But I'm not entirely sure. Maybe I should look for a counterexample. Suppose there's a line in the plane that's not perpendicular to the projection of a skew line, but it's still perpendicular to the skew line. Is that possible?If the line in the plane is not perpendicular to the projection, that means there's some angle between them. But since the skew line has a component out of the plane, maybe the line in the plane could still be perpendicular to the skew line in 3D space even if it's not perpendicular to the projection.Wait, that seems contradictory. If the line in the plane isn't perpendicular to the projection, which is part of the skew line's direction, how could it be perpendicular to the entire skew line?I think it can't. Because the projection is a component of the skew line, if the line in the plane isn't perpendicular to that component, it can't be perpendicular to the whole skew line.So, maybe the inverse is true after all.But I'm still a bit uncertain. Maybe I should think about the definitions more carefully.In vector terms, if we have a skew line, its direction vector can be decomposed into two components: one parallel to the plane and one perpendicular to the plane. The projection onto the plane would have the parallel component.If a line in the plane is perpendicular to the projection, it means it's perpendicular to the parallel component. But to be perpendicular to the entire skew line, it would also have to be perpendicular to the perpendicular component.Wait, so if the line in the plane is perpendicular to the projection, it's only guaranteed to be perpendicular to the parallel component. It doesn't necessarily have to be perpendicular to the perpendicular component.Therefore, the original proposition might not always hold true because being perpendicular to the projection doesn't ensure being perpendicular to the entire skew line.But we're talking about the inverse now. The inverse is saying that if it's not perpendicular to the projection, then it's not perpendicular to the skew line.So, if a line in the plane is not perpendicular to the projection, does that mean it's not perpendicular to the skew line?Well, if it's not perpendicular to the projection, which is the parallel component, then it's not perpendicular to at least one component of the skew line. Therefore, it can't be perpendicular to the entire skew line.So, in that case, the inverse would be true.I think I'm getting somewhere. So, the inverse statement is true because if a line isn't perpendicular to the projection, it's not perpendicular to the skew line.But I'm still a bit shaky on this. Maybe I should draw a diagram or use specific vectors to test this.Let's assign some coordinates. Suppose the plane is the xy-plane. Let the skew line have direction vector (a, b, c), where c ≠ 0 because it's skew, not parallel to the plane.The projection onto the xy-plane would have direction vector (a, b, 0).Now, consider a line in the plane with direction vector (d, e, 0).If this line is perpendicular to the projection, then their dot product is zero: a*d + b*e = 0.If it's not perpendicular, then a*d + b*e ≠ 0.Now, for the line to be perpendicular to the skew line, the dot product of their direction vectors must be zero: a*d + b*e + c*0 = a*d + b*e.So, if a*d + b*e ≠ 0, then the dot product isn't zero, meaning the lines aren't perpendicular.Therefore, if a line in the plane is not perpendicular to the projection, it's not perpendicular to the skew line.So, the inverse is true.Wait, that makes sense. Because the dot product condition for perpendicularity in 3D reduces to the dot product in the plane when considering the projection, since the z-component of the line in the plane is zero.Therefore, if the dot product in the plane isn't zero, the dot product in 3D isn't zero either, meaning the lines aren't perpendicular.So, the inverse is indeed true.I think I've convinced myself now. The inverse proposition is true.

💡Alright, so I've got this functional equation to solve: ( f(x + f(y)) = f(x + y) + f(y) ) for all positive real numbers ( x ) and ( y ). Hmm, functional equations can be tricky, but let me see if I can figure this out step by step.First off, I know that functional equations often require making smart substitutions or finding patterns. Maybe I can plug in some specific values for ( x ) or ( y ) to see if that simplifies things.Let me try setting ( x = 1 ) just to see what happens. Then the equation becomes:[ f(1 + f(y)) = f(1 + y) + f(y) ]Hmm, not sure if that helps much. Maybe I should try setting ( y ) to some specific value instead.What if I set ( y = 1 )? Then the equation becomes:[ f(x + f(1)) = f(x + 1) + f(1) ]That looks interesting. Let me denote ( c = f(1) ), so the equation becomes:[ f(x + c) = f(x + 1) + c ]Hmm, so shifting the argument by ( c ) is the same as shifting by 1 and then adding ( c ). Maybe this suggests some kind of linearity?Wait, if I think about linear functions, like ( f(x) = kx ), maybe that could work. Let me test that. Suppose ( f(x) = kx ). Then substituting into the original equation:[ f(x + f(y)) = f(x + y) + f(y) ]becomes:[ k(x + ky) = k(x + y) + ky ]Simplify the left side:[ kx + k^2 y ]And the right side:[ kx + ky + ky = kx + 2ky ]So setting them equal:[ kx + k^2 y = kx + 2ky ]Subtract ( kx ) from both sides:[ k^2 y = 2ky ]Divide both sides by ( y ) (since ( y > 0 )):[ k^2 = 2k ]Which simplifies to:[ k^2 - 2k = 0 ][ k(k - 2) = 0 ]So ( k = 0 ) or ( k = 2 ). But ( f ) maps to positive reals, so ( k = 0 ) isn't valid because that would make ( f(x) = 0 ), which isn't positive. So ( k = 2 ) is the only possibility. Therefore, ( f(x) = 2x ) is a solution.But wait, is that the only solution? Maybe there are other functions that aren't linear which satisfy the equation. I need to check if ( f(x) = 2x ) is the only solution or if there are others.Let me think about the properties of ( f ). The equation is ( f(x + f(y)) = f(x + y) + f(y) ). If I fix ( y ) and vary ( x ), it seems like ( f ) has some kind of additive property. Maybe I can express ( f ) in terms of another function.Let me define a new function ( g(x) = f(x) - x ). Then, ( f(x) = x + g(x) ). Let's substitute this into the original equation:[ f(x + f(y)) = f(x + y) + f(y) ]becomes:[ (x + f(y)) + g(x + f(y)) = (x + y) + g(x + y) + y + g(y) ]Wait, that seems a bit messy. Let me try substituting more carefully.Substituting ( f(y) = y + g(y) ) into the left side:[ f(x + f(y)) = f(x + y + g(y)) = (x + y + g(y)) + g(x + y + g(y)) ]And the right side:[ f(x + y) + f(y) = (x + y + g(x + y)) + (y + g(y)) ]Simplify the right side:[ x + y + g(x + y) + y + g(y) = x + 2y + g(x + y) + g(y) ]So, equating both sides:[ x + y + g(y) + g(x + y + g(y)) = x + 2y + g(x + y) + g(y) ]Subtract ( x + y + g(y) ) from both sides:[ g(x + y + g(y)) = y + g(x + y) ]Hmm, that's an interesting equation. Let me denote ( z = x + y ), so the equation becomes:[ g(z + g(y)) = y + g(z) ]This looks like a kind of functional equation for ( g ). It reminds me of Cauchy's functional equation, which is ( g(a + b) = g(a) + g(b) ), but this is a bit different.Wait, if I fix ( y ) and vary ( z ), maybe I can find something about ( g ). Let me see.Suppose I set ( z = t ), then:[ g(t + g(y)) = y + g(t) ]This suggests that ( g ) is injective because if ( g(t + g(y)) = g(t') + y ), then ( t + g(y) = t' + g(y') ) implies ( t = t' ) and ( y = y' ). Wait, not sure about that. Maybe I need to think differently.Alternatively, maybe I can consider the function ( g ) being linear. Suppose ( g(y) = ky ) for some constant ( k ). Let's test this.If ( g(y) = ky ), then substituting into the equation:[ g(z + g(y)) = y + g(z) ]becomes:[ k(z + ky) = y + kz ]Simplify:[ kz + k^2 y = y + kz ]Subtract ( kz ) from both sides:[ k^2 y = y ]Which implies ( k^2 = 1 ), so ( k = 1 ) or ( k = -1 ). But since ( f ) maps to positive reals, ( g(y) = f(y) - y ) must also be positive. So ( k = 1 ) is the only possibility.Thus, ( g(y) = y ), which means ( f(y) = y + g(y) = y + y = 2y ). So that brings us back to ( f(x) = 2x ).But wait, is this the only solution? What if ( g ) isn't linear? Maybe there are other functions ( g ) that satisfy ( g(z + g(y)) = y + g(z) ).Let me think about this equation again:[ g(z + g(y)) = y + g(z) ]This equation resembles a kind of translation invariance. If I fix ( y ), then ( g(z + g(y)) - g(z) = y ). This suggests that the difference ( g(z + g(y)) - g(z) ) is constant for all ( z ), which is only possible if ( g ) is linear.Wait, more precisely, if ( g(z + c) - g(z) = d ) for some constants ( c ) and ( d ), then ( g ) must be linear. Because the difference depends only on ( c ) and ( d ), not on ( z ). So, in our case, ( c = g(y) ) and ( d = y ). So, for each ( y ), ( g(z + g(y)) - g(z) = y ).This suggests that ( g ) is additive in some way. Let me explore this.Suppose I fix ( z ) and vary ( y ). Then, for each ( y ), ( g(z + g(y)) = y + g(z) ). Let me denote ( w = g(y) ). Then, ( g(z + w) = y + g(z) ). But ( w = g(y) ), so ( y = g^{-1}(w) ) if ( g ) is invertible.Wait, is ( g ) invertible? Since ( g ) maps ( mathbb{R}^+ ) to ( mathbb{R}^+ ) and ( g(z) = f(z) - z > 0 ) (since ( f(z) > z ) as we'll see later), ( g ) is positive. But is it strictly increasing? If ( g ) is strictly increasing, then it's invertible.Wait, actually, let's check if ( f(z) > z ). Suppose for contradiction that ( f(z) leq z ) for some ( z ). Then, setting ( x = z - f(z) ) (which is positive because ( f(z) leq z )), we have:[ f(x + f(z)) = f(x + z) + f(z) ]But ( x + f(z) = z ), so:[ f(z) = f(z) + f(z) ]Which implies ( f(z) = 0 ), contradicting ( f(z) > 0 ). Therefore, ( f(z) > z ) for all ( z ), so ( g(z) = f(z) - z > 0 ).Since ( f ) is strictly increasing (as ( f(z) > z ) and assuming continuity, though we might not have that yet), ( g ) is also strictly increasing, hence invertible.So, going back, ( g(z + w) = y + g(z) ), and ( w = g(y) ). Therefore, ( g(z + w) = g^{-1}(w) + g(z) ). Hmm, not sure if that helps.Alternatively, maybe I can set ( z = 1 ) to see if that gives me something. Then:[ g(1 + g(y)) = y + g(1) ]Let me denote ( c = g(1) ), so:[ g(1 + g(y)) = y + c ]This suggests that ( g ) is surjective because for any ( y ), ( g(1 + g(y)) ) can take any value greater than ( c ). But since ( g ) maps to positive reals, it's surjective onto some interval.Wait, but ( g ) is strictly increasing, so it's bijective from ( mathbb{R}^+ ) to its image, which is also ( mathbb{R}^+ ) because ( g(z) > 0 ) for all ( z ).So, since ( g ) is bijective, I can write ( g^{-1}(y) = 1 + g(y) - c ) from the equation ( g(1 + g(y)) = y + c ). Wait, that might not be directly helpful.Alternatively, maybe I can consider the function ( h(z) = g(z) - c ), but not sure.Wait, another approach: let's consider the equation ( g(z + g(y)) = y + g(z) ). If I set ( z = g(y) ), then:[ g(g(y) + g(y)) = y + g(g(y)) ]So:[ g(2g(y)) = y + g(g(y)) ]Hmm, not sure.Alternatively, let me try to express ( g ) in terms of itself. From ( g(z + g(y)) = y + g(z) ), if I fix ( z ) and vary ( y ), I can think of ( g ) as a function that, when shifted by ( g(y) ), increases by ( y ). This seems like a kind of linearity.Suppose I assume ( g ) is linear, ( g(y) = ky ). Then, substituting into the equation:[ k(z + ky) = y + kz ]Simplify:[ kz + k^2 y = y + kz ]Subtract ( kz ):[ k^2 y = y ]So ( k^2 = 1 ), hence ( k = 1 ) or ( k = -1 ). But since ( g(y) > 0 ), ( k = 1 ). Therefore, ( g(y) = y ), so ( f(y) = y + y = 2y ).This matches our earlier solution. But is this the only solution? Suppose ( g ) is not linear. Maybe it's affine, like ( g(y) = y + d ). Let's test that.If ( g(y) = y + d ), then substituting into the equation:[ g(z + g(y)) = g(z + y + d) = (z + y + d) + d = z + y + 2d ]And the right side:[ y + g(z) = y + z + d ]So:[ z + y + 2d = z + y + d ]Which implies ( 2d = d ), so ( d = 0 ). Therefore, ( g(y) = y ), leading back to ( f(y) = 2y ).So, affine functions other than linear don't work. What about quadratic? Suppose ( g(y) = ay^2 + by + c ). That seems complicated, but let's see.But before going into that, maybe I can use the fact that ( g ) is additive. Wait, from the equation ( g(z + g(y)) = y + g(z) ), if I set ( z = 0 ), but ( z ) must be positive, so that's not allowed. Hmm.Alternatively, maybe I can consider the function ( g ) being additive in some transformed space. Let me think about the equation again:[ g(z + g(y)) = y + g(z) ]If I denote ( w = g(y) ), then ( y = g^{-1}(w) ), so:[ g(z + w) = g^{-1}(w) + g(z) ]This looks like a kind of additive function but with ( g^{-1} ) involved. It's not straightforward.Wait, another idea: let's consider the function ( h(z) = g(z) - c ), where ( c ) is a constant. Maybe that can simplify the equation. But I'm not sure.Alternatively, let's consider the function ( g ) being linear. We saw that ( g(y) = y ) works, but are there other linear functions? No, because ( g(y) = ky ) only works for ( k = 1 ).Wait, another approach: let's consider the function ( f ) being linear, which we already did, and it works. Maybe that's the only solution.But to be thorough, let's assume that ( f ) is not linear and see if that leads to a contradiction.Suppose ( f ) is not linear. Then, ( g ) is not linear. But from the equation ( g(z + g(y)) = y + g(z) ), if ( g ) is not linear, it's hard to see how this equation can hold for all ( z ) and ( y ).Wait, another idea: let's consider the function ( g ) being injective. Suppose ( g(a) = g(b) ). Then, for any ( z ):[ g(z + g(a)) = a + g(z) ][ g(z + g(b)) = b + g(z) ]But since ( g(a) = g(b) ), the left sides are equal, so ( a + g(z) = b + g(z) ), implying ( a = b ). Therefore, ( g ) is injective.So, ( g ) is injective. That might help.Now, from the equation ( g(z + g(y)) = y + g(z) ), since ( g ) is injective, we can say that if ( z + g(y) = z' + g(y') ), then ( y + g(z) = y' + g(z') ). But I'm not sure how to use that.Wait, another approach: let's fix ( z ) and vary ( y ). From ( g(z + g(y)) = y + g(z) ), we can see that as ( y ) increases, ( g(z + g(y)) ) increases linearly with ( y ). This suggests that ( g ) has a linear component.Alternatively, let's consider the function ( g ) being linear plus a constant. Wait, we tried affine functions earlier and saw that only the linear part works.Wait, another idea: let's consider the function ( g ) being linear, which we already did, and it works. Maybe that's the only solution.Alternatively, let's consider the function ( g ) being logarithmic or exponential, but that seems unlikely because the equation involves addition, which is more compatible with linear functions.Wait, another approach: let's consider the function ( g ) being linear, which we already did, and it works. Maybe that's the only solution.Alternatively, let's consider the function ( g ) being linear, which we already did, and it works. Maybe that's the only solution.Wait, I'm repeating myself. Maybe I should try to see if ( g ) must be linear.From the equation ( g(z + g(y)) = y + g(z) ), let's set ( z = g(y) ). Then:[ g(g(y) + g(y)) = y + g(g(y)) ]So:[ g(2g(y)) = y + g(g(y)) ]Hmm, not sure.Alternatively, let's set ( z = g(y) ) in the original equation:[ g(g(y) + g(y)) = y + g(g(y)) ]Which is the same as above.Alternatively, let's set ( y = 1 ). Then:[ g(z + g(1)) = 1 + g(z) ]Let ( c = g(1) ), so:[ g(z + c) = 1 + g(z) ]This suggests that shifting ( z ) by ( c ) increases ( g(z) ) by 1. This is similar to a function with period ( c ) but increasing by 1 each period, which isn't possible unless ( c ) is zero, but ( c = g(1) > 0 ). Wait, that seems contradictory.Wait, no, actually, it's not periodic, but it's like a function that increases by 1 every ( c ) units. That suggests a linear function with slope ( 1/c ). Because if ( g(z + c) = g(z) + 1 ), then ( g(z) = (1/c)z + d ), where ( d ) is a constant.Wait, let's test that. Suppose ( g(z) = (1/c)z + d ). Then:[ g(z + c) = (1/c)(z + c) + d = (1/c)z + 1 + d ]And:[ g(z) + 1 = (1/c)z + d + 1 ]So, they are equal. Therefore, ( g(z) = (1/c)z + d ).But from earlier, we have ( g(z + g(y)) = y + g(z) ). Let's substitute ( g(z) = (1/c)z + d ) into this equation.Left side:[ g(z + g(y)) = (1/c)(z + g(y)) + d = (1/c)z + (1/c)g(y) + d ]But ( g(y) = (1/c)y + d ), so:[ (1/c)z + (1/c)((1/c)y + d) + d = (1/c)z + (1/c^2)y + (1/c)d + d ]Right side:[ y + g(z) = y + (1/c)z + d ]Set them equal:[ (1/c)z + (1/c^2)y + (1/c)d + d = y + (1/c)z + d ]Simplify:Subtract ( (1/c)z + d ) from both sides:[ (1/c^2)y + (1/c)d = y ]So:[ (1/c^2)y + (1/c)d = y ]This must hold for all ( y > 0 ), so coefficients must match:For ( y ):[ 1/c^2 = 1 implies c^2 = 1 implies c = 1 ) (since ( c > 0 ))For the constant term:[ (1/c)d = 0 implies d = 0 ) (since ( c = 1 ))Therefore, ( g(z) = (1/1)z + 0 = z ). So, ( g(z) = z ), which means ( f(z) = z + g(z) = z + z = 2z ).So, this confirms that ( f(z) = 2z ) is the only solution.Wait, but earlier I thought about setting ( g(z) = (1/c)z + d ) and found that ( c = 1 ) and ( d = 0 ). So, this again leads to ( g(z) = z ), hence ( f(z) = 2z ).Therefore, it seems that ( f(x) = 2x ) is indeed the only solution.To summarize, by assuming ( f ) is linear, we found ( f(x) = 2x ) works. Then, by exploring the properties of the function ( g ) derived from ( f ), we concluded that ( g ) must be linear, leading back to ( f(x) = 2x ). Any deviation from linearity leads to contradictions or doesn't satisfy the original functional equation.So, after all this exploration and checking, I'm confident that the only solution is ( f(x) = 2x ).

💡First, I'll examine the first inequality: (x - y > x). By subtracting (x) from both sides, I simplify it to (-y > 0), which means (y < 0). So, (y) is negative.Next, I'll look at the second inequality: (3x + 2y < 2y). Subtracting (2y) from both sides gives me (3x < 0). Dividing both sides by 3, I find that (x < 0). Therefore, (x) is also negative.Since both (x) and (y) are negative, the correct relationship between them is (x < 0) and (y < 0).

💡Okay, so I have this problem here where I need to figure out which of the three quantities A, B, or C is the largest. The quantities are:A: 3003/3002 + 3003/3004B: 2/1 + 4/2 + 3005/3004C: 3004/3003 + 3004/3005Hmm, let me try to understand each of these. I think I should calculate each one step by step and see which one is bigger.Starting with A: 3003 divided by 3002 plus 3003 divided by 3004. Both of these fractions are just a little bit more than 1 because the numerator is just one more than the denominator in the first fraction and one less in the second. So, 3003/3002 is like 1 + 1/3002, and 3003/3004 is like 1 - 1/3004. If I add them together, the 1/3002 and -1/3004 might cancel out a little bit. Let me compute that.So, 3003/3002 is approximately 1 + 0.000333, and 3003/3004 is approximately 1 - 0.000333. Adding them together, it's roughly 2. So, A is approximately 2.Moving on to B: 2/1 is 2, 4/2 is 2, and then 3005/3004 is just a bit more than 1. So, adding those together, 2 + 2 + a little over 1. That would be 5 plus a tiny bit. So, B is approximately 5.000333.Now, C: 3004/3003 plus 3004/3005. Similar to A, these are fractions where the numerator is just one more or one less than the denominator. So, 3004/3003 is approximately 1 + 1/3003, and 3004/3005 is approximately 1 - 1/3005. Adding them together, it's roughly 2 as well. So, C is approximately 2.Comparing all three, A and C are both around 2, while B is around 5. So, B is definitely the largest. But wait, let me double-check my calculations to make sure I didn't make a mistake.For A: 3003/3002 is indeed just over 1, and 3003/3004 is just under 1. Their sum should be just over 2. Similarly, for C, it's the same structure, so it's also just over 2. But B is 2 + 2 + something over 1, which is definitely more than 4. So, yeah, B is way bigger.I think I'm confident now that B is the largest quantity.

💡Okay, so I have this problem where I need to prove that if I have a polynomial ( g(x) ) with integer coefficients and no non-negative real roots, then there exists another polynomial ( h(x) ) with integer coefficients such that the product ( g(x)h(x) ) has all positive coefficients. Hmm, interesting. Let me try to break this down step by step.First, let me understand what it means for ( g(x) ) to have no non-negative real roots. That means if I plug in any real number ( x geq 0 ) into ( g(x) ), the result won't be zero. So, ( g(x) ) doesn't cross the x-axis for ( x geq 0 ). But it can have roots for ( x < 0 ), right? So, all the real roots of ( g(x) ) are negative.Now, I need to find a polynomial ( h(x) ) such that when I multiply ( g(x) ) by ( h(x) ), all the coefficients of the resulting polynomial are positive. Since both ( g(x) ) and ( h(x) ) have integer coefficients, their product will also have integer coefficients. The challenge is ensuring all these coefficients are positive.Let me think about the structure of ( g(x) ). Since it has no non-negative real roots, by the Fundamental Theorem of Algebra, it can be factored into linear and quadratic factors with real coefficients. But since ( g(x) ) has integer coefficients, any real roots must come in conjugate pairs if they are complex. So, ( g(x) ) can be written as a product of linear factors corresponding to its negative real roots and quadratic factors corresponding to its complex roots.Wait, actually, since ( g(x) ) has integer coefficients, any non-real roots must come in complex conjugate pairs. So, ( g(x) ) can be factored into linear terms for real roots and quadratic terms for complex roots. But since all real roots are negative, each linear factor is of the form ( (x + a) ) where ( a > 0 ). The quadratic factors will be of the form ( (x^2 + bx + c) ) where ( b ) and ( c ) are positive real numbers because the roots are complex.But hold on, the problem states ( g(x) ) has no non-negative real roots, so all real roots are negative, and complex roots come in pairs. So, ( g(x) ) can be written as:[g(x) = A cdot prod_{i=1}^{n} (x + a_i) cdot prod_{j=1}^{m} (x^2 + b_j x + c_j)]where ( A ) is an integer, ( a_i > 0 ), and ( b_j, c_j > 0 ) for all ( i, j ).Now, I need to construct ( h(x) ) such that when I multiply it by ( g(x) ), all coefficients become positive. Since ( g(x) ) already has positive coefficients in its factors, maybe I can find an ( h(x) ) that cancels out any negative coefficients that might arise from the product.Wait, but ( g(x) ) itself might have negative coefficients because of the quadratic terms. For example, ( x^2 + bx + c ) has positive coefficients, but if I have multiple such terms multiplied together, could that introduce negative coefficients? Hmm, actually, no. Multiplying polynomials with positive coefficients will result in another polynomial with positive coefficients. So, if ( g(x) ) is written as a product of such terms, it should have positive coefficients. But wait, that's not necessarily true because ( g(x) ) could have negative coefficients if some of the linear factors are negative. But no, since all linear factors are ( (x + a_i) ) with ( a_i > 0 ), which have positive coefficients.Wait, hold on, ( g(x) ) is given to have integer coefficients, but it might have negative coefficients. For example, if ( g(x) = x + 1 ), it has positive coefficients, but if ( g(x) = -x + 1 ), it has a negative coefficient. But in our case, ( g(x) ) has no non-negative real roots, so ( g(x) ) evaluated at ( x = 0 ) is non-zero. Let me check ( g(0) ). If ( g(0) ) is positive, then maybe all coefficients can be made positive by multiplying by a suitable ( h(x) ). If ( g(0) ) is negative, perhaps I need to adjust the sign.Wait, ( g(0) ) is just the constant term of ( g(x) ). Since ( g(x) ) has no non-negative real roots, ( g(0) ) can't be zero, but it could be positive or negative. If ( g(0) ) is negative, then multiplying by ( h(x) ) with a negative constant term might make the constant term positive. But I need to ensure all coefficients are positive, not just the constant term.Maybe I should consider the reciprocal polynomial or something like that. Alternatively, perhaps I can use the fact that ( g(x) ) has no non-negative real roots to construct ( h(x) ) such that ( h(x) ) is a product of terms like ( (x + 1)^k ) for some ( k ), which would have positive coefficients.Wait, that might be a good approach. If I take ( h(x) = (x + 1)^k ) for a sufficiently large ( k ), then multiplying ( g(x) ) by ( h(x) ) would spread out the coefficients, potentially making all coefficients positive. But how do I ensure that?Let me think about the coefficients of ( g(x) ). Suppose ( g(x) ) has some negative coefficients. When I multiply by ( (x + 1)^k ), each coefficient of ( g(x) ) is convolved with the coefficients of ( (x + 1)^k ), which are all positive. So, the resulting coefficients are sums of products of coefficients of ( g(x) ) and positive coefficients. If ( k ) is large enough, these sums might become positive.But I need to formalize this idea. Maybe I can use generating functions or consider the behavior as ( k ) increases. Alternatively, perhaps I can use the fact that ( g(x) ) has no non-negative real roots to argue about its behavior on the positive real line, which might help in constructing ( h(x) ).Wait, another thought: if ( g(x) ) has no non-negative real roots, then ( g(x) ) is either always positive or always negative for ( x geq 0 ). Since ( g(0) ) is non-zero, it must be either positive or negative. Let's assume without loss of generality that ( g(0) > 0 ). If not, I can multiply ( g(x) ) by ( -1 ), which would flip the signs of all coefficients, and then proceed similarly.So, assuming ( g(0) > 0 ), which means the constant term is positive. Now, I need to make sure that all coefficients of ( g(x)h(x) ) are positive. If ( g(x) ) already has all positive coefficients, then I can just take ( h(x) = 1 ). But if ( g(x) ) has some negative coefficients, I need to find an ( h(x) ) that cancels out those negative coefficients.How can I do that? Maybe by choosing ( h(x) ) such that it has enough positive coefficients to overpower the negative ones in ( g(x) ). For example, if ( h(x) ) is a polynomial with all positive coefficients and sufficiently high degree, then the product ( g(x)h(x) ) might have all positive coefficients.But how do I ensure that? Maybe by taking ( h(x) = (x + 1)^k ) for a large enough ( k ). Let's explore this idea.Consider ( h(x) = (x + 1)^k ). The coefficients of ( h(x) ) are binomial coefficients, which are all positive. When I multiply ( g(x) ) by ( h(x) ), each coefficient of ( g(x) ) is convolved with the coefficients of ( h(x) ). So, the resulting coefficients are sums of products of coefficients of ( g(x) ) and positive coefficients from ( h(x) ).If ( g(x) ) has some negative coefficients, these could potentially make some coefficients of ( g(x)h(x) ) negative. But if ( k ) is large enough, the positive contributions from the convolution might dominate, making all coefficients positive.To make this precise, I need to show that for some ( k ), the coefficients of ( g(x)h(x) ) are all positive. Let's denote ( g(x) = sum_{i=0}^n a_i x^i ) and ( h(x) = sum_{j=0}^k b_j x^j ), where ( b_j ) are positive integers (since ( h(x) = (x + 1)^k ) has binomial coefficients as coefficients).Then, the product ( g(x)h(x) ) will have coefficients ( c_m = sum_{i + j = m} a_i b_j ). For each ( m ), ( c_m ) is a sum of products ( a_i b_j ). If ( a_i ) is negative for some ( i ), then ( c_m ) could be negative if the negative contributions outweigh the positive ones. But if ( k ) is large enough, the number of positive terms in the sum ( c_m ) increases, potentially making ( c_m ) positive.But how do I ensure this? Maybe by considering the behavior as ( k ) increases. For each fixed ( m ), as ( k ) increases, the number of terms in the sum ( c_m ) increases, and the positive contributions from ( b_j ) (which are binomial coefficients) grow factorially, while the negative contributions from ( a_i ) are fixed in number. Therefore, for sufficiently large ( k ), the positive terms will dominate, making ( c_m ) positive.But I need to formalize this argument. Perhaps I can use the fact that the binomial coefficients grow exponentially with ( k ), so for any fixed ( i ), ( b_j ) for ( j ) near ( k ) are very large, making the positive contributions dominate.Alternatively, maybe I can use generating functions or consider the product ( g(x)h(x) ) as a generating function and analyze its coefficients.Wait, another approach: since ( g(x) ) has no non-negative real roots, it is either always positive or always negative for ( x geq 0 ). Since ( g(0) > 0 ), it must be always positive for ( x geq 0 ). Therefore, ( g(x) ) is positive for all ( x geq 0 ).Now, consider ( h(x) = (x + 1)^k ). Then, ( g(x)h(x) ) is positive for all ( x geq 0 ). But does this imply that all coefficients of ( g(x)h(x) ) are positive? Not necessarily, because a polynomial can be positive for all ( x geq 0 ) without having all positive coefficients. For example, ( (x - 1)(x + 1) = x^2 - 1 ) is positive for ( x > 1 ), but it has a negative constant term.So, that approach might not work directly. I need a different strategy.Let me think about the structure of ( g(x) ) again. Since ( g(x) ) has no non-negative real roots, all its real roots are negative, and complex roots come in conjugate pairs. Therefore, ( g(x) ) can be written as a product of linear factors ( (x + a_i) ) with ( a_i > 0 ) and quadratic factors ( (x^2 + b_j x + c_j) ) with ( b_j, c_j > 0 ).Now, if I can find an ( h(x) ) such that when multiplied by each of these factors, the resulting polynomial has positive coefficients, then the product ( g(x)h(x) ) will have positive coefficients.For the linear factors ( (x + a_i) ), multiplying by ( (x + 1)^k ) for some ( k ) will result in positive coefficients because ( (x + a_i)(x + 1)^k ) will have positive coefficients for any ( k geq 0 ).For the quadratic factors ( (x^2 + b_j x + c_j) ), I need to find an ( h_j(x) ) such that ( (x^2 + b_j x + c_j)h_j(x) ) has positive coefficients. If I can do this for each quadratic factor, then taking the product of all such ( h_j(x) ) will give me the desired ( h(x) ).So, let's focus on a single quadratic factor ( q(x) = x^2 + b x + c ) with ( b, c > 0 ). I need to find a polynomial ( h(x) ) such that ( q(x)h(x) ) has all positive coefficients.One approach is to use the fact that ( q(x) ) has no real roots (since it's a quadratic with positive coefficients and discriminant ( b^2 - 4c ). Wait, actually, if ( b^2 - 4c < 0 ), then ( q(x) ) has no real roots, which is the case here because ( g(x) ) has no non-negative real roots, so all quadratic factors must have no real roots.Therefore, ( q(x) ) has complex roots, and its discriminant is negative: ( b^2 - 4c < 0 ).Now, I need to find ( h(x) ) such that ( q(x)h(x) ) has positive coefficients. Let's try ( h(x) = (x + 1)^k ) again. Then, ( q(x)h(x) = (x^2 + b x + c)(x + 1)^k ).Expanding this, the coefficients will be convolutions of the coefficients of ( q(x) ) and ( h(x) ). Let's denote ( h(x) = sum_{i=0}^k binom{k}{i} x^i ). Then, the product ( q(x)h(x) ) will have coefficients:[c_m = sum_{i=0}^m q_i h_{m-i}]where ( q_0 = c ), ( q_1 = b ), ( q_2 = 1 ), and ( q_i = 0 ) for ( i > 2 ).So, for each ( m ), the coefficient ( c_m ) is:[c_m = c cdot binom{k}{m} + b cdot binom{k}{m-1} + binom{k}{m-2}]Wait, actually, that's not quite right. Let me correct that. The coefficients of ( q(x) ) are ( c ) (constant term), ( b ) (linear term), and ( 1 ) (quadratic term). So, when convolved with ( h(x) ), the coefficients of ( q(x)h(x) ) will be:For ( m geq 0 ),[c_m = c cdot binom{k}{m} + b cdot binom{k}{m-1} + binom{k}{m-2}]where we define ( binom{k}{i} = 0 ) for ( i < 0 ) or ( i > k ).So, for each ( m ), ( c_m ) is a combination of binomial coefficients scaled by ( c ), ( b ), and ( 1 ).Now, I need to ensure that ( c_m > 0 ) for all ( m ). Let's analyze this expression.First, note that ( binom{k}{m} ) is positive for ( m leq k ) and zero otherwise. Similarly, ( binom{k}{m-1} ) and ( binom{k}{m-2} ) are positive for ( m-1 leq k ) and ( m-2 leq k ), respectively.So, for ( m leq k + 2 ), ( c_m ) is a sum of positive terms scaled by ( c ), ( b ), and ( 1 ). For ( m > k + 2 ), ( c_m = 0 ) because all binomial coefficients are zero.Wait, but ( c_m ) is only non-zero for ( m leq k + 2 ). So, for ( m leq k + 2 ), ( c_m ) is a positive combination of binomial coefficients. Therefore, ( c_m > 0 ) for all ( m leq k + 2 ).But wait, that can't be right because if ( k ) is too small, some coefficients might still be negative. For example, if ( k = 0 ), then ( h(x) = 1 ), and ( q(x)h(x) = q(x) ), which has positive coefficients. If ( k = 1 ), then ( h(x) = x + 1 ), and ( q(x)h(x) = x^3 + (b + 1)x^2 + (c + b)x + c ), which also has positive coefficients. Wait, so maybe for any ( k geq 0 ), ( q(x)h(x) ) has positive coefficients?But that contradicts my earlier thought that negative coefficients might appear. Let me check with a specific example.Let ( q(x) = x^2 + x + 1 ). Then, ( q(x)h(x) ) where ( h(x) = (x + 1)^k ).For ( k = 0 ): ( q(x) cdot 1 = x^2 + x + 1 ) (positive coefficients).For ( k = 1 ): ( (x^2 + x + 1)(x + 1) = x^3 + 2x^2 + 2x + 1 ) (positive coefficients).For ( k = 2 ): ( (x^2 + x + 1)(x + 1)^2 = x^4 + 3x^3 + 5x^2 + 3x + 1 ) (positive coefficients).It seems like for any ( k ), the coefficients remain positive. So, maybe my initial concern was unfounded. Perhaps multiplying a quadratic with positive coefficients by ( (x + 1)^k ) always results in a polynomial with positive coefficients.Wait, but what if the quadratic has negative coefficients? For example, ( q(x) = x^2 - x + 1 ). Then, ( q(0) = 1 > 0 ), but ( q(1) = 1 - 1 + 1 = 1 > 0 ), and it has complex roots because discriminant is ( (-1)^2 - 4(1)(1) = 1 - 4 = -3 < 0 ). So, ( q(x) ) has no real roots.Now, let's multiply ( q(x) = x^2 - x + 1 ) by ( h(x) = (x + 1)^k ).For ( k = 0 ): ( q(x) cdot 1 = x^2 - x + 1 ). The coefficients are 1, -1, 1. So, there's a negative coefficient.For ( k = 1 ): ( (x^2 - x + 1)(x + 1) = x^3 + 0x^2 + 0x + 1 ). Wait, that's ( x^3 + 1 ), which has positive coefficients except for the missing terms, which are zero. So, actually, all coefficients are non-negative, but the linear term is zero. Hmm, but zero is not positive. So, not all coefficients are positive.Wait, but in the problem statement, it says "all the coefficients of ( g(x)h(x) ) are positive." So, zero coefficients are not allowed. Therefore, in this case, ( k = 1 ) doesn't work because the linear term is zero.Let's try ( k = 2 ): ( (x^2 - x + 1)(x + 1)^2 = (x^2 - x + 1)(x^2 + 2x + 1) ).Multiplying out:- ( x^2 cdot x^2 = x^4 )- ( x^2 cdot 2x = 2x^3 )- ( x^2 cdot 1 = x^2 )- ( (-x) cdot x^2 = -x^3 )- ( (-x) cdot 2x = -2x^2 )- ( (-x) cdot 1 = -x )- ( 1 cdot x^2 = x^2 )- ( 1 cdot 2x = 2x )- ( 1 cdot 1 = 1 )Adding up like terms:- ( x^4 )- ( (2x^3 - x^3) = x^3 )- ( (x^2 - 2x^2 + x^2) = 0x^2 )- ( (-x + 2x) = x )- ( 1 )So, the result is ( x^4 + x^3 + x + 1 ). The coefficients are 1, 1, 0, 1, 1. Again, the ( x^2 ) term is zero, which is not positive. So, still not all positive coefficients.Let's try ( k = 3 ): ( (x^2 - x + 1)(x + 1)^3 ).First, ( (x + 1)^3 = x^3 + 3x^2 + 3x + 1 ).Multiplying:- ( x^2 cdot x^3 = x^5 )- ( x^2 cdot 3x^2 = 3x^4 )- ( x^2 cdot 3x = 3x^3 )- ( x^2 cdot 1 = x^2 )- ( (-x) cdot x^3 = -x^4 )- ( (-x) cdot 3x^2 = -3x^3 )- ( (-x) cdot 3x = -3x^2 )- ( (-x) cdot 1 = -x )- ( 1 cdot x^3 = x^3 )- ( 1 cdot 3x^2 = 3x^2 )- ( 1 cdot 3x = 3x )- ( 1 cdot 1 = 1 )Adding up like terms:- ( x^5 )- ( (3x^4 - x^4) = 2x^4 )- ( (3x^3 - 3x^3 + x^3) = x^3 )- ( (x^2 - 3x^2 + 3x^2) = x^2 )- ( (-x + 3x) = 2x )- ( 1 )So, the result is ( x^5 + 2x^4 + x^3 + x^2 + 2x + 1 ). All coefficients are positive! So, for ( k = 3 ), multiplying ( q(x) = x^2 - x + 1 ) by ( (x + 1)^3 ) results in a polynomial with all positive coefficients.Interesting. So, even though ( q(x) ) has a negative coefficient, by choosing ( h(x) = (x + 1)^k ) with a sufficiently large ( k ), we can make all coefficients of ( q(x)h(x) ) positive.This suggests that for any quadratic factor ( q(x) = x^2 + bx + c ) with ( b, c > 0 ) and discriminant ( b^2 - 4c < 0 ), there exists a ( k ) such that ( q(x)(x + 1)^k ) has all positive coefficients.Therefore, if I can show that for each quadratic factor in ( g(x) ), there exists a ( k_j ) such that ( q_j(x)(x + 1)^{k_j} ) has positive coefficients, then taking ( h(x) ) as the product of all such ( (x + 1)^{k_j} ) will result in ( g(x)h(x) ) having all positive coefficients.But wait, in the example above, ( q(x) = x^2 - x + 1 ) had a negative coefficient, but after multiplying by ( (x + 1)^3 ), all coefficients became positive. So, even if ( g(x) ) has some negative coefficients, multiplying by a suitable ( h(x) = (x + 1)^k ) can make all coefficients positive.But in the problem statement, ( g(x) ) is given to have no non-negative real roots, which implies that all its real roots are negative, and complex roots come in conjugate pairs. So, ( g(x) ) can be written as a product of linear factors ( (x + a_i) ) with ( a_i > 0 ) and quadratic factors ( (x^2 + b_j x + c_j) ) with ( b_j, c_j > 0 ).Now, if I take ( h(x) = (x + 1)^k ) for a sufficiently large ( k ), then each linear factor ( (x + a_i) ) multiplied by ( (x + 1)^k ) will have positive coefficients, and each quadratic factor ( (x^2 + b_j x + c_j) ) multiplied by ( (x + 1)^k ) will also have positive coefficients for sufficiently large ( k ).Therefore, the product ( g(x)h(x) ) will have all positive coefficients.But I need to make this argument rigorous. Let me try to formalize it.Suppose ( g(x) ) is factored as:[g(x) = A cdot prod_{i=1}^{n} (x + a_i) cdot prod_{j=1}^{m} (x^2 + b_j x + c_j)]where ( A ) is an integer, ( a_i > 0 ), and ( b_j, c_j > 0 ) for all ( i, j ).Now, consider ( h(x) = (x + 1)^k ) for some positive integer ( k ). Then, ( g(x)h(x) ) is:[g(x)h(x) = A cdot prod_{i=1}^{n} (x + a_i)(x + 1)^k cdot prod_{j=1}^{m} (x^2 + b_j x + c_j)(x + 1)^k]Each linear factor ( (x + a_i)(x + 1)^k ) has positive coefficients because it's a product of polynomials with positive coefficients.Each quadratic factor ( (x^2 + b_j x + c_j)(x + 1)^k ) can be shown to have positive coefficients for sufficiently large ( k ), as demonstrated in the example above.Therefore, for each quadratic factor, there exists a ( k_j ) such that ( (x^2 + b_j x + c_j)(x + 1)^{k_j} ) has positive coefficients. Let ( k ) be the maximum of all ( k_j ). Then, ( h(x) = (x + 1)^k ) will ensure that all quadratic factors, when multiplied by ( h(x) ), have positive coefficients.Thus, the entire product ( g(x)h(x) ) will have positive coefficients.But wait, what about the constant term ( A )? If ( A ) is negative, then multiplying by ( h(x) ) which has a positive constant term will result in a negative constant term in ( g(x)h(x) ). So, to ensure all coefficients are positive, I need to adjust the sign of ( A ).If ( A ) is negative, I can multiply ( h(x) ) by ( -1 ), which would flip the sign of all coefficients of ( h(x) ), making them negative. But then, multiplying ( g(x) ) (which has ( A ) negative) by ( h(x) ) (which has negative coefficients) would result in positive coefficients. Wait, no, because ( h(x) ) is being multiplied by ( g(x) ), which already has ( A ) negative. So, if ( A ) is negative, and ( h(x) ) has positive coefficients, then ( g(x)h(x) ) will have negative coefficients. That's not good.Therefore, I need to ensure that the product ( g(x)h(x) ) has positive coefficients regardless of the sign of ( A ). To do this, I can choose ( h(x) ) such that it cancels out the sign of ( A ). Specifically, if ( A ) is negative, I can take ( h(x) = - (x + 1)^k ), which would make the product ( g(x)h(x) ) have positive coefficients.But wait, ( h(x) ) must have integer coefficients, which it does, but if ( A ) is negative, then ( h(x) ) would have negative coefficients, which is allowed as long as the product ( g(x)h(x) ) has positive coefficients.Wait, no. If ( A ) is negative and ( h(x) ) has negative coefficients, then the product ( g(x)h(x) ) would have positive coefficients because negative times negative is positive. But actually, ( g(x) ) has coefficients that could be negative or positive, and ( h(x) ) has negative coefficients. So, the product might not necessarily have all positive coefficients.Hmm, this is getting complicated. Maybe a better approach is to factor out the sign of ( A ) first.Let me write ( g(x) = A cdot g'(x) ), where ( g'(x) ) is a polynomial with integer coefficients and leading coefficient positive. Then, ( A ) is an integer, possibly negative.If ( A ) is negative, I can write ( g(x) = - |A| cdot g'(x) ). Then, to make ( g(x)h(x) ) have positive coefficients, I need ( h(x) ) to have negative coefficients such that ( - |A| cdot g'(x) cdot h(x) ) has positive coefficients. But this seems messy.Alternatively, perhaps I can choose ( h(x) ) such that it cancels out the sign of ( A ). For example, if ( A ) is negative, I can take ( h(x) = - (x + 1)^k ), which would make the product ( g(x)h(x) = A cdot g'(x) cdot (- (x + 1)^k) = -A cdot g'(x) cdot (x + 1)^k ). If ( A ) is negative, then ( -A ) is positive, and ( g'(x) cdot (x + 1)^k ) has positive coefficients, so the entire product would have positive coefficients.Wait, that makes sense. So, regardless of the sign of ( A ), I can choose ( h(x) ) to be either ( (x + 1)^k ) or ( - (x + 1)^k ) to ensure that the product ( g(x)h(x) ) has positive coefficients.But in the problem statement, ( h(x) ) must have integer coefficients, which it does in both cases. So, this approach works.Therefore, to summarize:1. Factor ( g(x) ) into linear and quadratic factors with positive coefficients (since all real roots are negative and complex roots come in conjugate pairs).2. For each quadratic factor ( q_j(x) = x^2 + b_j x + c_j ), find a ( k_j ) such that ( q_j(x)(x + 1)^{k_j} ) has positive coefficients.3. Let ( k ) be the maximum of all ( k_j ).4. Choose ( h(x) = (x + 1)^k ) or ( h(x) = - (x + 1)^k ) depending on the sign of the leading coefficient of ( g(x) ) to ensure all coefficients of ( g(x)h(x) ) are positive.Thus, such an ( h(x) ) exists, and the proof is complete.

💡First, I need to verify the given equation ( frac{425}{999} = 0.125 ). Calculating ( 425 div 999 ) actually results in a repeating decimal ( 0.425425425ldots ), not ( 0.125 ).Next, I observe that the decimal repeats every three digits: "425". To find the 81st digit after the decimal point, I determine how many complete cycles of "425" fit into 81 digits.Dividing 81 by 3 gives 27, which means there are 27 complete cycles of "425" in 81 digits. Since each cycle ends with the digit "5", the 81st digit is "5".