# Math Problems need hELP!!

• Sep 23rd 2007, 12:38 PM
lemontea
Math Problems need hELP!!
Can anyone help me with these two problems! THank You!!!

1) Evaluate:
http://img.photobucket.com/albums/v3...athproblem.jpg
2.) a and b are two integers, and a>b. The sum of a and b is 5, but their product is less than 1. What can you say about a and b?

What I found is:
a is greater than b
a+b=5 , but when the multiply, it is less than 1

I dont know what to do next...i tried forming numbers that is valid for the description above..but failed..can anyone help me..
• Sep 23rd 2007, 02:36 PM
Jhevon
Quote:

Originally Posted by lemontea
Can anyone help me with these two problems! THank You!!!

1) Evaluate:
http://img.photobucket.com/albums/v3...athproblem.jpg

don't be overwhelmed by this huge thing, just take your time and work from the bottom up. so work out $\displaystyle 1 + \frac 45$ and put it back into it's position. then work out $\displaystyle 1 + \frac 3{ \mbox {the answer you got before}}$, and put it back into it's position, then work out $\displaystyle 1 + \frac 2{\mbox {the answer you got before this}}$ and put it back in position, and so on and so forth

Quote:

2.) a and b are two integers, and a>b. The sum of a and b is 5, but their product is less than 1. What can you say about a and b?

What I found is:
a is greater than b
a+b=5 , but when the multiply, it is less than 1

I dont know what to do next...i tried forming numbers that is valid for the description above..but failed..can anyone help me..
we have that $\displaystyle a$ and $\displaystyle b$ are integers for $\displaystyle a > b$. furthermore, we are told that:

$\displaystyle a + b = 5$

and

$\displaystyle ab < 1$

what can we say about $\displaystyle a$ and $\displaystyle b$?

i am not exactly what you are looking for here, but notice that if both $\displaystyle a$ and $\displaystyle b$ are positive. then we could get their sum to be 5, but after that, getting their product to be less than 1 is impossible. after all, there are only a few possibilities with this scenario:

(1) a = 4, b = 1 ...........ab > 1
(2) a = 3, b = 2 ...........ab > 1
we cannot switch a and b since we must have a>b, and so, only these two possibilities work for a and b positive and in both cases, ab > 1. so this cannot be true

okay, can a and b both be negative then? obviously not, we can't add two negatives and get a positive, so that won't work.

what else is there? a and b are different signs. since a > b, a must be positive and b negative. can that work? yes! as long as a = |b| + 5, we can have the desired result.

example. a = 8, b = -3. then a + b = 8 - 3 = 5, and ab = -24 < 1.

Bingo! so this is a possible relationship between a and b. so we can say this about a and b, a can be positive, but if so, b must be negative, with a = |b| + 5

is there anything else we can say? sure, we didn't consider the integer zero. what if one is zero. to add and get 5 the other must be 5. obviously we must have a = 5 and b = 0 since a>b. and so we have a + b = 5 + 0 = 5, and ab = 5(0) = 0 < 1. Bingo!

so finally, here is what we can say about a and b. given these conditions, we must have one of two cases for all the conditions to be fulfilled.

(1) a = 5, b = 0

or

(2) a > 0, b < 0 with a = |b| + 5

and we're done

did you follow the thought process ok? of course you would not write all this out, but you should be thinking somewhere along these lines to answer the question