4.) 1225a is perfectly divisible by 1155 7*7*5*5*a is perfectly divisible by 11*

Question

4.) 1225a is perfectly divisible by 1155
7*7*5*5*a is perfectly divisible by 11*7*5*3

7*7*5*5*a/ 11*7*5*3 = 7*5*a/11*3
Thus a has to be a multiple of 33
Say it is 33 itself.
33=3*11. Since 3 and 11 are both prime we can say that 'a' is divisible by 2 primes '3' and '11'


5.) a and b are integers and x and y are rational numbers. A rational number is a number that can be represented in p/q form
ax+ by2
Since x is rational, ax will also be rational given that 'a' is an integer.
by2; since y is rational its square will also be rational. b*y2 will also be rational since 'b' is an integer and is being multiplied by a rational number 'y2'
Thus ax+by2 will also be a rational number since 'ax' and 'by2' are rational.

4.) Given that 1155 divides 1225a, find 2 primes that divide a. Explain briefly 5.) Prove: If a and b are integers and if x and y are rational numbers, then ax by is a rational number. Assume no roperties of rational numbers except the definition Con use use the Sum of integers

Explanation / Answer

problem 4th you solved absolutely right. And in the 5th one you did well and of course it is right, but we can prove this by other mathematical way. I am going through it..

scine x and y are rational number, so let X= c/d and Y = e/f, where c, d, e, f are integer and not equal to 0.

now aX + bY2 = a(c/d) + b(e/f)2 = ac/d + be2/f2

  = (acf2 + bde2)/(df2)

now here a*c*f*f and b*d*e*e and d*f*f is integer because a, b, c, d, e, f all are integers

there for acf2 + bde2 is integer, hence (acf2 + bde2)/(df2) is integer.

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