Question

Solve the following questions with clear explanations...

Points: 25 ☺

Answer 1

1sol)
Let 'a' be any positive integer and b=4.
Then by applying Euclid's division lemma we get,
a=4m+r , such that 0≤r<4
Then,
a=4m , a=4m+1 , a=4m+2 and a=4m+3.
Here odd positive integers are
a=4m+1 and a=4m+3
Now we have,
a=4m+1

a²=(4m+1)²

a²=16m²+1+8m

a²=8m(2m+1)+1

a²=8q+1 , where q=m(2m+1)
and,
a=4m+3

a²=(4m+3)²

a²=16m²+9+24m

a²=16m²+24m+8+1

a²=8(2m²+3m+1)+1

a²=8q+1 , where q=(2m²+3m+1)

Hence by above we can conclude that
square of any odd positive integer is of the form 8q+1 , where q is any integer.




2sol)
(i)
51/120=(3×17)/(2×2×2×3×5)

=17/(2³×5)
Here denominator is in the form of 2^n×5^m, where n=3 and m=1
Therefore decimal expansion of 51/120 is terminating.
51/120=17/(2³×5)×5²/5²

51/120=425/(2³×5³)

51/120=425/(2×5)³=425/10³

51/120=0.425

(ii)
637/7280=(7×7×13)/(2×2×2×2×5×7×13)
=7/(2⁴×5)
Here denominator is in the form of 2^n×5^m where n=4 and m=1.
Therefore decimal expansion of

637/7280 is terminating.

637/7280=7/(2⁴×5)×5³/5³

637/7280=875/(2⁴×5⁴)=875/10⁴

637/7280=0.0875




3sol)
(i)
47/(2³×5²)=47/(2³×5²)×5/5

47/(2³×5²)=235(2³×5³)

47/(2³×5²)=235/(2×5)³=235/(10)³

47/(2³×5²)=235/1000

47/(2³×5²)=0.235

Hence the decimal expansion of given fraction terminates after three places.

(ii)
359/(2×5⁴)=359/(2×5⁴)×2³/2³

359/(2×5⁴)=2872/(2⁴×5⁴)

359/(2×5⁴)=2872/(2×5)⁴

359/(2×5⁴)=0.2872

Hence the decimal expansion of given fraction terminates after four places.





4sol)
Let us assume to the contrary that 3+2√5 is a rational number, then there exists co-prime integers a and b such that,
3+2√5=a/b

2√5=a/b-3

2√5=(a-3b)/b

√5=(a-3b)/2b

(a-3b)/2b is a rational number.
Then √5 is also a rational number.
But we know that √5 is an irrational number.
This is a contradiction.
This contradiction has arisen as our assumption is wrong.
Hence 3+2√5 is an irrational number.

Answer 2

Answer:

The above answer is correct because...

Step-by-step explanation:

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