15. Use Euclid division lemma to show that the square of any positive integer cannot be ofthe form 5m 2 or 5m – 3 for some integer m. (3) About the author Emma
Answer: Let a be any positive integer By Euclid’s division lemma a=bm+r a=5m+r (when b=5) So, r can be 0,1,2,3,4 Case 1: ∴a=5m (when r=0) a 2 =25m 2 a 2 =5(5m 2 )=5q where q=5m 2 Case 2: when r=1 a=5m+1 a 2 =(5m+1) 2 =25m 2 +10m+1 a 2 =5(5m 2 +2m)+1 =5q+1 where q=5m 2 +2m Case 3: a=5m+2 a 2 =25m 2 +20m+4 a 2 =5(5m 2 +4m)+4 5q+4 where q=5m 2 +4m Case 4: a=5m+3 a 2 =25m 2 +30m+9 =25m 2 +30m+5+4 =5(5m 2 +6m+1)+4 =5q+4 where q=5m 2 +6m+1 Case 5: a=5m+4 a 2 =25m 2 +40m+16=25m 2 +40m+15+1 =5(5m 2 +8m+3)+1 =5q+1 where q=5m 2 +8m+3 From these cases, we see that square of any positive no can’t be of the form 5q+2,5q+3. Reply
Step-by-step explanation:
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Answer:
Let a be any positive integer
By Euclid’s division lemma
a=bm+r
a=5m+r (when b=5)
So, r can be 0,1,2,3,4
Case 1:
∴a=5m (when r=0)
a
2
=25m
2
a
2
=5(5m
2
)=5q
where q=5m
2
Case 2:
when r=1
a=5m+1
a
2
=(5m+1)
2
=25m
2
+10m+1
a
2
=5(5m
2
+2m)+1
=5q+1 where q=5m
2
+2m
Case 3:
a=5m+2
a
2
=25m
2
+20m+4
a
2
=5(5m
2
+4m)+4
5q+4
where q=5m
2
+4m
Case 4:
a=5m+3
a
2
=25m
2
+30m+9
=25m
2
+30m+5+4
=5(5m
2
+6m+1)+4
=5q+4
where q=5m
2
+6m+1
Case 5:
a=5m+4
a
2
=25m
2
+40m+16=25m
2
+40m+15+1
=5(5m
2
+8m+3)+1
=5q+1 where q=5m
2
+8m+3
From these cases, we see that square of any positive no can’t be of the form 5q+2,5q+3.