Prove that √5 is an irrational number.
Answers
Step-by-step explanation:
Let √5 is rational. That is, we can write √5=ab, where a and b are integers. In fact, we can choose a and b to have no common factors. For example, we might choose a=3 and b=2 rather than a=15 and b=10, if that worked. So from here on out, a and b are integers with no common factors such that√5=ab.
Since √5=ab , squaring both sides we get:
5=a2b2, which implies that:
5b2=a2.
This means that a2 is multiple of 5. Now recall that the square of multiple of 5 is a multiple of 5. If a were multiple of 5, that would mean a2 would be a multiple of 5. So a is divisible by 5. This means we can write a=5c for some integer c. With me so far? Now if we substitute 5c for a in the equation a2=2b2, we get (5c)2=5b2, which simplifies to 25c2=5b2 or 5c2=b2. Thus b2 is divisible by 5, so b is multiple of 5
But now we have reached a contradiction! We assumed that a and b have no common factors. But we have shown that a and b are both multiple of 5; that is, they are both divisible by 5.which is a contradiction that a,b have no common factors.
Hence √5– is not rational.
Therefore, √5 is irrational. (hence proved)
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▶Let us assume that _/5 is not an irrational number.
_/5 = p/q
...........Squaring on both sides............
=》5q^2 = p^2--------------(1)
5 divides p^2 = 5 also divides p--------(2).
Let,
p = 5m
p^2 = 25m^2
putting value of p^2 in (1),we get 5q^2 = 25m^2.
=》q^2 = 5m^2
5 divides q^2
= 5 divides q-------------(3)
From (2)
p is a multiple of 5 and from (3) q is also multiple of 5.
5 is a common factor of p and q.
▶This contradicts to our assupmtion.
So,there is no common factor of p and q.
Hence, _/5 is an irrational number.✔
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Hope you understand....✌
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