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# Mathematics induction question?

Prove by mathematical induction that n < 2n for all positive integers n

### 5 Answers

- BrendaLv 79 years agoFavorite Answer
n<2n

1)check for n=1

1<2*1

1<2 true

2)n=k

k<2k

3)n=k=k+1

k+1<2(k+1)

k+1<2k+2

remember k<2k by induction assumption

and of course 1<2 is always true

thus

k+1<2k+2 is true for all positive integers n

n<2n true

- DWReadLv 79 years ago
1 < 2·1, so the premise holds true for n = 1.

Suppose the premise is true for some value of n that is greater than 1. Then

n < 2n

Add 1 to both sides of the inequality.

n + 1 < 2n + 1

Since n is positive,

2n + 1 < 2n + 2

2n + 2 = 2(n + 1)

By transitivity,

n + 1 < 2(n + 1)

Therefore, the premise holds true for the case n + 1 as well.

- RaffaeleLv 79 years ago
property is valid for n = 1

1 < 2

suppose property holds for n

n < 2n

we must prove that it is true for n + 1, that is

n + 1 < 2(n + 1)

n + 1 < 2n + 2

n < 2n + 1

since n < 2n to a greater extent n < 2n + 1

in other words

if n < 2n < 2n + 1

for transitive property of < we have

n < 2n + 1

QED

- LuisLv 69 years ago
Dear Arjun,

n<2n

Since 2n contains the variable to solve for, move it to the left-hand side of the inequality by subtracting 2n from both sides.

n-2n<0

According to the distributive property, for any numbers a, b, and c, a(b+c)=ab+ac and (b+c)a=ba+ca. Here, n is a factor of both n and -2n.

(1-2)n<0

To add integers with different signs, subtract their absolute values and give the result the same sign as the integer with the greater absolute value. In this example, subtract the absolute values of 1 and -2 and give the result the same sign as the integer with the greater absolute value.

(-1)n<0

Remove the parentheses.

-n<0

Multiply each term in the inequality by -1. When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

-n*-1>0*-1

Multiply -n by -1 to get n.

n>0*-1

Multiply 0 by -1 to get 0.

n>0

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Source(s): Precalculus Solved! - How do you think about the answers? You can sign in to vote the answer.
- 9 years ago
Lemma: If a < b and c < d, then a + c < b + d.

Clearly, 1 < 2 so this holds for n = 1. Assume this holds for some n <= k. Thus, k < 2k. We need to show that k + 1 < 2(k + 1).

Apply this lemma (a = k, b = 2k, c = 1, and d = 2). Then k + 1 < 2k + 2 so then k + 1 < 2(k+1). Thus, this completes the proof.