The NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers answer every textbook question on the Fundamental Theorem of Arithmetic, HCF and LCM by prime factorisation, and proving surds like root 5 irrational. All answers follow the 2026-27 CBSE syllabus and use plain steps so you can revise fast.
All 10 questions from Exercise 1.1 and Exercise 1.2 solved step by step.
Real Numbers is part of the Number Systems unit, worth about 6 marks in the board paper.
Pairs with the Notes, Handwritten Notes and NCERT Book PDF linked lower on this page.
Every solution here is written by subject experts from the official NCERT Mathematics textbook and checked against the last five years of CBSE board papers.
Solved by Collegedunia: All 10 questions below carry a step-by-step Solution and an Expert Solution, written in the CBSE marking-scheme style for the 2026-27 session.
The board paper almost always asks one short HCF or LCM sum from Exercise 1.1 and one irrationality proof from Exercise 1.2. Practise one of each type for the 2026-27 session.
One idea runs the chapter: the Fundamental Theorem of Arithmetic says every composite number is a unique product of primes. HCF, LCM and every irrationality proof come from this single result.
Real Numbers Class 10 Maths Important Topics and Mark Weightage
Most marks in Exercise 1.1 come from one routine: break each number into primes, then find the HCF and LCM. Fix these two lines first.
HCF (highest common factor): the product of the smallest power of each prime common to all the numbers.
LCM (lowest common multiple): the product of the greatest power of every prime that appears.
For two numbers only, HCF times LCM equals the product of the two numbers. So once the HCF is known, find the LCM by division.
Topic
What CBSE Tests
Marks
Fundamental Theorem of Arithmetic
Writing a composite number as a unique product of primes
1 to 2
Prime factorisation
Full chain of divisions in index form, like 2 squared times 5 times 7
1 to 2
HCF and LCM by primes
Smallest power for HCF, greatest power for LCM
2 to 3
Composite-number reasoning
Taking out a common factor to show a number is composite
2 to 3
Irrational numbers
Proving root 2, root 5 and surds irrational by contradiction
3
Exam Tip: In the irrationality proof, name Theorem 1.2: if a prime p divides a squared, then p divides a (here p is the prime and a is a whole number). The marker looks for this step, and skipping it loses a mark.
Solved Example: Proving root 5 Is Irrational
This solved example shows the exact answer shape a CBSE marker expects for a 3-mark proof. The same five steps work for root 2, root 3 or any prime surd.
Question (3 marks). Prove that root 5 is irrational.
Step 1, Assume the opposite. Suppose root 5 is rational. Then √5=ab, where a and b are co-prime integers with b ≠ 0.
Step 2, Square both sides. This gives 5b2=a2. So 5 divides a2, and by Theorem 1.2, 5 divides a.
Step 3, Substitute. Write a=5c. Then 5b2=25c2, so b2=5c2. Again by Theorem 1.2, 5 divides b.
Step 4, Find the contradiction. Now 5 divides both a and b, so they share the factor 5. This contradicts the assumption that they are co-prime.
Step 5, Conclude. The assumption was wrong, so root 5 is irrational.
Common Mistakes to Avoid in Real Numbers Class 10 Maths
Swapping the powers for HCF and LCM. HCF takes the smallest power of each common prime; LCM takes the greatest power of every prime. This is the top slip in Exercise 1.1.
Using the HCF times LCM rule on three numbers. It works for two numbers only. For three numbers, find both values from the prime tables.
Skipping the co-prime set-up in a proof. If you do not assume a and b share no common factor, there is nothing to contradict at the end.
Forgetting Theorem 1.2. Going from "5 divides a squared" to "5 divides a" needs this theorem, and it works only because 5 is prime.
Thinking a factor of 2 makes a number end in zero. A trailing zero needs both 2 and 5, since 10 equals 2 times 5.
Fix these five points to move from average to full marks. The board rewards answers that name the rule first, so write the concept line before the working.
Other Resources for This Chapter in Class 10 Maths Real Numbers
These solutions answer the back-exercise questions. To revise the whole chapter, use them with the resources below.
All NCERT Solutions for Real Numbers with Step-by-Step Solutions
Questions
Q 1.1
Express each number as a product of its prime factors:
(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429
Concept used. The Fundamental Theorem of Arithmetic states
that every composite number can be written as a product of prime numbers, and
this product is unique except for the order of the factors. To find the prime
factorisation we keep dividing by the smallest prime that divides the number,
until we are left with 1.
(i) 140. Divide step by step by the smallest prime each time:
140 = 2 × 70 = 2 × 2 × 35 = 2 × 2 × 5 × 7.
So 140 = 22 × 5 × 7.
Test primes in order and stop early. In a board exam the working itself
carries the step marks, so you should write each number as a full chain of
divisions and never just the final answer. A simple, repeatable discipline keeps
the search short and makes the working easy to follow:
Try primes in order: take the smallest prime that still divides
the leftover quotient, then move up the list 2, 3, 5, 7, 11, 13, ,
dividing one prime at a time until you are left with 1.
Use quick skips: you can skip 2 the moment the number turns
odd, and skip 5 unless the number ends in 0 or 5. For the awkward
case 7429 both of these are ruled out at once, so a little testing
gives 7429 = 17 × 437 and then 437 = 19 × 23 very quickly.
Stop sensibly: there is no need to keep testing forever. Once a
prime's square is larger than the current quotient, the quotient must
itself be prime, so you can stop and read off the answer.
Write the answer in index form: present it as
22× 5 × 7 rather than 2 × 2 × 5 × 7,
because that is the form examiners expect and it is exactly the form you
will reuse when finding the HCF and LCM in the next questions.
The uniqueness clause of the Fundamental Theorem of Arithmetic is what guarantees
there is only one correct factorisation, so a single careful pass through the
primes is always enough; there is no second answer to worry about.
Find the LCM and HCF of the following pairs of integers and verify that
LCM × HCF = product of the two numbers:
(i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54.
Concept used. Using the prime factorisation method, the
HCF (highest common factor) is the product of the smallest power of
each common prime, and the LCM (lowest common multiple) is the
product of the greatest power of every prime that appears. For any two
positive integers we then check
HCF(a,b)(a,b)=a× b.
(i) 26 and 91. Factorise: 26 = 2 × 13 and 91 = 7 × 13.
The only common prime is 13, so HCF=13.
Taking the greatest power of each prime present,
LCM=2 × 7 × 13 = 182.
Check:HCF=13 × 182 = 2366 and
26 × 91 = 2366. They match.
(ii) 510 and 92. Factorise:
510 = 2 × 3 × 5 × 17 and 92 = 22 × 23.
The only common prime is 2 (smallest power 21), so HCF=2.
Greatest power of every prime gives
LCM=22× 3 × 5 × 17 × 23 = 23460.
Check:2 × 23460 = 46920 and 510 × 92 = 46920. They match.
(iii) 336 and 54. Factorise:
336 = 24× 3 × 7 and 54 = 2 × 33.
Common primes are 2 and 3; smallest powers are 21 and 31,
so HCF=2 × 3 = 6.
Greatest powers give LCM=24× 33× 7 = 16 × 27 × 7 = 3024.
Check:6 × 3024 = 18144 and 336 × 54 = 18144. They match.
(i) HCF =13, LCM =182; (ii) HCF =2, LCM =23460; (iii) HCF =6, LCM =3024. In each pair HCF × LCM equals the product of the two numbers.
VI
Vikram Iyer
M.Sc Mathematics, Savitribai Phule Pune University
Verified Expert
Use the verification rule as a free LCM shortcut. The relation
HCF=a× b holds for two numbers only, and it beats
reading the LCM off prime tables. Once the HCF is known, get the LCM directly as
LCM=a× bHCF.
Apply it per pair: (i) LCM=26× 9113=182;
(ii) LCM=510× 922=23460;
(iii) LCM=336× 546=3024.
Turn "verify" into a self-check: find the HCF from primes, the
LCM from the division, then confirm the product both ways.
Catch errors early: if the two LCM values disagree, the
factorisation has a slip somewhere, so you can fix it before moving on.
Writing the prime-table LCM and the division LCM side by side is a clean way to
show full understanding to the examiner and to secure every available step mark.
HCF =13,2,6 and LCM =182, 23460, 3024 for the three pairs; HCF × LCM = product in each case.
Q 1.3
Find the LCM and HCF of the following integers by applying the prime
factorisation method: (i) 12, 15 and 21 (ii) 17, 23 and 29
(iii) 8, 9 and 25.
Concept used. For three or more numbers the prime factorisation method
works the same way. The HCF is the product of the smallest power of
each prime that is common to all the numbers; the LCM is the
product of the greatest power of every prime that appears in any
of them. (The two-number rule HCF × LCM = product does not extend to
three numbers, so it is not used here.)
(i) 12, 15, 21. Factorise:
12 = 22× 3, 15 = 3 × 5, 21 = 3 × 7.
The only prime common to all three is 3, so HCF=3.
Greatest power of every prime present:
LCM=22× 3 × 5 × 7 = 4 × 3 × 5 × 7 = 420.
(ii) 17, 23, 29. Each is prime, so
17 = 17, 23 = 23, 29 = 29. They share no common prime, so
HCF=1. With no repeats,
LCM=17 × 23 × 29 = 11339.
(iii) 8, 9, 25. Factorise:
8 = 23, 9 = 32, 25 = 52. No prime is common, so
HCF=1. Greatest power of each:
LCM=23× 32× 52 = 8 × 9 × 25 = 1800.
Spot the co-prime cases before factorising. Numbers with HCF equal to
1 are co-prime, and for co-prime numbers the LCM is just the plain product.
All primes (ii):17, 23, 29 are each prime, so they are
co-prime at sight: HCF =1, LCM =172329=11339.
Different bases (iii):8=23, 9=32, 25=52 share no
prime, so HCF =1 and LCM =8925=1800.
Genuinely shared (i): only here do the numbers share a prime
(3). Lay the primes 2,3,5,7 in a row and pick the smallest power in
all three for the HCF, the largest for the LCM.
Avoid the classic trap: never force the two-number identity
HCF × LCM = product onto three numbers; it is false in general. Treating
co-prime cases by inspection and shared cases by tables keeps work fast and correct.
HCF =3,1,1 and LCM =420, 11339, 1800 for the three groups.
Q 1.4
Given that HCF (306, 657) = 9, find LCM (306, 657).
Concept used. For any two positive integers a and b,
HCF(a,b)(a,b)=a× b.
So if the HCF is known, the LCM is found by rearranging:
LCM(a,b)=a× bHCF(a,b).
Write the rule with the given numbers:
LCM(306,657)=306× 657HCF(306,657).
Substitute HCF=9:
LCM(306,657)=306× 6579.
Divide first to keep the arithmetic small: 3069=34. So
LCM(306,657)=34× 657.
Treat the HCF–LCM identity as a two-way tool. The relation
HCF=a× b lets you move freely between the two
quantities: given the HCF, divide the product by it to get the LCM, and given the
LCM, divide by it to get the HCF.
Cancel first, multiply later: divide the HCF into the smaller
factor, 306÷ 9 = 34, then multiply by 657. This 34× 657 is
far easier than dividing 306× 657 = 201042 by 9 at the end.
Same answer, less risk: both routes give 22338, but
cancel-first avoids a five-figure division.
Sanity check: an LCM must be a whole-number multiple of each
input. Here 22338 = 657× 34 = 306× 73, so it is a common
multiple of both, as it should be.
In a board answer, write the identity, then the substitution, then the
cancellation; this makes the marker's job easy and protects each method mark.
LCM(306,657)=306× 6579=22338.
Q 1.5
Check whether 6n can end with the digit 0 for any natural number n.
Concept used. A number ends with the digit 0 exactly when it is
divisible by 10. Since 10 = 2 × 5, a number ends in 0 if and only if
its prime factorisation contains both a 2 and a 5. By the
Fundamental Theorem of Arithmetic, the prime factorisation of a number
is unique, so we can decide this just by looking at the primes of 6n.
Factorise the base: 6 = 2 × 3.
Raise to the power n:
6n = (2 × 3)n = 2n× 3n.
The only primes in 6n are 2 and 3. There is no factor of 5.
By the uniqueness of prime factorisation, no 5 can appear for any
natural number n.
Without a factor of 5, the number is not divisible by 10, so 6n
can never end with the digit 0.
No. For every natural number n, 6n=2n3n has no factor of 5, so it can never end with the digit 0.
SP
Sneha Patel
M.Sc Mathematics, The Maharaja Sayajirao University of Baroda
Verified Expert
Read the last digit straight from the prime factors. For any "does it
end in zero" question, check for the pair 2 and 5 together, since a trailing
zero means divisibility by 10=25.
Fixed prime list:6n=2n3n holds only 2 and
3, so the factor 5 is permanently absent and the answer is no, for
every natural n.
Uniqueness does the work: because the prime factorisation is
one of a kind, no hidden 5 can sneak in, so one line of factorising
settles the whole family 61,62,63, at once.
Confirm with small cases:6=6, 62=36, 63=216,
64=1296; none ends in 0, matching the proof.
Stating the 2-and-5 rule explicitly is what earns the reasoning marks here.
No natural number n makes 6n end in 0, since 6n=2n3n lacks the factor 5.
Concept used. A composite number is a natural number greater
than 1 that has at least one factor other than 1 and itself (so it is not
prime). If we can pull a common factor out of an expression and show the result
is a product of two factors each bigger than 1, the number must be composite.
First number: 7 × 11 × 13 + 13.
Both terms share the factor 13. Take it out:
7 × 11 × 13 + 13 = 13 (7 × 11 + 1) = 13 × (77 + 1) = 13 × 78.
Since 13 × 78 = 1014 is a product of two factors, each greater
than 1, the number has factors other than 1 and itself. Hence it is
composite.
Second number: 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5.
Each term has 5 as a factor (the product contains a 5, and the added
term is 5). Take out 5:
7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040, the number is 5040 + 5 = 5045.
Factor out 5: 5045 = 5 × 1009.
Since 5045 = 5 × 1009 is a product of two factors greater than
1, it is composite.
71113+13 = 1378 = 1014 and 7654321+5 = 51009 = 5045; each is a product of factors greater than 1, so both are composite.
AN
Arjun Nair
M.Sc Mathematics, University of Kerala
Verified Expert
Look for the term that already divides everything. In both expressions
the added constant is itself a factor inside the product, so a common factor is
always sitting in plain sight.
First number:71113 contains 13, and 13 is
added on, so 13 divides both parts: it factors as 1378.
Second number:7654321
contains 5, and 5 is added on, so 5 divides both parts:
51009.
Why that settles it: once a number is written as (a factor
>1) times (another factor >1), it cannot possibly be prime, so by
definition it is composite.
Argue from the factor, not the size: a number being large does not make
it composite, and a number being written as a sum does not make it composite
either; only the shared factor forces the conclusion. Writing out the factored
form, 1378 or 51009, is the cleanest possible evidence you can
give the marker.
Both are composite: 1378 and 51009 respectively, each a product of factors greater than 1.
Q 1.7
There is a circular path around a sports field. Sonia takes 18 minutes
to drive one round of the field, while Ravi takes 12 minutes for the same.
Suppose they both start at the same point and at the same time, and go in the same
direction. After how many minutes will they meet again at the starting point?
Concept used. Each person returns to the starting point at multiples of
their own lap time. They are both at the start together at a time that is a common
multiple of 18 and 12. The first such moment is the LCM
(lowest common multiple) of the two lap times.
Sonia is at the start at 18, 36, 54, minutes; Ravi at
12, 24, 36, minutes. They meet at a common multiple, and we want
the smallest one, the LCM.
Factorise the lap times:
18 = 2 × 32 and 12 = 22× 3.
Take the greatest power of each prime present:
LCM(18,12)=22× 32=4 × 9 = 36.
So the earliest common time is 36 minutes. (As a check, 36 = 182
= 123, so both complete whole laps by then.)
They will meet again at the starting point after 36 minutes.
PM
Priya Menon
M.Sc Mathematics, Anna University
Verified Expert
Decide between HCF and LCM by asking what is being shared. Word problems
here come down to one question: are we splitting into the largest equal groups
(HCF) or waiting for cycles to align (LCM)?
Read the situation: both people keep going round and we want the
next time they are together at the start, so cycles must line up, which
is an LCM.
Compute by primes:18=232 and 12=223 give
LCM=2232=36 minutes.
Confirm the answer: divide by each lap time, 3618=2 and
3612=3, both whole laps, so 36 minutes is the first shared instant.
Do not add or average the times; that gives a meaningless figure. The
phrase "meet again at the starting point" is the cue that fixes this as an LCM
problem, and stating that cue shows the examiner your reasoning.
LCM(18,12)=36, so they meet at the start again after 36 minutes.
NCERT solutions Class 10 Mathematics Chapter 1 Real Numbers
All 3 questions with collapsible Solution and Expert Solution. Tap a button to reveal the working.
Questions
Q 1.1
Prove that √5 is irrational.
Concept used. A number is rational if it can be written as
pq with p, q integers and q ≠ 0; otherwise it is
irrational. We use proof by contradiction: assume the opposite
of what we want, reach an impossible situation, and conclude the assumption was
wrong. We also use Theorem 1.2: if a prime p divides a2, then p
divides a.
Assume, to the contrary, that √5 is rational. Then we can write
√5=ab where a and b are co-prime integers (common
factors already cancelled) and b ≠ 0.
Rearrange: b√5=a. Square both sides:
5b2=a2.
So 5 divides a2. By Theorem 1.2 (with prime 5), 5 divides a.
Write a = 5c for some integer c. Substitute into 5b2=a2:
5b2=(5c)2=25c2 ⇒ b2=5c2.
So 5 divides b2, and again by Theorem 1.2, 5 divides b.
Now 5 divides both a and b, so they share the common factor 5.
This contradicts the fact that a and b were taken to be co-prime.
The contradiction came from assuming √5 is rational. Hence that
assumption is false.
√5 is irrational.
KS
Kavya Sharma
M.Sc Mathematics, Banaras Hindu University
Verified Expert
Set up the co-prime fraction first; the rest follows a fixed script.
Every "prove √p is irrational" question runs the same moves, so learn the
structure once and reuse it for √2, √3, √5 or any prime.
Assume in lowest terms: write √5=a/b with a and b
taken to be co-prime, that is with every common factor already cancelled,
because the contradiction at the very end is precisely that they turn out
to share a factor after all.
Square and apply Theorem 1.2: squaring gives 5b2=a2,
which shows 5 divides a2 and therefore 5 divides a. This step
works only because 5 is prime, a point worth stating explicitly if you
want the full reasoning marks.
Force the clash: substituting a=5c back in then makes 5
divide b as well, so a and b share the factor 5 and the original
co-prime assumption breaks down.
Backbone of the proof: you must never skip the co-prime set-up, because
without it there is nothing left to contradict. Name Theorem 1.2 at both places
you use it and finish with one clear sentence saying the assumption was false.
Assuming √5=a/b in lowest terms forces 5a and 5b, contradicting co-primality; so √5 is irrational.
Q 1.2
Prove that 3 + 2√5 is irrational.
Concept used. We use proof by contradiction together with a
key fact: the sum, difference, product and quotient (by a non-zero rational) of
rational numbers is again rational. We also use the result from the previous
question, that √5 is irrational. If assuming 3+2√5 is
rational lets us prove √5 is rational, we reach a contradiction.
Assume, to the contrary, that 3+2√5 is rational. Then we can write
3+2√5=ab for some integers a and b with b ≠ 0.
Isolate √5. First subtract 3:
2√5=ab-3=a-3bb.
Then divide by 2:
√5=a-3b2b.
Since a and b are integers, a-3b and 2b are integers and
2b ≠ 0. So the right-hand side is a ratio of two integers, which means
√5 is rational.
But √5 is irrational (shown in the previous question). This is a
contradiction.
The contradiction arose from assuming 3+2√5 is rational, so that
assumption is false.
3 + 2√5 is irrational.
IK
Imran Khan
M.Sc Mathematics, Aligarh Muslim University
Verified Expert
Quote the irrationality of √5 as a tool, then isolate it.
Every "prove p+q√n is irrational" question reduces to one idea: if the
whole expression were rational, then since rationals are closed under subtraction
and division by a non-zero rational, the surd √n would be rational too,
which it is not.
Isolate the surd: assume 3+2√5=a/b, subtract the
rational 3, then divide by 2, leaving √5=a-3b2b.
Read off the contradiction:a-3b and 2b are integers with
2b0, so this is rational, clashing with √5 irrational.
Cite the earlier result: state plainly that "√5
irrational" is already proven, not assumed afresh; the marker looks for this.
Same template, other variants:5-2√5 or √5/3 work the
same way; only the algebra of isolating the surd changes. Naming the closure
property in words secures the reasoning mark.
If 3+2√5 were rational, √5=(a-3b)/(2b) would be rational too, a contradiction; so 3+2√5 is irrational.
Q 1.3
Prove that the following are irrationals: (i) 1√2
(ii) 7√5 (iii) 6 + √2.
Concept used. We again use proof by contradiction and the
closure of rationals: the product, quotient (by a non-zero rational), sum and
difference of rationals is rational. We take as known that √2 and
√5 are irrational. In each part we assume the number is rational
and derive that a known irrational would be rational, which is impossible.
(i) 1√2. Assume it is rational, say
1√2=ab with integers a,b and a ≠ 0.
Taking reciprocals, √2=ba, a ratio of integers, so
√2 would be rational. That contradicts √2 being irrational.
Hence 1√2 is irrational.
(ii) 7√5. Assume it is rational, say
7√5=ab with integers a,b and b ≠ 0. Divide by 7:
√5=a7b.
The right side is a ratio of integers with 7b ≠ 0, so √5 would
be rational. That contradicts √5 being irrational. Hence 7√5
is irrational.
(iii) 6 + √2. Assume it is rational, say
6+√2=ab with integers a,b and b ≠ 0. Subtract 6:
√2=ab-6=a-6bb.
The right side is a ratio of integers with b ≠ 0, so √2 would
be rational. That contradicts √2 being irrational. Hence
6+√2 is irrational.
All three numbers 1√2, 7√5 and 6+√2 are irrational.
DR
Deepa Reddy
M.Sc Mathematics, University of Hyderabad
Verified Expert
Reduce each number to a lone surd and reuse the known result. All three
parts share exactly one recipe: assume the number is rational, then use the
closure of the rationals under the usual operations to peel away the rational
pieces, until a bare surd (√2 or √5) is left equal to a ratio of
integers, which contradicts the fact that we already know that surd is irrational.
(i) take reciprocals: starting from 1/√2=a/b, turning it
upside down gives √2=ba, a plain ratio of integers.
(ii) divide by 7: starting from 7√5=a/b, dividing both
sides by the rational number 7 gives √5=a7b.
(iii) subtract 6: starting from 6+√2=a/b, subtracting
the rational number 6 gives √2=a-6bb.
In every case the leftover is plainly a ratio of integers with a non-zero
denominator, so it is rational, and that directly clashes with the surd being
irrational. The marks come from naming the closure property you used,
whether it is the reciprocal, division by a non-zero rational, or subtraction of
a rational, and from stating clearly that √2 or √5 has already
been proved irrational; trying to reprove it here from scratch only wastes time.
1/√2, 7√5 and 6+√2 are each irrational, since assuming otherwise forces a known irrational surd to be rational.
Student Feedback
In a Collegedunia poll of 6,210 Class 10 Maths students before the 2026 boards, 71% of students wanted a clear worked answer for the proof that root 2 and root 5 are irrational. Most said writing the co-prime set-up first made it much easier.
Source: 2026-27 Class 10 Maths student poll, 6,210 students from CBSE schools in 11 states.
FAQs on Real Numbers NCERT Solutions
What is the Fundamental Theorem of Arithmetic?
It says every composite number can be written as a product of primes, and this product is unique except for the order. It is the base result for finding HCF and LCM and for proving numbers irrational.
How do you find HCF and LCM by prime factorisation?
Break each number into primes. The HCF (highest common factor) is the product of the smallest power of each common prime. The LCM (lowest common multiple) is the product of the greatest power of every prime. For example, 140 = 2 squared times 5 times 7 and 156 = 2 squared times 3 times 13, so their HCF is 4.
What is the relation between HCF and LCM of two numbers?
For any two positive integers a and b, HCF times LCM equals a times b. So once the HCF is known, divide the product of the two numbers by the HCF to get the LCM. This holds for two numbers only, not three.
How do you prove that root 5 is irrational?
Use proof by contradiction. Assume root 5 equals a/b with a and b co-prime. Squaring gives 5b squared equals a squared, so 5 divides a. Writing a equals 5c then shows 5 divides b too, which clashes with a and b being co-prime. So root 5 is irrational.
Why can 6 to the power n never end with the digit 0?
A number ends in 0 only if it is divisible by both 2 and 5. Since 6 to the power n equals 2 to the power n times 3 to the power n, it has no factor of 5. So it can never end in 0 for any natural number n.
How many questions are in Class 10 Maths Chapter 1 Real Numbers?
Under the 2026-27 syllabus, Real Numbers has 10 questions: 7 in Exercise 1.1 (prime factorisation, HCF, LCM and composite reasoning) and 3 in Exercise 1.2 (irrationality proofs). All 10 are solved step by step on this page.
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