165.395 862 948 928 595 415 054 587 647 318 840 026 856 11 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 165.395 862 948 928 595 415 054 587 647 318 840 026 856 11(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
165.395 862 948 928 595 415 054 587 647 318 840 026 856 11(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 165.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 165 ÷ 2 = 82 + 1;
  • 82 ÷ 2 = 41 + 0;
  • 41 ÷ 2 = 20 + 1;
  • 20 ÷ 2 = 10 + 0;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

2. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

165(10) =

1010 0101(2)


3. Convert to binary (base 2) the fractional part: 0.395 862 948 928 595 415 054 587 647 318 840 026 856 11.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.395 862 948 928 595 415 054 587 647 318 840 026 856 11 × 2 = 0 + 0.791 725 897 857 190 830 109 175 294 637 680 053 712 22;
  • 2) 0.791 725 897 857 190 830 109 175 294 637 680 053 712 22 × 2 = 1 + 0.583 451 795 714 381 660 218 350 589 275 360 107 424 44;
  • 3) 0.583 451 795 714 381 660 218 350 589 275 360 107 424 44 × 2 = 1 + 0.166 903 591 428 763 320 436 701 178 550 720 214 848 88;
  • 4) 0.166 903 591 428 763 320 436 701 178 550 720 214 848 88 × 2 = 0 + 0.333 807 182 857 526 640 873 402 357 101 440 429 697 76;
  • 5) 0.333 807 182 857 526 640 873 402 357 101 440 429 697 76 × 2 = 0 + 0.667 614 365 715 053 281 746 804 714 202 880 859 395 52;
  • 6) 0.667 614 365 715 053 281 746 804 714 202 880 859 395 52 × 2 = 1 + 0.335 228 731 430 106 563 493 609 428 405 761 718 791 04;
  • 7) 0.335 228 731 430 106 563 493 609 428 405 761 718 791 04 × 2 = 0 + 0.670 457 462 860 213 126 987 218 856 811 523 437 582 08;
  • 8) 0.670 457 462 860 213 126 987 218 856 811 523 437 582 08 × 2 = 1 + 0.340 914 925 720 426 253 974 437 713 623 046 875 164 16;
  • 9) 0.340 914 925 720 426 253 974 437 713 623 046 875 164 16 × 2 = 0 + 0.681 829 851 440 852 507 948 875 427 246 093 750 328 32;
  • 10) 0.681 829 851 440 852 507 948 875 427 246 093 750 328 32 × 2 = 1 + 0.363 659 702 881 705 015 897 750 854 492 187 500 656 64;
  • 11) 0.363 659 702 881 705 015 897 750 854 492 187 500 656 64 × 2 = 0 + 0.727 319 405 763 410 031 795 501 708 984 375 001 313 28;
  • 12) 0.727 319 405 763 410 031 795 501 708 984 375 001 313 28 × 2 = 1 + 0.454 638 811 526 820 063 591 003 417 968 750 002 626 56;
  • 13) 0.454 638 811 526 820 063 591 003 417 968 750 002 626 56 × 2 = 0 + 0.909 277 623 053 640 127 182 006 835 937 500 005 253 12;
  • 14) 0.909 277 623 053 640 127 182 006 835 937 500 005 253 12 × 2 = 1 + 0.818 555 246 107 280 254 364 013 671 875 000 010 506 24;
  • 15) 0.818 555 246 107 280 254 364 013 671 875 000 010 506 24 × 2 = 1 + 0.637 110 492 214 560 508 728 027 343 750 000 021 012 48;
  • 16) 0.637 110 492 214 560 508 728 027 343 750 000 021 012 48 × 2 = 1 + 0.274 220 984 429 121 017 456 054 687 500 000 042 024 96;
  • 17) 0.274 220 984 429 121 017 456 054 687 500 000 042 024 96 × 2 = 0 + 0.548 441 968 858 242 034 912 109 375 000 000 084 049 92;
  • 18) 0.548 441 968 858 242 034 912 109 375 000 000 084 049 92 × 2 = 1 + 0.096 883 937 716 484 069 824 218 750 000 000 168 099 84;
  • 19) 0.096 883 937 716 484 069 824 218 750 000 000 168 099 84 × 2 = 0 + 0.193 767 875 432 968 139 648 437 500 000 000 336 199 68;
  • 20) 0.193 767 875 432 968 139 648 437 500 000 000 336 199 68 × 2 = 0 + 0.387 535 750 865 936 279 296 875 000 000 000 672 399 36;
  • 21) 0.387 535 750 865 936 279 296 875 000 000 000 672 399 36 × 2 = 0 + 0.775 071 501 731 872 558 593 750 000 000 001 344 798 72;
  • 22) 0.775 071 501 731 872 558 593 750 000 000 001 344 798 72 × 2 = 1 + 0.550 143 003 463 745 117 187 500 000 000 002 689 597 44;
  • 23) 0.550 143 003 463 745 117 187 500 000 000 002 689 597 44 × 2 = 1 + 0.100 286 006 927 490 234 375 000 000 000 005 379 194 88;
  • 24) 0.100 286 006 927 490 234 375 000 000 000 005 379 194 88 × 2 = 0 + 0.200 572 013 854 980 468 750 000 000 000 010 758 389 76;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


4. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.395 862 948 928 595 415 054 587 647 318 840 026 856 11(10) =

0.0110 0101 0101 0111 0100 0110(2)

5. Positive number before normalization:

165.395 862 948 928 595 415 054 587 647 318 840 026 856 11(10) =

1010 0101.0110 0101 0101 0111 0100 0110(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 7 positions to the left, so that only one non zero digit remains to the left of it:


165.395 862 948 928 595 415 054 587 647 318 840 026 856 11(10) =

1010 0101.0110 0101 0101 0111 0100 0110(2) =

1010 0101.0110 0101 0101 0111 0100 0110(2) × 20 =

1.0100 1010 1100 1010 1010 1110 1000 110(2) × 27


7. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)

Exponent (unadjusted): 7

Mantissa (not normalized):
1.0100 1010 1100 1010 1010 1110 1000 110


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

7 + 2(8-1) - 1 =

(7 + 127)(10) =

134(10)


9. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 134 ÷ 2 = 67 + 0;
  • 67 ÷ 2 = 33 + 1;
  • 33 ÷ 2 = 16 + 1;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

134(10) =

1000 0110(2)


11. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 23 bits, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 010 0101 0110 0101 0101 0111 0100 0110 =

010 0101 0110 0101 0101 0111


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
1000 0110

Mantissa (23 bits) =
010 0101 0110 0101 0101 0111


Decimal number 165.395 862 948 928 595 415 054 587 647 318 840 026 856 11 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 1000 0110 - 010 0101 0110 0101 0101 0111

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111