1.969 740 035 797 184 500 623 368 800 833 883 27 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.969 740 035 797 184 500 623 368 800 833 883 27(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
1.969 740 035 797 184 500 623 368 800 833 883 27(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: 1.
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;
  • 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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.969 740 035 797 184 500 623 368 800 833 883 27.

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) 0.969 740 035 797 184 500 623 368 800 833 883 27 × 2 = 1 + 0.939 480 071 594 369 001 246 737 601 667 766 54;
  • 2) 0.939 480 071 594 369 001 246 737 601 667 766 54 × 2 = 1 + 0.878 960 143 188 738 002 493 475 203 335 533 08;
  • 3) 0.878 960 143 188 738 002 493 475 203 335 533 08 × 2 = 1 + 0.757 920 286 377 476 004 986 950 406 671 066 16;
  • 4) 0.757 920 286 377 476 004 986 950 406 671 066 16 × 2 = 1 + 0.515 840 572 754 952 009 973 900 813 342 132 32;
  • 5) 0.515 840 572 754 952 009 973 900 813 342 132 32 × 2 = 1 + 0.031 681 145 509 904 019 947 801 626 684 264 64;
  • 6) 0.031 681 145 509 904 019 947 801 626 684 264 64 × 2 = 0 + 0.063 362 291 019 808 039 895 603 253 368 529 28;
  • 7) 0.063 362 291 019 808 039 895 603 253 368 529 28 × 2 = 0 + 0.126 724 582 039 616 079 791 206 506 737 058 56;
  • 8) 0.126 724 582 039 616 079 791 206 506 737 058 56 × 2 = 0 + 0.253 449 164 079 232 159 582 413 013 474 117 12;
  • 9) 0.253 449 164 079 232 159 582 413 013 474 117 12 × 2 = 0 + 0.506 898 328 158 464 319 164 826 026 948 234 24;
  • 10) 0.506 898 328 158 464 319 164 826 026 948 234 24 × 2 = 1 + 0.013 796 656 316 928 638 329 652 053 896 468 48;
  • 11) 0.013 796 656 316 928 638 329 652 053 896 468 48 × 2 = 0 + 0.027 593 312 633 857 276 659 304 107 792 936 96;
  • 12) 0.027 593 312 633 857 276 659 304 107 792 936 96 × 2 = 0 + 0.055 186 625 267 714 553 318 608 215 585 873 92;
  • 13) 0.055 186 625 267 714 553 318 608 215 585 873 92 × 2 = 0 + 0.110 373 250 535 429 106 637 216 431 171 747 84;
  • 14) 0.110 373 250 535 429 106 637 216 431 171 747 84 × 2 = 0 + 0.220 746 501 070 858 213 274 432 862 343 495 68;
  • 15) 0.220 746 501 070 858 213 274 432 862 343 495 68 × 2 = 0 + 0.441 493 002 141 716 426 548 865 724 686 991 36;
  • 16) 0.441 493 002 141 716 426 548 865 724 686 991 36 × 2 = 0 + 0.882 986 004 283 432 853 097 731 449 373 982 72;
  • 17) 0.882 986 004 283 432 853 097 731 449 373 982 72 × 2 = 1 + 0.765 972 008 566 865 706 195 462 898 747 965 44;
  • 18) 0.765 972 008 566 865 706 195 462 898 747 965 44 × 2 = 1 + 0.531 944 017 133 731 412 390 925 797 495 930 88;
  • 19) 0.531 944 017 133 731 412 390 925 797 495 930 88 × 2 = 1 + 0.063 888 034 267 462 824 781 851 594 991 861 76;
  • 20) 0.063 888 034 267 462 824 781 851 594 991 861 76 × 2 = 0 + 0.127 776 068 534 925 649 563 703 189 983 723 52;
  • 21) 0.127 776 068 534 925 649 563 703 189 983 723 52 × 2 = 0 + 0.255 552 137 069 851 299 127 406 379 967 447 04;
  • 22) 0.255 552 137 069 851 299 127 406 379 967 447 04 × 2 = 0 + 0.511 104 274 139 702 598 254 812 759 934 894 08;
  • 23) 0.511 104 274 139 702 598 254 812 759 934 894 08 × 2 = 1 + 0.022 208 548 279 405 196 509 625 519 869 788 16;
  • 24) 0.022 208 548 279 405 196 509 625 519 869 788 16 × 2 = 0 + 0.044 417 096 558 810 393 019 251 039 739 576 32;

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:


0.969 740 035 797 184 500 623 368 800 833 883 27(10) =

0.1111 1000 0100 0000 1110 0010(2)

5. Positive number before normalization:

1.969 740 035 797 184 500 623 368 800 833 883 27(10) =

1.1111 1000 0100 0000 1110 0010(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.969 740 035 797 184 500 623 368 800 833 883 27(10) =

1.1111 1000 0100 0000 1110 0010(2) =

1.1111 1000 0100 0000 1110 0010(2) × 20


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): 0

Mantissa (not normalized):
1.1111 1000 0100 0000 1110 0010


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

0 + 2(8-1) - 1 =

(0 + 127)(10) =

127(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;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

127(10) =

0111 1111(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. 111 1100 0010 0000 0111 0001 0 =

111 1100 0010 0000 0111 0001


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) =
0111 1111

Mantissa (23 bits) =
111 1100 0010 0000 0111 0001


Decimal number 1.969 740 035 797 184 500 623 368 800 833 883 27 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0111 1111 - 111 1100 0010 0000 0111 0001

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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