-0.000 000 000 742 147 676 14 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 14(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
-0.000 000 000 742 147 676 14(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. Start with the positive version of the number:

|-0.000 000 000 742 147 676 14| = 0.000 000 000 742 147 676 14


2. First, convert to binary (in base 2) the integer part: 0.
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;
  • 0 ÷ 2 = 0 + 0;

3. 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.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 742 147 676 14.

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) 0.000 000 000 742 147 676 14 × 2 = 0 + 0.000 000 001 484 295 352 28;
  • 2) 0.000 000 001 484 295 352 28 × 2 = 0 + 0.000 000 002 968 590 704 56;
  • 3) 0.000 000 002 968 590 704 56 × 2 = 0 + 0.000 000 005 937 181 409 12;
  • 4) 0.000 000 005 937 181 409 12 × 2 = 0 + 0.000 000 011 874 362 818 24;
  • 5) 0.000 000 011 874 362 818 24 × 2 = 0 + 0.000 000 023 748 725 636 48;
  • 6) 0.000 000 023 748 725 636 48 × 2 = 0 + 0.000 000 047 497 451 272 96;
  • 7) 0.000 000 047 497 451 272 96 × 2 = 0 + 0.000 000 094 994 902 545 92;
  • 8) 0.000 000 094 994 902 545 92 × 2 = 0 + 0.000 000 189 989 805 091 84;
  • 9) 0.000 000 189 989 805 091 84 × 2 = 0 + 0.000 000 379 979 610 183 68;
  • 10) 0.000 000 379 979 610 183 68 × 2 = 0 + 0.000 000 759 959 220 367 36;
  • 11) 0.000 000 759 959 220 367 36 × 2 = 0 + 0.000 001 519 918 440 734 72;
  • 12) 0.000 001 519 918 440 734 72 × 2 = 0 + 0.000 003 039 836 881 469 44;
  • 13) 0.000 003 039 836 881 469 44 × 2 = 0 + 0.000 006 079 673 762 938 88;
  • 14) 0.000 006 079 673 762 938 88 × 2 = 0 + 0.000 012 159 347 525 877 76;
  • 15) 0.000 012 159 347 525 877 76 × 2 = 0 + 0.000 024 318 695 051 755 52;
  • 16) 0.000 024 318 695 051 755 52 × 2 = 0 + 0.000 048 637 390 103 511 04;
  • 17) 0.000 048 637 390 103 511 04 × 2 = 0 + 0.000 097 274 780 207 022 08;
  • 18) 0.000 097 274 780 207 022 08 × 2 = 0 + 0.000 194 549 560 414 044 16;
  • 19) 0.000 194 549 560 414 044 16 × 2 = 0 + 0.000 389 099 120 828 088 32;
  • 20) 0.000 389 099 120 828 088 32 × 2 = 0 + 0.000 778 198 241 656 176 64;
  • 21) 0.000 778 198 241 656 176 64 × 2 = 0 + 0.001 556 396 483 312 353 28;
  • 22) 0.001 556 396 483 312 353 28 × 2 = 0 + 0.003 112 792 966 624 706 56;
  • 23) 0.003 112 792 966 624 706 56 × 2 = 0 + 0.006 225 585 933 249 413 12;
  • 24) 0.006 225 585 933 249 413 12 × 2 = 0 + 0.012 451 171 866 498 826 24;
  • 25) 0.012 451 171 866 498 826 24 × 2 = 0 + 0.024 902 343 732 997 652 48;
  • 26) 0.024 902 343 732 997 652 48 × 2 = 0 + 0.049 804 687 465 995 304 96;
  • 27) 0.049 804 687 465 995 304 96 × 2 = 0 + 0.099 609 374 931 990 609 92;
  • 28) 0.099 609 374 931 990 609 92 × 2 = 0 + 0.199 218 749 863 981 219 84;
  • 29) 0.199 218 749 863 981 219 84 × 2 = 0 + 0.398 437 499 727 962 439 68;
  • 30) 0.398 437 499 727 962 439 68 × 2 = 0 + 0.796 874 999 455 924 879 36;
  • 31) 0.796 874 999 455 924 879 36 × 2 = 1 + 0.593 749 998 911 849 758 72;
  • 32) 0.593 749 998 911 849 758 72 × 2 = 1 + 0.187 499 997 823 699 517 44;
  • 33) 0.187 499 997 823 699 517 44 × 2 = 0 + 0.374 999 995 647 399 034 88;
  • 34) 0.374 999 995 647 399 034 88 × 2 = 0 + 0.749 999 991 294 798 069 76;
  • 35) 0.749 999 991 294 798 069 76 × 2 = 1 + 0.499 999 982 589 596 139 52;
  • 36) 0.499 999 982 589 596 139 52 × 2 = 0 + 0.999 999 965 179 192 279 04;
  • 37) 0.999 999 965 179 192 279 04 × 2 = 1 + 0.999 999 930 358 384 558 08;
  • 38) 0.999 999 930 358 384 558 08 × 2 = 1 + 0.999 999 860 716 769 116 16;
  • 39) 0.999 999 860 716 769 116 16 × 2 = 1 + 0.999 999 721 433 538 232 32;
  • 40) 0.999 999 721 433 538 232 32 × 2 = 1 + 0.999 999 442 867 076 464 64;
  • 41) 0.999 999 442 867 076 464 64 × 2 = 1 + 0.999 998 885 734 152 929 28;
  • 42) 0.999 998 885 734 152 929 28 × 2 = 1 + 0.999 997 771 468 305 858 56;
  • 43) 0.999 997 771 468 305 858 56 × 2 = 1 + 0.999 995 542 936 611 717 12;
  • 44) 0.999 995 542 936 611 717 12 × 2 = 1 + 0.999 991 085 873 223 434 24;
  • 45) 0.999 991 085 873 223 434 24 × 2 = 1 + 0.999 982 171 746 446 868 48;
  • 46) 0.999 982 171 746 446 868 48 × 2 = 1 + 0.999 964 343 492 893 736 96;
  • 47) 0.999 964 343 492 893 736 96 × 2 = 1 + 0.999 928 686 985 787 473 92;
  • 48) 0.999 928 686 985 787 473 92 × 2 = 1 + 0.999 857 373 971 574 947 84;
  • 49) 0.999 857 373 971 574 947 84 × 2 = 1 + 0.999 714 747 943 149 895 68;
  • 50) 0.999 714 747 943 149 895 68 × 2 = 1 + 0.999 429 495 886 299 791 36;
  • 51) 0.999 429 495 886 299 791 36 × 2 = 1 + 0.998 858 991 772 599 582 72;
  • 52) 0.998 858 991 772 599 582 72 × 2 = 1 + 0.997 717 983 545 199 165 44;
  • 53) 0.997 717 983 545 199 165 44 × 2 = 1 + 0.995 435 967 090 398 330 88;
  • 54) 0.995 435 967 090 398 330 88 × 2 = 1 + 0.990 871 934 180 796 661 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).


5. 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:


0.000 000 000 742 147 676 14(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 14(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 742 147 676 14(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(2) × 2-31


8. 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 1 (a negative number)

Exponent (unadjusted): -31

Mantissa (not normalized):
1.1001 0111 1111 1111 1111 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(10)


10. 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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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

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


Exponent (adjusted) =

96(10) =

0110 0000(2)


12. 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, only if necessary (not the case here).


Mantissa (normalized) =

1. 100 1011 1111 1111 1111 1111 =

100 1011 1111 1111 1111 1111


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

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

Exponent (8 bits) =
0110 0000

Mantissa (23 bits) =
100 1011 1111 1111 1111 1111


Decimal number -0.000 000 000 742 147 676 14 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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