-0.000 000 000 742 147 674 82 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 674 82(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
-0.000 000 000 742 147 674 82(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:

|-0.000 000 000 742 147 674 82| = 0.000 000 000 742 147 674 82


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: 0.000 000 000 742 147 674 82.

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) 0.000 000 000 742 147 674 82 × 2 = 0 + 0.000 000 001 484 295 349 64;
  • 2) 0.000 000 001 484 295 349 64 × 2 = 0 + 0.000 000 002 968 590 699 28;
  • 3) 0.000 000 002 968 590 699 28 × 2 = 0 + 0.000 000 005 937 181 398 56;
  • 4) 0.000 000 005 937 181 398 56 × 2 = 0 + 0.000 000 011 874 362 797 12;
  • 5) 0.000 000 011 874 362 797 12 × 2 = 0 + 0.000 000 023 748 725 594 24;
  • 6) 0.000 000 023 748 725 594 24 × 2 = 0 + 0.000 000 047 497 451 188 48;
  • 7) 0.000 000 047 497 451 188 48 × 2 = 0 + 0.000 000 094 994 902 376 96;
  • 8) 0.000 000 094 994 902 376 96 × 2 = 0 + 0.000 000 189 989 804 753 92;
  • 9) 0.000 000 189 989 804 753 92 × 2 = 0 + 0.000 000 379 979 609 507 84;
  • 10) 0.000 000 379 979 609 507 84 × 2 = 0 + 0.000 000 759 959 219 015 68;
  • 11) 0.000 000 759 959 219 015 68 × 2 = 0 + 0.000 001 519 918 438 031 36;
  • 12) 0.000 001 519 918 438 031 36 × 2 = 0 + 0.000 003 039 836 876 062 72;
  • 13) 0.000 003 039 836 876 062 72 × 2 = 0 + 0.000 006 079 673 752 125 44;
  • 14) 0.000 006 079 673 752 125 44 × 2 = 0 + 0.000 012 159 347 504 250 88;
  • 15) 0.000 012 159 347 504 250 88 × 2 = 0 + 0.000 024 318 695 008 501 76;
  • 16) 0.000 024 318 695 008 501 76 × 2 = 0 + 0.000 048 637 390 017 003 52;
  • 17) 0.000 048 637 390 017 003 52 × 2 = 0 + 0.000 097 274 780 034 007 04;
  • 18) 0.000 097 274 780 034 007 04 × 2 = 0 + 0.000 194 549 560 068 014 08;
  • 19) 0.000 194 549 560 068 014 08 × 2 = 0 + 0.000 389 099 120 136 028 16;
  • 20) 0.000 389 099 120 136 028 16 × 2 = 0 + 0.000 778 198 240 272 056 32;
  • 21) 0.000 778 198 240 272 056 32 × 2 = 0 + 0.001 556 396 480 544 112 64;
  • 22) 0.001 556 396 480 544 112 64 × 2 = 0 + 0.003 112 792 961 088 225 28;
  • 23) 0.003 112 792 961 088 225 28 × 2 = 0 + 0.006 225 585 922 176 450 56;
  • 24) 0.006 225 585 922 176 450 56 × 2 = 0 + 0.012 451 171 844 352 901 12;
  • 25) 0.012 451 171 844 352 901 12 × 2 = 0 + 0.024 902 343 688 705 802 24;
  • 26) 0.024 902 343 688 705 802 24 × 2 = 0 + 0.049 804 687 377 411 604 48;
  • 27) 0.049 804 687 377 411 604 48 × 2 = 0 + 0.099 609 374 754 823 208 96;
  • 28) 0.099 609 374 754 823 208 96 × 2 = 0 + 0.199 218 749 509 646 417 92;
  • 29) 0.199 218 749 509 646 417 92 × 2 = 0 + 0.398 437 499 019 292 835 84;
  • 30) 0.398 437 499 019 292 835 84 × 2 = 0 + 0.796 874 998 038 585 671 68;
  • 31) 0.796 874 998 038 585 671 68 × 2 = 1 + 0.593 749 996 077 171 343 36;
  • 32) 0.593 749 996 077 171 343 36 × 2 = 1 + 0.187 499 992 154 342 686 72;
  • 33) 0.187 499 992 154 342 686 72 × 2 = 0 + 0.374 999 984 308 685 373 44;
  • 34) 0.374 999 984 308 685 373 44 × 2 = 0 + 0.749 999 968 617 370 746 88;
  • 35) 0.749 999 968 617 370 746 88 × 2 = 1 + 0.499 999 937 234 741 493 76;
  • 36) 0.499 999 937 234 741 493 76 × 2 = 0 + 0.999 999 874 469 482 987 52;
  • 37) 0.999 999 874 469 482 987 52 × 2 = 1 + 0.999 999 748 938 965 975 04;
  • 38) 0.999 999 748 938 965 975 04 × 2 = 1 + 0.999 999 497 877 931 950 08;
  • 39) 0.999 999 497 877 931 950 08 × 2 = 1 + 0.999 998 995 755 863 900 16;
  • 40) 0.999 998 995 755 863 900 16 × 2 = 1 + 0.999 997 991 511 727 800 32;
  • 41) 0.999 997 991 511 727 800 32 × 2 = 1 + 0.999 995 983 023 455 600 64;
  • 42) 0.999 995 983 023 455 600 64 × 2 = 1 + 0.999 991 966 046 911 201 28;
  • 43) 0.999 991 966 046 911 201 28 × 2 = 1 + 0.999 983 932 093 822 402 56;
  • 44) 0.999 983 932 093 822 402 56 × 2 = 1 + 0.999 967 864 187 644 805 12;
  • 45) 0.999 967 864 187 644 805 12 × 2 = 1 + 0.999 935 728 375 289 610 24;
  • 46) 0.999 935 728 375 289 610 24 × 2 = 1 + 0.999 871 456 750 579 220 48;
  • 47) 0.999 871 456 750 579 220 48 × 2 = 1 + 0.999 742 913 501 158 440 96;
  • 48) 0.999 742 913 501 158 440 96 × 2 = 1 + 0.999 485 827 002 316 881 92;
  • 49) 0.999 485 827 002 316 881 92 × 2 = 1 + 0.998 971 654 004 633 763 84;
  • 50) 0.998 971 654 004 633 763 84 × 2 = 1 + 0.997 943 308 009 267 527 68;
  • 51) 0.997 943 308 009 267 527 68 × 2 = 1 + 0.995 886 616 018 535 055 36;
  • 52) 0.995 886 616 018 535 055 36 × 2 = 1 + 0.991 773 232 037 070 110 72;
  • 53) 0.991 773 232 037 070 110 72 × 2 = 1 + 0.983 546 464 074 140 221 44;
  • 54) 0.983 546 464 074 140 221 44 × 2 = 1 + 0.967 092 928 148 280 442 88;

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:


0.000 000 000 742 147 674 82(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 674 82(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:


0.000 000 000 742 147 674 82(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 -0.000 000 000 742 147 674 82 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