-0.000 000 000 742 147 676 646 716 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 716 3(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
-0.000 000 000 742 147 676 646 716 3(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:

|-0.000 000 000 742 147 676 646 716 3| = 0.000 000 000 742 147 676 646 716 3


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: 0.000 000 000 742 147 676 646 716 3.

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) 0.000 000 000 742 147 676 646 716 3 × 2 = 0 + 0.000 000 001 484 295 353 293 432 6;
  • 2) 0.000 000 001 484 295 353 293 432 6 × 2 = 0 + 0.000 000 002 968 590 706 586 865 2;
  • 3) 0.000 000 002 968 590 706 586 865 2 × 2 = 0 + 0.000 000 005 937 181 413 173 730 4;
  • 4) 0.000 000 005 937 181 413 173 730 4 × 2 = 0 + 0.000 000 011 874 362 826 347 460 8;
  • 5) 0.000 000 011 874 362 826 347 460 8 × 2 = 0 + 0.000 000 023 748 725 652 694 921 6;
  • 6) 0.000 000 023 748 725 652 694 921 6 × 2 = 0 + 0.000 000 047 497 451 305 389 843 2;
  • 7) 0.000 000 047 497 451 305 389 843 2 × 2 = 0 + 0.000 000 094 994 902 610 779 686 4;
  • 8) 0.000 000 094 994 902 610 779 686 4 × 2 = 0 + 0.000 000 189 989 805 221 559 372 8;
  • 9) 0.000 000 189 989 805 221 559 372 8 × 2 = 0 + 0.000 000 379 979 610 443 118 745 6;
  • 10) 0.000 000 379 979 610 443 118 745 6 × 2 = 0 + 0.000 000 759 959 220 886 237 491 2;
  • 11) 0.000 000 759 959 220 886 237 491 2 × 2 = 0 + 0.000 001 519 918 441 772 474 982 4;
  • 12) 0.000 001 519 918 441 772 474 982 4 × 2 = 0 + 0.000 003 039 836 883 544 949 964 8;
  • 13) 0.000 003 039 836 883 544 949 964 8 × 2 = 0 + 0.000 006 079 673 767 089 899 929 6;
  • 14) 0.000 006 079 673 767 089 899 929 6 × 2 = 0 + 0.000 012 159 347 534 179 799 859 2;
  • 15) 0.000 012 159 347 534 179 799 859 2 × 2 = 0 + 0.000 024 318 695 068 359 599 718 4;
  • 16) 0.000 024 318 695 068 359 599 718 4 × 2 = 0 + 0.000 048 637 390 136 719 199 436 8;
  • 17) 0.000 048 637 390 136 719 199 436 8 × 2 = 0 + 0.000 097 274 780 273 438 398 873 6;
  • 18) 0.000 097 274 780 273 438 398 873 6 × 2 = 0 + 0.000 194 549 560 546 876 797 747 2;
  • 19) 0.000 194 549 560 546 876 797 747 2 × 2 = 0 + 0.000 389 099 121 093 753 595 494 4;
  • 20) 0.000 389 099 121 093 753 595 494 4 × 2 = 0 + 0.000 778 198 242 187 507 190 988 8;
  • 21) 0.000 778 198 242 187 507 190 988 8 × 2 = 0 + 0.001 556 396 484 375 014 381 977 6;
  • 22) 0.001 556 396 484 375 014 381 977 6 × 2 = 0 + 0.003 112 792 968 750 028 763 955 2;
  • 23) 0.003 112 792 968 750 028 763 955 2 × 2 = 0 + 0.006 225 585 937 500 057 527 910 4;
  • 24) 0.006 225 585 937 500 057 527 910 4 × 2 = 0 + 0.012 451 171 875 000 115 055 820 8;
  • 25) 0.012 451 171 875 000 115 055 820 8 × 2 = 0 + 0.024 902 343 750 000 230 111 641 6;
  • 26) 0.024 902 343 750 000 230 111 641 6 × 2 = 0 + 0.049 804 687 500 000 460 223 283 2;
  • 27) 0.049 804 687 500 000 460 223 283 2 × 2 = 0 + 0.099 609 375 000 000 920 446 566 4;
  • 28) 0.099 609 375 000 000 920 446 566 4 × 2 = 0 + 0.199 218 750 000 001 840 893 132 8;
  • 29) 0.199 218 750 000 001 840 893 132 8 × 2 = 0 + 0.398 437 500 000 003 681 786 265 6;
  • 30) 0.398 437 500 000 003 681 786 265 6 × 2 = 0 + 0.796 875 000 000 007 363 572 531 2;
  • 31) 0.796 875 000 000 007 363 572 531 2 × 2 = 1 + 0.593 750 000 000 014 727 145 062 4;
  • 32) 0.593 750 000 000 014 727 145 062 4 × 2 = 1 + 0.187 500 000 000 029 454 290 124 8;
  • 33) 0.187 500 000 000 029 454 290 124 8 × 2 = 0 + 0.375 000 000 000 058 908 580 249 6;
  • 34) 0.375 000 000 000 058 908 580 249 6 × 2 = 0 + 0.750 000 000 000 117 817 160 499 2;
  • 35) 0.750 000 000 000 117 817 160 499 2 × 2 = 1 + 0.500 000 000 000 235 634 320 998 4;
  • 36) 0.500 000 000 000 235 634 320 998 4 × 2 = 1 + 0.000 000 000 000 471 268 641 996 8;
  • 37) 0.000 000 000 000 471 268 641 996 8 × 2 = 0 + 0.000 000 000 000 942 537 283 993 6;
  • 38) 0.000 000 000 000 942 537 283 993 6 × 2 = 0 + 0.000 000 000 001 885 074 567 987 2;
  • 39) 0.000 000 000 001 885 074 567 987 2 × 2 = 0 + 0.000 000 000 003 770 149 135 974 4;
  • 40) 0.000 000 000 003 770 149 135 974 4 × 2 = 0 + 0.000 000 000 007 540 298 271 948 8;
  • 41) 0.000 000 000 007 540 298 271 948 8 × 2 = 0 + 0.000 000 000 015 080 596 543 897 6;
  • 42) 0.000 000 000 015 080 596 543 897 6 × 2 = 0 + 0.000 000 000 030 161 193 087 795 2;
  • 43) 0.000 000 000 030 161 193 087 795 2 × 2 = 0 + 0.000 000 000 060 322 386 175 590 4;
  • 44) 0.000 000 000 060 322 386 175 590 4 × 2 = 0 + 0.000 000 000 120 644 772 351 180 8;
  • 45) 0.000 000 000 120 644 772 351 180 8 × 2 = 0 + 0.000 000 000 241 289 544 702 361 6;
  • 46) 0.000 000 000 241 289 544 702 361 6 × 2 = 0 + 0.000 000 000 482 579 089 404 723 2;
  • 47) 0.000 000 000 482 579 089 404 723 2 × 2 = 0 + 0.000 000 000 965 158 178 809 446 4;
  • 48) 0.000 000 000 965 158 178 809 446 4 × 2 = 0 + 0.000 000 001 930 316 357 618 892 8;
  • 49) 0.000 000 001 930 316 357 618 892 8 × 2 = 0 + 0.000 000 003 860 632 715 237 785 6;
  • 50) 0.000 000 003 860 632 715 237 785 6 × 2 = 0 + 0.000 000 007 721 265 430 475 571 2;
  • 51) 0.000 000 007 721 265 430 475 571 2 × 2 = 0 + 0.000 000 015 442 530 860 951 142 4;
  • 52) 0.000 000 015 442 530 860 951 142 4 × 2 = 0 + 0.000 000 030 885 061 721 902 284 8;
  • 53) 0.000 000 030 885 061 721 902 284 8 × 2 = 0 + 0.000 000 061 770 123 443 804 569 6;
  • 54) 0.000 000 061 770 123 443 804 569 6 × 2 = 0 + 0.000 000 123 540 246 887 609 139 2;

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:


0.000 000 000 742 147 676 646 716 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 646 716 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 676 646 716 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 1000 0000 0000 0000 000


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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 676 646 716 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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