-0.000 000 000 742 147 676 646 62 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 62(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 62(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 62| = 0.000 000 000 742 147 676 646 62


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 62.

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 62 × 2 = 0 + 0.000 000 001 484 295 353 293 24;
  • 2) 0.000 000 001 484 295 353 293 24 × 2 = 0 + 0.000 000 002 968 590 706 586 48;
  • 3) 0.000 000 002 968 590 706 586 48 × 2 = 0 + 0.000 000 005 937 181 413 172 96;
  • 4) 0.000 000 005 937 181 413 172 96 × 2 = 0 + 0.000 000 011 874 362 826 345 92;
  • 5) 0.000 000 011 874 362 826 345 92 × 2 = 0 + 0.000 000 023 748 725 652 691 84;
  • 6) 0.000 000 023 748 725 652 691 84 × 2 = 0 + 0.000 000 047 497 451 305 383 68;
  • 7) 0.000 000 047 497 451 305 383 68 × 2 = 0 + 0.000 000 094 994 902 610 767 36;
  • 8) 0.000 000 094 994 902 610 767 36 × 2 = 0 + 0.000 000 189 989 805 221 534 72;
  • 9) 0.000 000 189 989 805 221 534 72 × 2 = 0 + 0.000 000 379 979 610 443 069 44;
  • 10) 0.000 000 379 979 610 443 069 44 × 2 = 0 + 0.000 000 759 959 220 886 138 88;
  • 11) 0.000 000 759 959 220 886 138 88 × 2 = 0 + 0.000 001 519 918 441 772 277 76;
  • 12) 0.000 001 519 918 441 772 277 76 × 2 = 0 + 0.000 003 039 836 883 544 555 52;
  • 13) 0.000 003 039 836 883 544 555 52 × 2 = 0 + 0.000 006 079 673 767 089 111 04;
  • 14) 0.000 006 079 673 767 089 111 04 × 2 = 0 + 0.000 012 159 347 534 178 222 08;
  • 15) 0.000 012 159 347 534 178 222 08 × 2 = 0 + 0.000 024 318 695 068 356 444 16;
  • 16) 0.000 024 318 695 068 356 444 16 × 2 = 0 + 0.000 048 637 390 136 712 888 32;
  • 17) 0.000 048 637 390 136 712 888 32 × 2 = 0 + 0.000 097 274 780 273 425 776 64;
  • 18) 0.000 097 274 780 273 425 776 64 × 2 = 0 + 0.000 194 549 560 546 851 553 28;
  • 19) 0.000 194 549 560 546 851 553 28 × 2 = 0 + 0.000 389 099 121 093 703 106 56;
  • 20) 0.000 389 099 121 093 703 106 56 × 2 = 0 + 0.000 778 198 242 187 406 213 12;
  • 21) 0.000 778 198 242 187 406 213 12 × 2 = 0 + 0.001 556 396 484 374 812 426 24;
  • 22) 0.001 556 396 484 374 812 426 24 × 2 = 0 + 0.003 112 792 968 749 624 852 48;
  • 23) 0.003 112 792 968 749 624 852 48 × 2 = 0 + 0.006 225 585 937 499 249 704 96;
  • 24) 0.006 225 585 937 499 249 704 96 × 2 = 0 + 0.012 451 171 874 998 499 409 92;
  • 25) 0.012 451 171 874 998 499 409 92 × 2 = 0 + 0.024 902 343 749 996 998 819 84;
  • 26) 0.024 902 343 749 996 998 819 84 × 2 = 0 + 0.049 804 687 499 993 997 639 68;
  • 27) 0.049 804 687 499 993 997 639 68 × 2 = 0 + 0.099 609 374 999 987 995 279 36;
  • 28) 0.099 609 374 999 987 995 279 36 × 2 = 0 + 0.199 218 749 999 975 990 558 72;
  • 29) 0.199 218 749 999 975 990 558 72 × 2 = 0 + 0.398 437 499 999 951 981 117 44;
  • 30) 0.398 437 499 999 951 981 117 44 × 2 = 0 + 0.796 874 999 999 903 962 234 88;
  • 31) 0.796 874 999 999 903 962 234 88 × 2 = 1 + 0.593 749 999 999 807 924 469 76;
  • 32) 0.593 749 999 999 807 924 469 76 × 2 = 1 + 0.187 499 999 999 615 848 939 52;
  • 33) 0.187 499 999 999 615 848 939 52 × 2 = 0 + 0.374 999 999 999 231 697 879 04;
  • 34) 0.374 999 999 999 231 697 879 04 × 2 = 0 + 0.749 999 999 998 463 395 758 08;
  • 35) 0.749 999 999 998 463 395 758 08 × 2 = 1 + 0.499 999 999 996 926 791 516 16;
  • 36) 0.499 999 999 996 926 791 516 16 × 2 = 0 + 0.999 999 999 993 853 583 032 32;
  • 37) 0.999 999 999 993 853 583 032 32 × 2 = 1 + 0.999 999 999 987 707 166 064 64;
  • 38) 0.999 999 999 987 707 166 064 64 × 2 = 1 + 0.999 999 999 975 414 332 129 28;
  • 39) 0.999 999 999 975 414 332 129 28 × 2 = 1 + 0.999 999 999 950 828 664 258 56;
  • 40) 0.999 999 999 950 828 664 258 56 × 2 = 1 + 0.999 999 999 901 657 328 517 12;
  • 41) 0.999 999 999 901 657 328 517 12 × 2 = 1 + 0.999 999 999 803 314 657 034 24;
  • 42) 0.999 999 999 803 314 657 034 24 × 2 = 1 + 0.999 999 999 606 629 314 068 48;
  • 43) 0.999 999 999 606 629 314 068 48 × 2 = 1 + 0.999 999 999 213 258 628 136 96;
  • 44) 0.999 999 999 213 258 628 136 96 × 2 = 1 + 0.999 999 998 426 517 256 273 92;
  • 45) 0.999 999 998 426 517 256 273 92 × 2 = 1 + 0.999 999 996 853 034 512 547 84;
  • 46) 0.999 999 996 853 034 512 547 84 × 2 = 1 + 0.999 999 993 706 069 025 095 68;
  • 47) 0.999 999 993 706 069 025 095 68 × 2 = 1 + 0.999 999 987 412 138 050 191 36;
  • 48) 0.999 999 987 412 138 050 191 36 × 2 = 1 + 0.999 999 974 824 276 100 382 72;
  • 49) 0.999 999 974 824 276 100 382 72 × 2 = 1 + 0.999 999 949 648 552 200 765 44;
  • 50) 0.999 999 949 648 552 200 765 44 × 2 = 1 + 0.999 999 899 297 104 401 530 88;
  • 51) 0.999 999 899 297 104 401 530 88 × 2 = 1 + 0.999 999 798 594 208 803 061 76;
  • 52) 0.999 999 798 594 208 803 061 76 × 2 = 1 + 0.999 999 597 188 417 606 123 52;
  • 53) 0.999 999 597 188 417 606 123 52 × 2 = 1 + 0.999 999 194 376 835 212 247 04;
  • 54) 0.999 999 194 376 835 212 247 04 × 2 = 1 + 0.999 998 388 753 670 424 494 08;

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 62(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 62(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:


0.000 000 000 742 147 676 646 62(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 -0.000 000 000 742 147 676 646 62 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