-0.000 000 000 742 147 676 639 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 639 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
-0.000 000 000 742 147 676 639 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:

|-0.000 000 000 742 147 676 639 3| = 0.000 000 000 742 147 676 639 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: 0.000 000 000 742 147 676 639 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) 0.000 000 000 742 147 676 639 3 × 2 = 0 + 0.000 000 001 484 295 353 278 6;
  • 2) 0.000 000 001 484 295 353 278 6 × 2 = 0 + 0.000 000 002 968 590 706 557 2;
  • 3) 0.000 000 002 968 590 706 557 2 × 2 = 0 + 0.000 000 005 937 181 413 114 4;
  • 4) 0.000 000 005 937 181 413 114 4 × 2 = 0 + 0.000 000 011 874 362 826 228 8;
  • 5) 0.000 000 011 874 362 826 228 8 × 2 = 0 + 0.000 000 023 748 725 652 457 6;
  • 6) 0.000 000 023 748 725 652 457 6 × 2 = 0 + 0.000 000 047 497 451 304 915 2;
  • 7) 0.000 000 047 497 451 304 915 2 × 2 = 0 + 0.000 000 094 994 902 609 830 4;
  • 8) 0.000 000 094 994 902 609 830 4 × 2 = 0 + 0.000 000 189 989 805 219 660 8;
  • 9) 0.000 000 189 989 805 219 660 8 × 2 = 0 + 0.000 000 379 979 610 439 321 6;
  • 10) 0.000 000 379 979 610 439 321 6 × 2 = 0 + 0.000 000 759 959 220 878 643 2;
  • 11) 0.000 000 759 959 220 878 643 2 × 2 = 0 + 0.000 001 519 918 441 757 286 4;
  • 12) 0.000 001 519 918 441 757 286 4 × 2 = 0 + 0.000 003 039 836 883 514 572 8;
  • 13) 0.000 003 039 836 883 514 572 8 × 2 = 0 + 0.000 006 079 673 767 029 145 6;
  • 14) 0.000 006 079 673 767 029 145 6 × 2 = 0 + 0.000 012 159 347 534 058 291 2;
  • 15) 0.000 012 159 347 534 058 291 2 × 2 = 0 + 0.000 024 318 695 068 116 582 4;
  • 16) 0.000 024 318 695 068 116 582 4 × 2 = 0 + 0.000 048 637 390 136 233 164 8;
  • 17) 0.000 048 637 390 136 233 164 8 × 2 = 0 + 0.000 097 274 780 272 466 329 6;
  • 18) 0.000 097 274 780 272 466 329 6 × 2 = 0 + 0.000 194 549 560 544 932 659 2;
  • 19) 0.000 194 549 560 544 932 659 2 × 2 = 0 + 0.000 389 099 121 089 865 318 4;
  • 20) 0.000 389 099 121 089 865 318 4 × 2 = 0 + 0.000 778 198 242 179 730 636 8;
  • 21) 0.000 778 198 242 179 730 636 8 × 2 = 0 + 0.001 556 396 484 359 461 273 6;
  • 22) 0.001 556 396 484 359 461 273 6 × 2 = 0 + 0.003 112 792 968 718 922 547 2;
  • 23) 0.003 112 792 968 718 922 547 2 × 2 = 0 + 0.006 225 585 937 437 845 094 4;
  • 24) 0.006 225 585 937 437 845 094 4 × 2 = 0 + 0.012 451 171 874 875 690 188 8;
  • 25) 0.012 451 171 874 875 690 188 8 × 2 = 0 + 0.024 902 343 749 751 380 377 6;
  • 26) 0.024 902 343 749 751 380 377 6 × 2 = 0 + 0.049 804 687 499 502 760 755 2;
  • 27) 0.049 804 687 499 502 760 755 2 × 2 = 0 + 0.099 609 374 999 005 521 510 4;
  • 28) 0.099 609 374 999 005 521 510 4 × 2 = 0 + 0.199 218 749 998 011 043 020 8;
  • 29) 0.199 218 749 998 011 043 020 8 × 2 = 0 + 0.398 437 499 996 022 086 041 6;
  • 30) 0.398 437 499 996 022 086 041 6 × 2 = 0 + 0.796 874 999 992 044 172 083 2;
  • 31) 0.796 874 999 992 044 172 083 2 × 2 = 1 + 0.593 749 999 984 088 344 166 4;
  • 32) 0.593 749 999 984 088 344 166 4 × 2 = 1 + 0.187 499 999 968 176 688 332 8;
  • 33) 0.187 499 999 968 176 688 332 8 × 2 = 0 + 0.374 999 999 936 353 376 665 6;
  • 34) 0.374 999 999 936 353 376 665 6 × 2 = 0 + 0.749 999 999 872 706 753 331 2;
  • 35) 0.749 999 999 872 706 753 331 2 × 2 = 1 + 0.499 999 999 745 413 506 662 4;
  • 36) 0.499 999 999 745 413 506 662 4 × 2 = 0 + 0.999 999 999 490 827 013 324 8;
  • 37) 0.999 999 999 490 827 013 324 8 × 2 = 1 + 0.999 999 998 981 654 026 649 6;
  • 38) 0.999 999 998 981 654 026 649 6 × 2 = 1 + 0.999 999 997 963 308 053 299 2;
  • 39) 0.999 999 997 963 308 053 299 2 × 2 = 1 + 0.999 999 995 926 616 106 598 4;
  • 40) 0.999 999 995 926 616 106 598 4 × 2 = 1 + 0.999 999 991 853 232 213 196 8;
  • 41) 0.999 999 991 853 232 213 196 8 × 2 = 1 + 0.999 999 983 706 464 426 393 6;
  • 42) 0.999 999 983 706 464 426 393 6 × 2 = 1 + 0.999 999 967 412 928 852 787 2;
  • 43) 0.999 999 967 412 928 852 787 2 × 2 = 1 + 0.999 999 934 825 857 705 574 4;
  • 44) 0.999 999 934 825 857 705 574 4 × 2 = 1 + 0.999 999 869 651 715 411 148 8;
  • 45) 0.999 999 869 651 715 411 148 8 × 2 = 1 + 0.999 999 739 303 430 822 297 6;
  • 46) 0.999 999 739 303 430 822 297 6 × 2 = 1 + 0.999 999 478 606 861 644 595 2;
  • 47) 0.999 999 478 606 861 644 595 2 × 2 = 1 + 0.999 998 957 213 723 289 190 4;
  • 48) 0.999 998 957 213 723 289 190 4 × 2 = 1 + 0.999 997 914 427 446 578 380 8;
  • 49) 0.999 997 914 427 446 578 380 8 × 2 = 1 + 0.999 995 828 854 893 156 761 6;
  • 50) 0.999 995 828 854 893 156 761 6 × 2 = 1 + 0.999 991 657 709 786 313 523 2;
  • 51) 0.999 991 657 709 786 313 523 2 × 2 = 1 + 0.999 983 315 419 572 627 046 4;
  • 52) 0.999 983 315 419 572 627 046 4 × 2 = 1 + 0.999 966 630 839 145 254 092 8;
  • 53) 0.999 966 630 839 145 254 092 8 × 2 = 1 + 0.999 933 261 678 290 508 185 6;
  • 54) 0.999 933 261 678 290 508 185 6 × 2 = 1 + 0.999 866 523 356 581 016 371 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:


0.000 000 000 742 147 676 639 3(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 639 3(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:


0.000 000 000 742 147 676 639 3(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 -0.000 000 000 742 147 676 639 3 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