-0.000 000 000 742 147 676 646 690 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 690 6(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
-0.000 000 000 742 147 676 646 690 6(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:

|-0.000 000 000 742 147 676 646 690 6| = 0.000 000 000 742 147 676 646 690 6


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: 0.000 000 000 742 147 676 646 690 6.

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) 0.000 000 000 742 147 676 646 690 6 × 2 = 0 + 0.000 000 001 484 295 353 293 381 2;
  • 2) 0.000 000 001 484 295 353 293 381 2 × 2 = 0 + 0.000 000 002 968 590 706 586 762 4;
  • 3) 0.000 000 002 968 590 706 586 762 4 × 2 = 0 + 0.000 000 005 937 181 413 173 524 8;
  • 4) 0.000 000 005 937 181 413 173 524 8 × 2 = 0 + 0.000 000 011 874 362 826 347 049 6;
  • 5) 0.000 000 011 874 362 826 347 049 6 × 2 = 0 + 0.000 000 023 748 725 652 694 099 2;
  • 6) 0.000 000 023 748 725 652 694 099 2 × 2 = 0 + 0.000 000 047 497 451 305 388 198 4;
  • 7) 0.000 000 047 497 451 305 388 198 4 × 2 = 0 + 0.000 000 094 994 902 610 776 396 8;
  • 8) 0.000 000 094 994 902 610 776 396 8 × 2 = 0 + 0.000 000 189 989 805 221 552 793 6;
  • 9) 0.000 000 189 989 805 221 552 793 6 × 2 = 0 + 0.000 000 379 979 610 443 105 587 2;
  • 10) 0.000 000 379 979 610 443 105 587 2 × 2 = 0 + 0.000 000 759 959 220 886 211 174 4;
  • 11) 0.000 000 759 959 220 886 211 174 4 × 2 = 0 + 0.000 001 519 918 441 772 422 348 8;
  • 12) 0.000 001 519 918 441 772 422 348 8 × 2 = 0 + 0.000 003 039 836 883 544 844 697 6;
  • 13) 0.000 003 039 836 883 544 844 697 6 × 2 = 0 + 0.000 006 079 673 767 089 689 395 2;
  • 14) 0.000 006 079 673 767 089 689 395 2 × 2 = 0 + 0.000 012 159 347 534 179 378 790 4;
  • 15) 0.000 012 159 347 534 179 378 790 4 × 2 = 0 + 0.000 024 318 695 068 358 757 580 8;
  • 16) 0.000 024 318 695 068 358 757 580 8 × 2 = 0 + 0.000 048 637 390 136 717 515 161 6;
  • 17) 0.000 048 637 390 136 717 515 161 6 × 2 = 0 + 0.000 097 274 780 273 435 030 323 2;
  • 18) 0.000 097 274 780 273 435 030 323 2 × 2 = 0 + 0.000 194 549 560 546 870 060 646 4;
  • 19) 0.000 194 549 560 546 870 060 646 4 × 2 = 0 + 0.000 389 099 121 093 740 121 292 8;
  • 20) 0.000 389 099 121 093 740 121 292 8 × 2 = 0 + 0.000 778 198 242 187 480 242 585 6;
  • 21) 0.000 778 198 242 187 480 242 585 6 × 2 = 0 + 0.001 556 396 484 374 960 485 171 2;
  • 22) 0.001 556 396 484 374 960 485 171 2 × 2 = 0 + 0.003 112 792 968 749 920 970 342 4;
  • 23) 0.003 112 792 968 749 920 970 342 4 × 2 = 0 + 0.006 225 585 937 499 841 940 684 8;
  • 24) 0.006 225 585 937 499 841 940 684 8 × 2 = 0 + 0.012 451 171 874 999 683 881 369 6;
  • 25) 0.012 451 171 874 999 683 881 369 6 × 2 = 0 + 0.024 902 343 749 999 367 762 739 2;
  • 26) 0.024 902 343 749 999 367 762 739 2 × 2 = 0 + 0.049 804 687 499 998 735 525 478 4;
  • 27) 0.049 804 687 499 998 735 525 478 4 × 2 = 0 + 0.099 609 374 999 997 471 050 956 8;
  • 28) 0.099 609 374 999 997 471 050 956 8 × 2 = 0 + 0.199 218 749 999 994 942 101 913 6;
  • 29) 0.199 218 749 999 994 942 101 913 6 × 2 = 0 + 0.398 437 499 999 989 884 203 827 2;
  • 30) 0.398 437 499 999 989 884 203 827 2 × 2 = 0 + 0.796 874 999 999 979 768 407 654 4;
  • 31) 0.796 874 999 999 979 768 407 654 4 × 2 = 1 + 0.593 749 999 999 959 536 815 308 8;
  • 32) 0.593 749 999 999 959 536 815 308 8 × 2 = 1 + 0.187 499 999 999 919 073 630 617 6;
  • 33) 0.187 499 999 999 919 073 630 617 6 × 2 = 0 + 0.374 999 999 999 838 147 261 235 2;
  • 34) 0.374 999 999 999 838 147 261 235 2 × 2 = 0 + 0.749 999 999 999 676 294 522 470 4;
  • 35) 0.749 999 999 999 676 294 522 470 4 × 2 = 1 + 0.499 999 999 999 352 589 044 940 8;
  • 36) 0.499 999 999 999 352 589 044 940 8 × 2 = 0 + 0.999 999 999 998 705 178 089 881 6;
  • 37) 0.999 999 999 998 705 178 089 881 6 × 2 = 1 + 0.999 999 999 997 410 356 179 763 2;
  • 38) 0.999 999 999 997 410 356 179 763 2 × 2 = 1 + 0.999 999 999 994 820 712 359 526 4;
  • 39) 0.999 999 999 994 820 712 359 526 4 × 2 = 1 + 0.999 999 999 989 641 424 719 052 8;
  • 40) 0.999 999 999 989 641 424 719 052 8 × 2 = 1 + 0.999 999 999 979 282 849 438 105 6;
  • 41) 0.999 999 999 979 282 849 438 105 6 × 2 = 1 + 0.999 999 999 958 565 698 876 211 2;
  • 42) 0.999 999 999 958 565 698 876 211 2 × 2 = 1 + 0.999 999 999 917 131 397 752 422 4;
  • 43) 0.999 999 999 917 131 397 752 422 4 × 2 = 1 + 0.999 999 999 834 262 795 504 844 8;
  • 44) 0.999 999 999 834 262 795 504 844 8 × 2 = 1 + 0.999 999 999 668 525 591 009 689 6;
  • 45) 0.999 999 999 668 525 591 009 689 6 × 2 = 1 + 0.999 999 999 337 051 182 019 379 2;
  • 46) 0.999 999 999 337 051 182 019 379 2 × 2 = 1 + 0.999 999 998 674 102 364 038 758 4;
  • 47) 0.999 999 998 674 102 364 038 758 4 × 2 = 1 + 0.999 999 997 348 204 728 077 516 8;
  • 48) 0.999 999 997 348 204 728 077 516 8 × 2 = 1 + 0.999 999 994 696 409 456 155 033 6;
  • 49) 0.999 999 994 696 409 456 155 033 6 × 2 = 1 + 0.999 999 989 392 818 912 310 067 2;
  • 50) 0.999 999 989 392 818 912 310 067 2 × 2 = 1 + 0.999 999 978 785 637 824 620 134 4;
  • 51) 0.999 999 978 785 637 824 620 134 4 × 2 = 1 + 0.999 999 957 571 275 649 240 268 8;
  • 52) 0.999 999 957 571 275 649 240 268 8 × 2 = 1 + 0.999 999 915 142 551 298 480 537 6;
  • 53) 0.999 999 915 142 551 298 480 537 6 × 2 = 1 + 0.999 999 830 285 102 596 961 075 2;
  • 54) 0.999 999 830 285 102 596 961 075 2 × 2 = 1 + 0.999 999 660 570 205 193 922 150 4;

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:


0.000 000 000 742 147 676 646 690 6(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 690 6(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:


0.000 000 000 742 147 676 646 690 6(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 -0.000 000 000 742 147 676 646 690 6 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