-0.000 000 000 742 147 676 646 17 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 17(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 17(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 17| = 0.000 000 000 742 147 676 646 17


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 17.

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 17 × 2 = 0 + 0.000 000 001 484 295 353 292 34;
  • 2) 0.000 000 001 484 295 353 292 34 × 2 = 0 + 0.000 000 002 968 590 706 584 68;
  • 3) 0.000 000 002 968 590 706 584 68 × 2 = 0 + 0.000 000 005 937 181 413 169 36;
  • 4) 0.000 000 005 937 181 413 169 36 × 2 = 0 + 0.000 000 011 874 362 826 338 72;
  • 5) 0.000 000 011 874 362 826 338 72 × 2 = 0 + 0.000 000 023 748 725 652 677 44;
  • 6) 0.000 000 023 748 725 652 677 44 × 2 = 0 + 0.000 000 047 497 451 305 354 88;
  • 7) 0.000 000 047 497 451 305 354 88 × 2 = 0 + 0.000 000 094 994 902 610 709 76;
  • 8) 0.000 000 094 994 902 610 709 76 × 2 = 0 + 0.000 000 189 989 805 221 419 52;
  • 9) 0.000 000 189 989 805 221 419 52 × 2 = 0 + 0.000 000 379 979 610 442 839 04;
  • 10) 0.000 000 379 979 610 442 839 04 × 2 = 0 + 0.000 000 759 959 220 885 678 08;
  • 11) 0.000 000 759 959 220 885 678 08 × 2 = 0 + 0.000 001 519 918 441 771 356 16;
  • 12) 0.000 001 519 918 441 771 356 16 × 2 = 0 + 0.000 003 039 836 883 542 712 32;
  • 13) 0.000 003 039 836 883 542 712 32 × 2 = 0 + 0.000 006 079 673 767 085 424 64;
  • 14) 0.000 006 079 673 767 085 424 64 × 2 = 0 + 0.000 012 159 347 534 170 849 28;
  • 15) 0.000 012 159 347 534 170 849 28 × 2 = 0 + 0.000 024 318 695 068 341 698 56;
  • 16) 0.000 024 318 695 068 341 698 56 × 2 = 0 + 0.000 048 637 390 136 683 397 12;
  • 17) 0.000 048 637 390 136 683 397 12 × 2 = 0 + 0.000 097 274 780 273 366 794 24;
  • 18) 0.000 097 274 780 273 366 794 24 × 2 = 0 + 0.000 194 549 560 546 733 588 48;
  • 19) 0.000 194 549 560 546 733 588 48 × 2 = 0 + 0.000 389 099 121 093 467 176 96;
  • 20) 0.000 389 099 121 093 467 176 96 × 2 = 0 + 0.000 778 198 242 186 934 353 92;
  • 21) 0.000 778 198 242 186 934 353 92 × 2 = 0 + 0.001 556 396 484 373 868 707 84;
  • 22) 0.001 556 396 484 373 868 707 84 × 2 = 0 + 0.003 112 792 968 747 737 415 68;
  • 23) 0.003 112 792 968 747 737 415 68 × 2 = 0 + 0.006 225 585 937 495 474 831 36;
  • 24) 0.006 225 585 937 495 474 831 36 × 2 = 0 + 0.012 451 171 874 990 949 662 72;
  • 25) 0.012 451 171 874 990 949 662 72 × 2 = 0 + 0.024 902 343 749 981 899 325 44;
  • 26) 0.024 902 343 749 981 899 325 44 × 2 = 0 + 0.049 804 687 499 963 798 650 88;
  • 27) 0.049 804 687 499 963 798 650 88 × 2 = 0 + 0.099 609 374 999 927 597 301 76;
  • 28) 0.099 609 374 999 927 597 301 76 × 2 = 0 + 0.199 218 749 999 855 194 603 52;
  • 29) 0.199 218 749 999 855 194 603 52 × 2 = 0 + 0.398 437 499 999 710 389 207 04;
  • 30) 0.398 437 499 999 710 389 207 04 × 2 = 0 + 0.796 874 999 999 420 778 414 08;
  • 31) 0.796 874 999 999 420 778 414 08 × 2 = 1 + 0.593 749 999 998 841 556 828 16;
  • 32) 0.593 749 999 998 841 556 828 16 × 2 = 1 + 0.187 499 999 997 683 113 656 32;
  • 33) 0.187 499 999 997 683 113 656 32 × 2 = 0 + 0.374 999 999 995 366 227 312 64;
  • 34) 0.374 999 999 995 366 227 312 64 × 2 = 0 + 0.749 999 999 990 732 454 625 28;
  • 35) 0.749 999 999 990 732 454 625 28 × 2 = 1 + 0.499 999 999 981 464 909 250 56;
  • 36) 0.499 999 999 981 464 909 250 56 × 2 = 0 + 0.999 999 999 962 929 818 501 12;
  • 37) 0.999 999 999 962 929 818 501 12 × 2 = 1 + 0.999 999 999 925 859 637 002 24;
  • 38) 0.999 999 999 925 859 637 002 24 × 2 = 1 + 0.999 999 999 851 719 274 004 48;
  • 39) 0.999 999 999 851 719 274 004 48 × 2 = 1 + 0.999 999 999 703 438 548 008 96;
  • 40) 0.999 999 999 703 438 548 008 96 × 2 = 1 + 0.999 999 999 406 877 096 017 92;
  • 41) 0.999 999 999 406 877 096 017 92 × 2 = 1 + 0.999 999 998 813 754 192 035 84;
  • 42) 0.999 999 998 813 754 192 035 84 × 2 = 1 + 0.999 999 997 627 508 384 071 68;
  • 43) 0.999 999 997 627 508 384 071 68 × 2 = 1 + 0.999 999 995 255 016 768 143 36;
  • 44) 0.999 999 995 255 016 768 143 36 × 2 = 1 + 0.999 999 990 510 033 536 286 72;
  • 45) 0.999 999 990 510 033 536 286 72 × 2 = 1 + 0.999 999 981 020 067 072 573 44;
  • 46) 0.999 999 981 020 067 072 573 44 × 2 = 1 + 0.999 999 962 040 134 145 146 88;
  • 47) 0.999 999 962 040 134 145 146 88 × 2 = 1 + 0.999 999 924 080 268 290 293 76;
  • 48) 0.999 999 924 080 268 290 293 76 × 2 = 1 + 0.999 999 848 160 536 580 587 52;
  • 49) 0.999 999 848 160 536 580 587 52 × 2 = 1 + 0.999 999 696 321 073 161 175 04;
  • 50) 0.999 999 696 321 073 161 175 04 × 2 = 1 + 0.999 999 392 642 146 322 350 08;
  • 51) 0.999 999 392 642 146 322 350 08 × 2 = 1 + 0.999 998 785 284 292 644 700 16;
  • 52) 0.999 998 785 284 292 644 700 16 × 2 = 1 + 0.999 997 570 568 585 289 400 32;
  • 53) 0.999 997 570 568 585 289 400 32 × 2 = 1 + 0.999 995 141 137 170 578 800 64;
  • 54) 0.999 995 141 137 170 578 800 64 × 2 = 1 + 0.999 990 282 274 341 157 601 28;

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 17(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 17(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 17(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 17 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