-0.000 000 000 742 147 676 646 415 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 415(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
-0.000 000 000 742 147 676 646 415(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:

|-0.000 000 000 742 147 676 646 415| = 0.000 000 000 742 147 676 646 415


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: 0.000 000 000 742 147 676 646 415.

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) 0.000 000 000 742 147 676 646 415 × 2 = 0 + 0.000 000 001 484 295 353 292 83;
  • 2) 0.000 000 001 484 295 353 292 83 × 2 = 0 + 0.000 000 002 968 590 706 585 66;
  • 3) 0.000 000 002 968 590 706 585 66 × 2 = 0 + 0.000 000 005 937 181 413 171 32;
  • 4) 0.000 000 005 937 181 413 171 32 × 2 = 0 + 0.000 000 011 874 362 826 342 64;
  • 5) 0.000 000 011 874 362 826 342 64 × 2 = 0 + 0.000 000 023 748 725 652 685 28;
  • 6) 0.000 000 023 748 725 652 685 28 × 2 = 0 + 0.000 000 047 497 451 305 370 56;
  • 7) 0.000 000 047 497 451 305 370 56 × 2 = 0 + 0.000 000 094 994 902 610 741 12;
  • 8) 0.000 000 094 994 902 610 741 12 × 2 = 0 + 0.000 000 189 989 805 221 482 24;
  • 9) 0.000 000 189 989 805 221 482 24 × 2 = 0 + 0.000 000 379 979 610 442 964 48;
  • 10) 0.000 000 379 979 610 442 964 48 × 2 = 0 + 0.000 000 759 959 220 885 928 96;
  • 11) 0.000 000 759 959 220 885 928 96 × 2 = 0 + 0.000 001 519 918 441 771 857 92;
  • 12) 0.000 001 519 918 441 771 857 92 × 2 = 0 + 0.000 003 039 836 883 543 715 84;
  • 13) 0.000 003 039 836 883 543 715 84 × 2 = 0 + 0.000 006 079 673 767 087 431 68;
  • 14) 0.000 006 079 673 767 087 431 68 × 2 = 0 + 0.000 012 159 347 534 174 863 36;
  • 15) 0.000 012 159 347 534 174 863 36 × 2 = 0 + 0.000 024 318 695 068 349 726 72;
  • 16) 0.000 024 318 695 068 349 726 72 × 2 = 0 + 0.000 048 637 390 136 699 453 44;
  • 17) 0.000 048 637 390 136 699 453 44 × 2 = 0 + 0.000 097 274 780 273 398 906 88;
  • 18) 0.000 097 274 780 273 398 906 88 × 2 = 0 + 0.000 194 549 560 546 797 813 76;
  • 19) 0.000 194 549 560 546 797 813 76 × 2 = 0 + 0.000 389 099 121 093 595 627 52;
  • 20) 0.000 389 099 121 093 595 627 52 × 2 = 0 + 0.000 778 198 242 187 191 255 04;
  • 21) 0.000 778 198 242 187 191 255 04 × 2 = 0 + 0.001 556 396 484 374 382 510 08;
  • 22) 0.001 556 396 484 374 382 510 08 × 2 = 0 + 0.003 112 792 968 748 765 020 16;
  • 23) 0.003 112 792 968 748 765 020 16 × 2 = 0 + 0.006 225 585 937 497 530 040 32;
  • 24) 0.006 225 585 937 497 530 040 32 × 2 = 0 + 0.012 451 171 874 995 060 080 64;
  • 25) 0.012 451 171 874 995 060 080 64 × 2 = 0 + 0.024 902 343 749 990 120 161 28;
  • 26) 0.024 902 343 749 990 120 161 28 × 2 = 0 + 0.049 804 687 499 980 240 322 56;
  • 27) 0.049 804 687 499 980 240 322 56 × 2 = 0 + 0.099 609 374 999 960 480 645 12;
  • 28) 0.099 609 374 999 960 480 645 12 × 2 = 0 + 0.199 218 749 999 920 961 290 24;
  • 29) 0.199 218 749 999 920 961 290 24 × 2 = 0 + 0.398 437 499 999 841 922 580 48;
  • 30) 0.398 437 499 999 841 922 580 48 × 2 = 0 + 0.796 874 999 999 683 845 160 96;
  • 31) 0.796 874 999 999 683 845 160 96 × 2 = 1 + 0.593 749 999 999 367 690 321 92;
  • 32) 0.593 749 999 999 367 690 321 92 × 2 = 1 + 0.187 499 999 998 735 380 643 84;
  • 33) 0.187 499 999 998 735 380 643 84 × 2 = 0 + 0.374 999 999 997 470 761 287 68;
  • 34) 0.374 999 999 997 470 761 287 68 × 2 = 0 + 0.749 999 999 994 941 522 575 36;
  • 35) 0.749 999 999 994 941 522 575 36 × 2 = 1 + 0.499 999 999 989 883 045 150 72;
  • 36) 0.499 999 999 989 883 045 150 72 × 2 = 0 + 0.999 999 999 979 766 090 301 44;
  • 37) 0.999 999 999 979 766 090 301 44 × 2 = 1 + 0.999 999 999 959 532 180 602 88;
  • 38) 0.999 999 999 959 532 180 602 88 × 2 = 1 + 0.999 999 999 919 064 361 205 76;
  • 39) 0.999 999 999 919 064 361 205 76 × 2 = 1 + 0.999 999 999 838 128 722 411 52;
  • 40) 0.999 999 999 838 128 722 411 52 × 2 = 1 + 0.999 999 999 676 257 444 823 04;
  • 41) 0.999 999 999 676 257 444 823 04 × 2 = 1 + 0.999 999 999 352 514 889 646 08;
  • 42) 0.999 999 999 352 514 889 646 08 × 2 = 1 + 0.999 999 998 705 029 779 292 16;
  • 43) 0.999 999 998 705 029 779 292 16 × 2 = 1 + 0.999 999 997 410 059 558 584 32;
  • 44) 0.999 999 997 410 059 558 584 32 × 2 = 1 + 0.999 999 994 820 119 117 168 64;
  • 45) 0.999 999 994 820 119 117 168 64 × 2 = 1 + 0.999 999 989 640 238 234 337 28;
  • 46) 0.999 999 989 640 238 234 337 28 × 2 = 1 + 0.999 999 979 280 476 468 674 56;
  • 47) 0.999 999 979 280 476 468 674 56 × 2 = 1 + 0.999 999 958 560 952 937 349 12;
  • 48) 0.999 999 958 560 952 937 349 12 × 2 = 1 + 0.999 999 917 121 905 874 698 24;
  • 49) 0.999 999 917 121 905 874 698 24 × 2 = 1 + 0.999 999 834 243 811 749 396 48;
  • 50) 0.999 999 834 243 811 749 396 48 × 2 = 1 + 0.999 999 668 487 623 498 792 96;
  • 51) 0.999 999 668 487 623 498 792 96 × 2 = 1 + 0.999 999 336 975 246 997 585 92;
  • 52) 0.999 999 336 975 246 997 585 92 × 2 = 1 + 0.999 998 673 950 493 995 171 84;
  • 53) 0.999 998 673 950 493 995 171 84 × 2 = 1 + 0.999 997 347 900 987 990 343 68;
  • 54) 0.999 997 347 900 987 990 343 68 × 2 = 1 + 0.999 994 695 801 975 980 687 36;

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:


0.000 000 000 742 147 676 646 415(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 415(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:


0.000 000 000 742 147 676 646 415(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 -0.000 000 000 742 147 676 646 415 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 646 374 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 minutes

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