-0.000 000 000 742 147 676 357 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 357(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
-0.000 000 000 742 147 676 357(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:

|-0.000 000 000 742 147 676 357| = 0.000 000 000 742 147 676 357


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: 0.000 000 000 742 147 676 357.

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) 0.000 000 000 742 147 676 357 × 2 = 0 + 0.000 000 001 484 295 352 714;
  • 2) 0.000 000 001 484 295 352 714 × 2 = 0 + 0.000 000 002 968 590 705 428;
  • 3) 0.000 000 002 968 590 705 428 × 2 = 0 + 0.000 000 005 937 181 410 856;
  • 4) 0.000 000 005 937 181 410 856 × 2 = 0 + 0.000 000 011 874 362 821 712;
  • 5) 0.000 000 011 874 362 821 712 × 2 = 0 + 0.000 000 023 748 725 643 424;
  • 6) 0.000 000 023 748 725 643 424 × 2 = 0 + 0.000 000 047 497 451 286 848;
  • 7) 0.000 000 047 497 451 286 848 × 2 = 0 + 0.000 000 094 994 902 573 696;
  • 8) 0.000 000 094 994 902 573 696 × 2 = 0 + 0.000 000 189 989 805 147 392;
  • 9) 0.000 000 189 989 805 147 392 × 2 = 0 + 0.000 000 379 979 610 294 784;
  • 10) 0.000 000 379 979 610 294 784 × 2 = 0 + 0.000 000 759 959 220 589 568;
  • 11) 0.000 000 759 959 220 589 568 × 2 = 0 + 0.000 001 519 918 441 179 136;
  • 12) 0.000 001 519 918 441 179 136 × 2 = 0 + 0.000 003 039 836 882 358 272;
  • 13) 0.000 003 039 836 882 358 272 × 2 = 0 + 0.000 006 079 673 764 716 544;
  • 14) 0.000 006 079 673 764 716 544 × 2 = 0 + 0.000 012 159 347 529 433 088;
  • 15) 0.000 012 159 347 529 433 088 × 2 = 0 + 0.000 024 318 695 058 866 176;
  • 16) 0.000 024 318 695 058 866 176 × 2 = 0 + 0.000 048 637 390 117 732 352;
  • 17) 0.000 048 637 390 117 732 352 × 2 = 0 + 0.000 097 274 780 235 464 704;
  • 18) 0.000 097 274 780 235 464 704 × 2 = 0 + 0.000 194 549 560 470 929 408;
  • 19) 0.000 194 549 560 470 929 408 × 2 = 0 + 0.000 389 099 120 941 858 816;
  • 20) 0.000 389 099 120 941 858 816 × 2 = 0 + 0.000 778 198 241 883 717 632;
  • 21) 0.000 778 198 241 883 717 632 × 2 = 0 + 0.001 556 396 483 767 435 264;
  • 22) 0.001 556 396 483 767 435 264 × 2 = 0 + 0.003 112 792 967 534 870 528;
  • 23) 0.003 112 792 967 534 870 528 × 2 = 0 + 0.006 225 585 935 069 741 056;
  • 24) 0.006 225 585 935 069 741 056 × 2 = 0 + 0.012 451 171 870 139 482 112;
  • 25) 0.012 451 171 870 139 482 112 × 2 = 0 + 0.024 902 343 740 278 964 224;
  • 26) 0.024 902 343 740 278 964 224 × 2 = 0 + 0.049 804 687 480 557 928 448;
  • 27) 0.049 804 687 480 557 928 448 × 2 = 0 + 0.099 609 374 961 115 856 896;
  • 28) 0.099 609 374 961 115 856 896 × 2 = 0 + 0.199 218 749 922 231 713 792;
  • 29) 0.199 218 749 922 231 713 792 × 2 = 0 + 0.398 437 499 844 463 427 584;
  • 30) 0.398 437 499 844 463 427 584 × 2 = 0 + 0.796 874 999 688 926 855 168;
  • 31) 0.796 874 999 688 926 855 168 × 2 = 1 + 0.593 749 999 377 853 710 336;
  • 32) 0.593 749 999 377 853 710 336 × 2 = 1 + 0.187 499 998 755 707 420 672;
  • 33) 0.187 499 998 755 707 420 672 × 2 = 0 + 0.374 999 997 511 414 841 344;
  • 34) 0.374 999 997 511 414 841 344 × 2 = 0 + 0.749 999 995 022 829 682 688;
  • 35) 0.749 999 995 022 829 682 688 × 2 = 1 + 0.499 999 990 045 659 365 376;
  • 36) 0.499 999 990 045 659 365 376 × 2 = 0 + 0.999 999 980 091 318 730 752;
  • 37) 0.999 999 980 091 318 730 752 × 2 = 1 + 0.999 999 960 182 637 461 504;
  • 38) 0.999 999 960 182 637 461 504 × 2 = 1 + 0.999 999 920 365 274 923 008;
  • 39) 0.999 999 920 365 274 923 008 × 2 = 1 + 0.999 999 840 730 549 846 016;
  • 40) 0.999 999 840 730 549 846 016 × 2 = 1 + 0.999 999 681 461 099 692 032;
  • 41) 0.999 999 681 461 099 692 032 × 2 = 1 + 0.999 999 362 922 199 384 064;
  • 42) 0.999 999 362 922 199 384 064 × 2 = 1 + 0.999 998 725 844 398 768 128;
  • 43) 0.999 998 725 844 398 768 128 × 2 = 1 + 0.999 997 451 688 797 536 256;
  • 44) 0.999 997 451 688 797 536 256 × 2 = 1 + 0.999 994 903 377 595 072 512;
  • 45) 0.999 994 903 377 595 072 512 × 2 = 1 + 0.999 989 806 755 190 145 024;
  • 46) 0.999 989 806 755 190 145 024 × 2 = 1 + 0.999 979 613 510 380 290 048;
  • 47) 0.999 979 613 510 380 290 048 × 2 = 1 + 0.999 959 227 020 760 580 096;
  • 48) 0.999 959 227 020 760 580 096 × 2 = 1 + 0.999 918 454 041 521 160 192;
  • 49) 0.999 918 454 041 521 160 192 × 2 = 1 + 0.999 836 908 083 042 320 384;
  • 50) 0.999 836 908 083 042 320 384 × 2 = 1 + 0.999 673 816 166 084 640 768;
  • 51) 0.999 673 816 166 084 640 768 × 2 = 1 + 0.999 347 632 332 169 281 536;
  • 52) 0.999 347 632 332 169 281 536 × 2 = 1 + 0.998 695 264 664 338 563 072;
  • 53) 0.998 695 264 664 338 563 072 × 2 = 1 + 0.997 390 529 328 677 126 144;
  • 54) 0.997 390 529 328 677 126 144 × 2 = 1 + 0.994 781 058 657 354 252 288;

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:


0.000 000 000 742 147 676 357(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 357(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:


0.000 000 000 742 147 676 357(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 -0.000 000 000 742 147 676 357 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 288 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: Reputation: Bridge Between Company's Actions and Market Value. NotebookLM YouTube Video of 6.55 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