-0.000 000 000 742 147 417 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 417(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
-0.000 000 000 742 147 417(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:

|-0.000 000 000 742 147 417| = 0.000 000 000 742 147 417


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: 0.000 000 000 742 147 417.

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) 0.000 000 000 742 147 417 × 2 = 0 + 0.000 000 001 484 294 834;
  • 2) 0.000 000 001 484 294 834 × 2 = 0 + 0.000 000 002 968 589 668;
  • 3) 0.000 000 002 968 589 668 × 2 = 0 + 0.000 000 005 937 179 336;
  • 4) 0.000 000 005 937 179 336 × 2 = 0 + 0.000 000 011 874 358 672;
  • 5) 0.000 000 011 874 358 672 × 2 = 0 + 0.000 000 023 748 717 344;
  • 6) 0.000 000 023 748 717 344 × 2 = 0 + 0.000 000 047 497 434 688;
  • 7) 0.000 000 047 497 434 688 × 2 = 0 + 0.000 000 094 994 869 376;
  • 8) 0.000 000 094 994 869 376 × 2 = 0 + 0.000 000 189 989 738 752;
  • 9) 0.000 000 189 989 738 752 × 2 = 0 + 0.000 000 379 979 477 504;
  • 10) 0.000 000 379 979 477 504 × 2 = 0 + 0.000 000 759 958 955 008;
  • 11) 0.000 000 759 958 955 008 × 2 = 0 + 0.000 001 519 917 910 016;
  • 12) 0.000 001 519 917 910 016 × 2 = 0 + 0.000 003 039 835 820 032;
  • 13) 0.000 003 039 835 820 032 × 2 = 0 + 0.000 006 079 671 640 064;
  • 14) 0.000 006 079 671 640 064 × 2 = 0 + 0.000 012 159 343 280 128;
  • 15) 0.000 012 159 343 280 128 × 2 = 0 + 0.000 024 318 686 560 256;
  • 16) 0.000 024 318 686 560 256 × 2 = 0 + 0.000 048 637 373 120 512;
  • 17) 0.000 048 637 373 120 512 × 2 = 0 + 0.000 097 274 746 241 024;
  • 18) 0.000 097 274 746 241 024 × 2 = 0 + 0.000 194 549 492 482 048;
  • 19) 0.000 194 549 492 482 048 × 2 = 0 + 0.000 389 098 984 964 096;
  • 20) 0.000 389 098 984 964 096 × 2 = 0 + 0.000 778 197 969 928 192;
  • 21) 0.000 778 197 969 928 192 × 2 = 0 + 0.001 556 395 939 856 384;
  • 22) 0.001 556 395 939 856 384 × 2 = 0 + 0.003 112 791 879 712 768;
  • 23) 0.003 112 791 879 712 768 × 2 = 0 + 0.006 225 583 759 425 536;
  • 24) 0.006 225 583 759 425 536 × 2 = 0 + 0.012 451 167 518 851 072;
  • 25) 0.012 451 167 518 851 072 × 2 = 0 + 0.024 902 335 037 702 144;
  • 26) 0.024 902 335 037 702 144 × 2 = 0 + 0.049 804 670 075 404 288;
  • 27) 0.049 804 670 075 404 288 × 2 = 0 + 0.099 609 340 150 808 576;
  • 28) 0.099 609 340 150 808 576 × 2 = 0 + 0.199 218 680 301 617 152;
  • 29) 0.199 218 680 301 617 152 × 2 = 0 + 0.398 437 360 603 234 304;
  • 30) 0.398 437 360 603 234 304 × 2 = 0 + 0.796 874 721 206 468 608;
  • 31) 0.796 874 721 206 468 608 × 2 = 1 + 0.593 749 442 412 937 216;
  • 32) 0.593 749 442 412 937 216 × 2 = 1 + 0.187 498 884 825 874 432;
  • 33) 0.187 498 884 825 874 432 × 2 = 0 + 0.374 997 769 651 748 864;
  • 34) 0.374 997 769 651 748 864 × 2 = 0 + 0.749 995 539 303 497 728;
  • 35) 0.749 995 539 303 497 728 × 2 = 1 + 0.499 991 078 606 995 456;
  • 36) 0.499 991 078 606 995 456 × 2 = 0 + 0.999 982 157 213 990 912;
  • 37) 0.999 982 157 213 990 912 × 2 = 1 + 0.999 964 314 427 981 824;
  • 38) 0.999 964 314 427 981 824 × 2 = 1 + 0.999 928 628 855 963 648;
  • 39) 0.999 928 628 855 963 648 × 2 = 1 + 0.999 857 257 711 927 296;
  • 40) 0.999 857 257 711 927 296 × 2 = 1 + 0.999 714 515 423 854 592;
  • 41) 0.999 714 515 423 854 592 × 2 = 1 + 0.999 429 030 847 709 184;
  • 42) 0.999 429 030 847 709 184 × 2 = 1 + 0.998 858 061 695 418 368;
  • 43) 0.998 858 061 695 418 368 × 2 = 1 + 0.997 716 123 390 836 736;
  • 44) 0.997 716 123 390 836 736 × 2 = 1 + 0.995 432 246 781 673 472;
  • 45) 0.995 432 246 781 673 472 × 2 = 1 + 0.990 864 493 563 346 944;
  • 46) 0.990 864 493 563 346 944 × 2 = 1 + 0.981 728 987 126 693 888;
  • 47) 0.981 728 987 126 693 888 × 2 = 1 + 0.963 457 974 253 387 776;
  • 48) 0.963 457 974 253 387 776 × 2 = 1 + 0.926 915 948 506 775 552;
  • 49) 0.926 915 948 506 775 552 × 2 = 1 + 0.853 831 897 013 551 104;
  • 50) 0.853 831 897 013 551 104 × 2 = 1 + 0.707 663 794 027 102 208;
  • 51) 0.707 663 794 027 102 208 × 2 = 1 + 0.415 327 588 054 204 416;
  • 52) 0.415 327 588 054 204 416 × 2 = 0 + 0.830 655 176 108 408 832;
  • 53) 0.830 655 176 108 408 832 × 2 = 1 + 0.661 310 352 216 817 664;
  • 54) 0.661 310 352 216 817 664 × 2 = 1 + 0.322 620 704 433 635 328;

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:


0.000 000 000 742 147 417(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 417(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 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:


0.000 000 000 742 147 417(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(2) × 20 =

1.1001 0111 1111 1111 1111 011(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 011


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 1011 =

100 1011 1111 1111 1111 1011


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 1011


Decimal number -0.000 000 000 742 147 417 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1011

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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