-0.000 000 000 742 147 676 594 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 594(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
-0.000 000 000 742 147 676 594(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:

|-0.000 000 000 742 147 676 594| = 0.000 000 000 742 147 676 594


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: 0.000 000 000 742 147 676 594.

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) 0.000 000 000 742 147 676 594 × 2 = 0 + 0.000 000 001 484 295 353 188;
  • 2) 0.000 000 001 484 295 353 188 × 2 = 0 + 0.000 000 002 968 590 706 376;
  • 3) 0.000 000 002 968 590 706 376 × 2 = 0 + 0.000 000 005 937 181 412 752;
  • 4) 0.000 000 005 937 181 412 752 × 2 = 0 + 0.000 000 011 874 362 825 504;
  • 5) 0.000 000 011 874 362 825 504 × 2 = 0 + 0.000 000 023 748 725 651 008;
  • 6) 0.000 000 023 748 725 651 008 × 2 = 0 + 0.000 000 047 497 451 302 016;
  • 7) 0.000 000 047 497 451 302 016 × 2 = 0 + 0.000 000 094 994 902 604 032;
  • 8) 0.000 000 094 994 902 604 032 × 2 = 0 + 0.000 000 189 989 805 208 064;
  • 9) 0.000 000 189 989 805 208 064 × 2 = 0 + 0.000 000 379 979 610 416 128;
  • 10) 0.000 000 379 979 610 416 128 × 2 = 0 + 0.000 000 759 959 220 832 256;
  • 11) 0.000 000 759 959 220 832 256 × 2 = 0 + 0.000 001 519 918 441 664 512;
  • 12) 0.000 001 519 918 441 664 512 × 2 = 0 + 0.000 003 039 836 883 329 024;
  • 13) 0.000 003 039 836 883 329 024 × 2 = 0 + 0.000 006 079 673 766 658 048;
  • 14) 0.000 006 079 673 766 658 048 × 2 = 0 + 0.000 012 159 347 533 316 096;
  • 15) 0.000 012 159 347 533 316 096 × 2 = 0 + 0.000 024 318 695 066 632 192;
  • 16) 0.000 024 318 695 066 632 192 × 2 = 0 + 0.000 048 637 390 133 264 384;
  • 17) 0.000 048 637 390 133 264 384 × 2 = 0 + 0.000 097 274 780 266 528 768;
  • 18) 0.000 097 274 780 266 528 768 × 2 = 0 + 0.000 194 549 560 533 057 536;
  • 19) 0.000 194 549 560 533 057 536 × 2 = 0 + 0.000 389 099 121 066 115 072;
  • 20) 0.000 389 099 121 066 115 072 × 2 = 0 + 0.000 778 198 242 132 230 144;
  • 21) 0.000 778 198 242 132 230 144 × 2 = 0 + 0.001 556 396 484 264 460 288;
  • 22) 0.001 556 396 484 264 460 288 × 2 = 0 + 0.003 112 792 968 528 920 576;
  • 23) 0.003 112 792 968 528 920 576 × 2 = 0 + 0.006 225 585 937 057 841 152;
  • 24) 0.006 225 585 937 057 841 152 × 2 = 0 + 0.012 451 171 874 115 682 304;
  • 25) 0.012 451 171 874 115 682 304 × 2 = 0 + 0.024 902 343 748 231 364 608;
  • 26) 0.024 902 343 748 231 364 608 × 2 = 0 + 0.049 804 687 496 462 729 216;
  • 27) 0.049 804 687 496 462 729 216 × 2 = 0 + 0.099 609 374 992 925 458 432;
  • 28) 0.099 609 374 992 925 458 432 × 2 = 0 + 0.199 218 749 985 850 916 864;
  • 29) 0.199 218 749 985 850 916 864 × 2 = 0 + 0.398 437 499 971 701 833 728;
  • 30) 0.398 437 499 971 701 833 728 × 2 = 0 + 0.796 874 999 943 403 667 456;
  • 31) 0.796 874 999 943 403 667 456 × 2 = 1 + 0.593 749 999 886 807 334 912;
  • 32) 0.593 749 999 886 807 334 912 × 2 = 1 + 0.187 499 999 773 614 669 824;
  • 33) 0.187 499 999 773 614 669 824 × 2 = 0 + 0.374 999 999 547 229 339 648;
  • 34) 0.374 999 999 547 229 339 648 × 2 = 0 + 0.749 999 999 094 458 679 296;
  • 35) 0.749 999 999 094 458 679 296 × 2 = 1 + 0.499 999 998 188 917 358 592;
  • 36) 0.499 999 998 188 917 358 592 × 2 = 0 + 0.999 999 996 377 834 717 184;
  • 37) 0.999 999 996 377 834 717 184 × 2 = 1 + 0.999 999 992 755 669 434 368;
  • 38) 0.999 999 992 755 669 434 368 × 2 = 1 + 0.999 999 985 511 338 868 736;
  • 39) 0.999 999 985 511 338 868 736 × 2 = 1 + 0.999 999 971 022 677 737 472;
  • 40) 0.999 999 971 022 677 737 472 × 2 = 1 + 0.999 999 942 045 355 474 944;
  • 41) 0.999 999 942 045 355 474 944 × 2 = 1 + 0.999 999 884 090 710 949 888;
  • 42) 0.999 999 884 090 710 949 888 × 2 = 1 + 0.999 999 768 181 421 899 776;
  • 43) 0.999 999 768 181 421 899 776 × 2 = 1 + 0.999 999 536 362 843 799 552;
  • 44) 0.999 999 536 362 843 799 552 × 2 = 1 + 0.999 999 072 725 687 599 104;
  • 45) 0.999 999 072 725 687 599 104 × 2 = 1 + 0.999 998 145 451 375 198 208;
  • 46) 0.999 998 145 451 375 198 208 × 2 = 1 + 0.999 996 290 902 750 396 416;
  • 47) 0.999 996 290 902 750 396 416 × 2 = 1 + 0.999 992 581 805 500 792 832;
  • 48) 0.999 992 581 805 500 792 832 × 2 = 1 + 0.999 985 163 611 001 585 664;
  • 49) 0.999 985 163 611 001 585 664 × 2 = 1 + 0.999 970 327 222 003 171 328;
  • 50) 0.999 970 327 222 003 171 328 × 2 = 1 + 0.999 940 654 444 006 342 656;
  • 51) 0.999 940 654 444 006 342 656 × 2 = 1 + 0.999 881 308 888 012 685 312;
  • 52) 0.999 881 308 888 012 685 312 × 2 = 1 + 0.999 762 617 776 025 370 624;
  • 53) 0.999 762 617 776 025 370 624 × 2 = 1 + 0.999 525 235 552 050 741 248;
  • 54) 0.999 525 235 552 050 741 248 × 2 = 1 + 0.999 050 471 104 101 482 496;

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:


0.000 000 000 742 147 676 594(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 594(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:


0.000 000 000 742 147 676 594(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 -0.000 000 000 742 147 676 594 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 542 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 Art of Strategic Ambiguity and Concealed Intentions. NotebookLM YouTube Video of 6.59 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