-0.000 000 000 742 147 676 620 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 620 4(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
-0.000 000 000 742 147 676 620 4(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:

|-0.000 000 000 742 147 676 620 4| = 0.000 000 000 742 147 676 620 4


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: 0.000 000 000 742 147 676 620 4.

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) 0.000 000 000 742 147 676 620 4 × 2 = 0 + 0.000 000 001 484 295 353 240 8;
  • 2) 0.000 000 001 484 295 353 240 8 × 2 = 0 + 0.000 000 002 968 590 706 481 6;
  • 3) 0.000 000 002 968 590 706 481 6 × 2 = 0 + 0.000 000 005 937 181 412 963 2;
  • 4) 0.000 000 005 937 181 412 963 2 × 2 = 0 + 0.000 000 011 874 362 825 926 4;
  • 5) 0.000 000 011 874 362 825 926 4 × 2 = 0 + 0.000 000 023 748 725 651 852 8;
  • 6) 0.000 000 023 748 725 651 852 8 × 2 = 0 + 0.000 000 047 497 451 303 705 6;
  • 7) 0.000 000 047 497 451 303 705 6 × 2 = 0 + 0.000 000 094 994 902 607 411 2;
  • 8) 0.000 000 094 994 902 607 411 2 × 2 = 0 + 0.000 000 189 989 805 214 822 4;
  • 9) 0.000 000 189 989 805 214 822 4 × 2 = 0 + 0.000 000 379 979 610 429 644 8;
  • 10) 0.000 000 379 979 610 429 644 8 × 2 = 0 + 0.000 000 759 959 220 859 289 6;
  • 11) 0.000 000 759 959 220 859 289 6 × 2 = 0 + 0.000 001 519 918 441 718 579 2;
  • 12) 0.000 001 519 918 441 718 579 2 × 2 = 0 + 0.000 003 039 836 883 437 158 4;
  • 13) 0.000 003 039 836 883 437 158 4 × 2 = 0 + 0.000 006 079 673 766 874 316 8;
  • 14) 0.000 006 079 673 766 874 316 8 × 2 = 0 + 0.000 012 159 347 533 748 633 6;
  • 15) 0.000 012 159 347 533 748 633 6 × 2 = 0 + 0.000 024 318 695 067 497 267 2;
  • 16) 0.000 024 318 695 067 497 267 2 × 2 = 0 + 0.000 048 637 390 134 994 534 4;
  • 17) 0.000 048 637 390 134 994 534 4 × 2 = 0 + 0.000 097 274 780 269 989 068 8;
  • 18) 0.000 097 274 780 269 989 068 8 × 2 = 0 + 0.000 194 549 560 539 978 137 6;
  • 19) 0.000 194 549 560 539 978 137 6 × 2 = 0 + 0.000 389 099 121 079 956 275 2;
  • 20) 0.000 389 099 121 079 956 275 2 × 2 = 0 + 0.000 778 198 242 159 912 550 4;
  • 21) 0.000 778 198 242 159 912 550 4 × 2 = 0 + 0.001 556 396 484 319 825 100 8;
  • 22) 0.001 556 396 484 319 825 100 8 × 2 = 0 + 0.003 112 792 968 639 650 201 6;
  • 23) 0.003 112 792 968 639 650 201 6 × 2 = 0 + 0.006 225 585 937 279 300 403 2;
  • 24) 0.006 225 585 937 279 300 403 2 × 2 = 0 + 0.012 451 171 874 558 600 806 4;
  • 25) 0.012 451 171 874 558 600 806 4 × 2 = 0 + 0.024 902 343 749 117 201 612 8;
  • 26) 0.024 902 343 749 117 201 612 8 × 2 = 0 + 0.049 804 687 498 234 403 225 6;
  • 27) 0.049 804 687 498 234 403 225 6 × 2 = 0 + 0.099 609 374 996 468 806 451 2;
  • 28) 0.099 609 374 996 468 806 451 2 × 2 = 0 + 0.199 218 749 992 937 612 902 4;
  • 29) 0.199 218 749 992 937 612 902 4 × 2 = 0 + 0.398 437 499 985 875 225 804 8;
  • 30) 0.398 437 499 985 875 225 804 8 × 2 = 0 + 0.796 874 999 971 750 451 609 6;
  • 31) 0.796 874 999 971 750 451 609 6 × 2 = 1 + 0.593 749 999 943 500 903 219 2;
  • 32) 0.593 749 999 943 500 903 219 2 × 2 = 1 + 0.187 499 999 887 001 806 438 4;
  • 33) 0.187 499 999 887 001 806 438 4 × 2 = 0 + 0.374 999 999 774 003 612 876 8;
  • 34) 0.374 999 999 774 003 612 876 8 × 2 = 0 + 0.749 999 999 548 007 225 753 6;
  • 35) 0.749 999 999 548 007 225 753 6 × 2 = 1 + 0.499 999 999 096 014 451 507 2;
  • 36) 0.499 999 999 096 014 451 507 2 × 2 = 0 + 0.999 999 998 192 028 903 014 4;
  • 37) 0.999 999 998 192 028 903 014 4 × 2 = 1 + 0.999 999 996 384 057 806 028 8;
  • 38) 0.999 999 996 384 057 806 028 8 × 2 = 1 + 0.999 999 992 768 115 612 057 6;
  • 39) 0.999 999 992 768 115 612 057 6 × 2 = 1 + 0.999 999 985 536 231 224 115 2;
  • 40) 0.999 999 985 536 231 224 115 2 × 2 = 1 + 0.999 999 971 072 462 448 230 4;
  • 41) 0.999 999 971 072 462 448 230 4 × 2 = 1 + 0.999 999 942 144 924 896 460 8;
  • 42) 0.999 999 942 144 924 896 460 8 × 2 = 1 + 0.999 999 884 289 849 792 921 6;
  • 43) 0.999 999 884 289 849 792 921 6 × 2 = 1 + 0.999 999 768 579 699 585 843 2;
  • 44) 0.999 999 768 579 699 585 843 2 × 2 = 1 + 0.999 999 537 159 399 171 686 4;
  • 45) 0.999 999 537 159 399 171 686 4 × 2 = 1 + 0.999 999 074 318 798 343 372 8;
  • 46) 0.999 999 074 318 798 343 372 8 × 2 = 1 + 0.999 998 148 637 596 686 745 6;
  • 47) 0.999 998 148 637 596 686 745 6 × 2 = 1 + 0.999 996 297 275 193 373 491 2;
  • 48) 0.999 996 297 275 193 373 491 2 × 2 = 1 + 0.999 992 594 550 386 746 982 4;
  • 49) 0.999 992 594 550 386 746 982 4 × 2 = 1 + 0.999 985 189 100 773 493 964 8;
  • 50) 0.999 985 189 100 773 493 964 8 × 2 = 1 + 0.999 970 378 201 546 987 929 6;
  • 51) 0.999 970 378 201 546 987 929 6 × 2 = 1 + 0.999 940 756 403 093 975 859 2;
  • 52) 0.999 940 756 403 093 975 859 2 × 2 = 1 + 0.999 881 512 806 187 951 718 4;
  • 53) 0.999 881 512 806 187 951 718 4 × 2 = 1 + 0.999 763 025 612 375 903 436 8;
  • 54) 0.999 763 025 612 375 903 436 8 × 2 = 1 + 0.999 526 051 224 751 806 873 6;

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:


0.000 000 000 742 147 676 620 4(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 620 4(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:


0.000 000 000 742 147 676 620 4(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 -0.000 000 000 742 147 676 620 4 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 611 4 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: Are you using communication by design, or by default? NotebookLM YouTube Video of 6.33 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