2.858 648 867 222 626 824 684 404 97 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.858 648 867 222 626 824 684 404 97(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
2.858 648 867 222 626 824 684 404 97(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. First, convert to binary (in base 2) the integer part: 2.
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;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

2. 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.

2(10) =

10(2)


3. Convert to binary (base 2) the fractional part: 0.858 648 867 222 626 824 684 404 97.

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) 0.858 648 867 222 626 824 684 404 97 × 2 = 1 + 0.717 297 734 445 253 649 368 809 94;
  • 2) 0.717 297 734 445 253 649 368 809 94 × 2 = 1 + 0.434 595 468 890 507 298 737 619 88;
  • 3) 0.434 595 468 890 507 298 737 619 88 × 2 = 0 + 0.869 190 937 781 014 597 475 239 76;
  • 4) 0.869 190 937 781 014 597 475 239 76 × 2 = 1 + 0.738 381 875 562 029 194 950 479 52;
  • 5) 0.738 381 875 562 029 194 950 479 52 × 2 = 1 + 0.476 763 751 124 058 389 900 959 04;
  • 6) 0.476 763 751 124 058 389 900 959 04 × 2 = 0 + 0.953 527 502 248 116 779 801 918 08;
  • 7) 0.953 527 502 248 116 779 801 918 08 × 2 = 1 + 0.907 055 004 496 233 559 603 836 16;
  • 8) 0.907 055 004 496 233 559 603 836 16 × 2 = 1 + 0.814 110 008 992 467 119 207 672 32;
  • 9) 0.814 110 008 992 467 119 207 672 32 × 2 = 1 + 0.628 220 017 984 934 238 415 344 64;
  • 10) 0.628 220 017 984 934 238 415 344 64 × 2 = 1 + 0.256 440 035 969 868 476 830 689 28;
  • 11) 0.256 440 035 969 868 476 830 689 28 × 2 = 0 + 0.512 880 071 939 736 953 661 378 56;
  • 12) 0.512 880 071 939 736 953 661 378 56 × 2 = 1 + 0.025 760 143 879 473 907 322 757 12;
  • 13) 0.025 760 143 879 473 907 322 757 12 × 2 = 0 + 0.051 520 287 758 947 814 645 514 24;
  • 14) 0.051 520 287 758 947 814 645 514 24 × 2 = 0 + 0.103 040 575 517 895 629 291 028 48;
  • 15) 0.103 040 575 517 895 629 291 028 48 × 2 = 0 + 0.206 081 151 035 791 258 582 056 96;
  • 16) 0.206 081 151 035 791 258 582 056 96 × 2 = 0 + 0.412 162 302 071 582 517 164 113 92;
  • 17) 0.412 162 302 071 582 517 164 113 92 × 2 = 0 + 0.824 324 604 143 165 034 328 227 84;
  • 18) 0.824 324 604 143 165 034 328 227 84 × 2 = 1 + 0.648 649 208 286 330 068 656 455 68;
  • 19) 0.648 649 208 286 330 068 656 455 68 × 2 = 1 + 0.297 298 416 572 660 137 312 911 36;
  • 20) 0.297 298 416 572 660 137 312 911 36 × 2 = 0 + 0.594 596 833 145 320 274 625 822 72;
  • 21) 0.594 596 833 145 320 274 625 822 72 × 2 = 1 + 0.189 193 666 290 640 549 251 645 44;
  • 22) 0.189 193 666 290 640 549 251 645 44 × 2 = 0 + 0.378 387 332 581 281 098 503 290 88;
  • 23) 0.378 387 332 581 281 098 503 290 88 × 2 = 0 + 0.756 774 665 162 562 197 006 581 76;
  • 24) 0.756 774 665 162 562 197 006 581 76 × 2 = 1 + 0.513 549 330 325 124 394 013 163 52;

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).


4. 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:


0.858 648 867 222 626 824 684 404 97(10) =

0.1101 1011 1101 0000 0110 1001(2)

5. Positive number before normalization:

2.858 648 867 222 626 824 684 404 97(10) =

10.1101 1011 1101 0000 0110 1001(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


2.858 648 867 222 626 824 684 404 97(10) =

10.1101 1011 1101 0000 0110 1001(2) =

10.1101 1011 1101 0000 0110 1001(2) × 20 =

1.0110 1101 1110 1000 0011 0100 1(2) × 21


7. 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 0 (a positive number)

Exponent (unadjusted): 1

Mantissa (not normalized):
1.0110 1101 1110 1000 0011 0100 1


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

1 + 2(8-1) - 1 =

(1 + 127)(10) =

128(10)


9. 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;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

128(10) =

1000 0000(2)


11. 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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 011 0110 1111 0100 0001 1010 01 =

011 0110 1111 0100 0001 1010


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
1000 0000

Mantissa (23 bits) =
011 0110 1111 0100 0001 1010


Decimal number 2.858 648 867 222 626 824 684 404 97 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 1000 0000 - 011 0110 1111 0100 0001 1010

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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